Definite Integrals
Fundamental Theorem Of Integral Calculus
Let f(x) be a continuous function defined as a noninterval [a,b], and let the antiderivative of f(x) be F(x). Then, the definite integral of f(x) over [a,b], denoted by \(\int_a^b f(x) d x\), is given by
\(\int_a^b f(x) d x=[F(x)]_a^b=F(b)-F(a)\)Note Here, a and b are respectively known as the lower limit and the upper limit of the integral.
The value of a definite integral is unique, for if \(\int f(x) d x=F(x)+C\) then
⇒ \(\int_a^b f(x) d x=[F(x)+C]_a^b=\{F(b)+C\}-\{F(a)+C\}=F(b)-F(a) .\)
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Solved Examples
Example 1 Evaluate:
- \(\int_2^4 \frac{d x}{x}\)
- \(\int_4^9 \sqrt{x} d x\)
- \(\int_0^2 \sqrt{6 x+4} d x\)
- \(\int_0^1 \frac{d x}{\sqrt{5 x+3}}\)
- \(\int_1^{\sqrt{2}} \frac{d x}{x\left(\sqrt{x^2-1}\right)}\)
- \(\int_0^\pi \sin 5 x d x\)
- \(\int_0^{\pi / 2} \cos ^2 x d x\)
- \(\int_0^{\pi / 4} \tan ^2 x d x\)
- \(\int_0^{\pi / 4} \sin 2 x \sin 3 x d x\)
Solution
(1) \(\int_2^4 \frac{d x}{x}=[\log x]_2^4=(\log 4-\log 2)=(2 \log 2-\log 2)=\log 2\)
(2) \(\int_4^9 \sqrt{x} d x=\left[\frac{2}{3} x^{3 / 2}\right]_4^9=\frac{2}{3} \cdot\left[(9)^{3 / 2}-(4)^{3 / 2}\right]=\frac{38}{3} .\)
(3) \(\int_0^2 \sqrt{6 x+4} d x=\left[\frac{2}{3} \cdot \frac{(6 x+4)^{3 / 2}}{6}\right]_0^2=\frac{1}{9} \cdot\left[(16)^{3 / 2}-(4)^{3 / 2}\right]=\frac{56}{9}\)
(4) \(\int_0^1 \frac{1}{\sqrt{5 x+3}} d x=\int_0^1(5 x+3)^{-1 / 2} d x=\left[2 \cdot \frac{(5 x+3)^{1 / 2}}{5}\right]_0^1=\frac{2}{5}(\sqrt{8}-\sqrt{3}) .\)
(5) \(\int_1^{\sqrt{2}} \frac{d x}{x\left(\sqrt{x^2-1}\right)}=\left[\sec ^{-1} x\right]_1^{\sqrt{2}}=\left[\sec ^{-1}(\sqrt{2})-\sec ^{-1}(1)\right]=\left(\frac{\pi}{4}-0\right)=\frac{\pi}{4}\)
(6) \(\int_0^\pi \sin 5 x d x=\left[\frac{-\cos 5 x}{5}\right]_0^\pi=-\frac{1}{5}[\cos 5 \pi-\cos 0]=\frac{2}{5}\)
(7) \(\int_0^{\pi / 2} \cos ^2 x d x=\frac{1}{2} \int_0^{\pi / 2}(1+\cos 2 x) d x=\frac{1}{2}\left[x+\frac{\sin 2 x}{2}\right]_0^{\pi / 2}=\frac{\pi}{4}\)
(8) \(\int_0^{\pi / 4} \tan ^2 x d x=\int_0^{\pi / 4}\left(\sec ^2 x-1\right) d x=[\tan x-x]_0^{\pi / 4}=\left(1-\frac{\pi}{4}\right)\)
(9) \(\int_0^{\pi / 4} \sin 2 x \sin 3 x d x=\frac{1}{2} \int_0^{\pi / 4}(2 \sin 2 x \sin 3 x) d x\)
= \(\frac{1}{2} \int_0^{\pi / 4}(\cos x-\cos 5 x) d x=\frac{1}{2}\left[\sin x-\frac{\sin 5 x}{5}\right]_0^{\pi / 4}\)
= \(\frac{1}{2}\left[\left(\sin \frac{\pi}{4}-\frac{\sin (5 \pi / 4)}{5}\right)\right]=\frac{3 \sqrt{2}}{10}\)
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Example 2 Evaluate:
- \(\int_0^{\pi / 4} \sqrt{1+\sin 2 x} d x\)
- \(\int_0^{\pi / 2} \sqrt{1+\cos 2 x} d x\)
Solution
(1) \(\int_0^{\pi / 4} \sqrt{1+\sin 2 x} d x=\int_0^{\pi / 4} \sqrt{\cos ^2 x+\sin ^2 x+2 \sin x \cos x} d x\)
= \(\int_0^{\pi / 4}(\cos x+\sin x) d x=[\sin x-\cos x]_0^{\pi / 4}=1 .\)
(2) \(\int_0^{\pi / 2} \sqrt{1+\cos 2 x} d x=\int_0^{\pi / 2} \sqrt{2 \cos ^2 x} d x\)
= \(\sqrt{2} \int_0^{\pi / 2} \cos x d x=\sqrt{2}[\sin x]_0^{\pi / 2}=\sqrt{2}\)
Example 3 Evaluate:
- \(\int_0^{\pi / 2} \cos ^3 x d x\)
- \(\int_0^{\pi / 2} \sin ^4 x d x\)
Solution
(1) \(\int_0^{\pi / 2} \cos ^3 x d x=\int_0^{\pi / 2}\left(\frac{3 \cos x+\cos 3 x}{4}\right) d x\) [∵ cos 3x = 4 cos3x – 3 cos x]
= \(\frac{3}{4} \cdot \int_0^{\pi / 2} \cos x d x+\frac{1}{4} \cdot \int_0^{\pi / 2} \cos 3 x d x\)
= \(\frac{3}{4} \cdot[\sin x]_0^{\pi / 2}+\frac{1}{4} \cdot\left[\frac{\sin 3 x}{3}\right]_0^{\pi / 2}=\left(\frac{3}{4}-\frac{1}{12}\right)=\frac{8}{12}=\frac{2}{3}\).
(2) \(\int_0^{\pi / 2} \sin ^4 x d x=\frac{1}{4} \int_0^{\pi / 2}\left(2 \sin ^2 x\right)^2 d x\)
= \(\frac{1}{4} \cdot \int_0^{\pi / 2}(1-\cos 2 x)^2 d x=\frac{1}{4} \cdot \int_0^{\pi / 2}\left(1-2 \cos 2 x+\cos ^2 2 x\right) d x\)
= \(\frac{1}{4} \cdot \int_0^{\pi / 2}\left[1-2 \cos 2 x+\frac{(1+\cos 4 x)}{2}\right] d x\)
= \(\frac{1}{4} \cdot \int_0^{\pi / 2}\left(\frac{3}{2}-2 \cos 2 x+\frac{1}{2} \cos 4 x\right) d x\)
= \(\frac{3}{8} \cdot \int_0^{\pi / 2} d x-\frac{1}{2} \int_0^{\pi / 2} \cos 2 x d x+\frac{1}{8} \int_0^{\pi / 2} \cos 4 x d x\)
= \(\frac{3}{8} \cdot[x]_0^{\pi / 2}-\frac{1}{2} \cdot\left[\frac{\sin 2 x}{2}\right]_0^{\pi / 2}+\frac{1}{8} \cdot\left[\frac{\sin 4 x}{4}\right]_0^{\pi / 2}\)
= \(\left(\frac{3 \pi}{16}-0+0\right)=\frac{3 \pi}{16} .\)
Example 4 Evaluate:
- \(\int_0^4 \frac{d x}{\sqrt{x^2+2 x+3}}\)
- \(\int_0^1 \frac{d x}{\left(1+x+x^2\right)}\)
Solution
(1) \(\int_0^4 \frac{d x}{\sqrt{x^2+2 x+3}}=\int_0^4 \frac{d x}{\sqrt{(x+1)^2+(\sqrt{2})^2}}\)
= \(\left[\log \left|(x+1)+\sqrt{x^2+2 x+3}\right|\right]_0^4\)
= {log |5 + √27| – log |1 + √3|}.
(2) \(\int_0^1 \frac{d x}{\left(1+x+x^2\right)}=\int_0^1 \frac{d x}{\left[\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}\right]}=\int_0^1 \frac{d x}{\left[\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\right]}\)
= \(\left[\frac{2}{\sqrt{3}} \tan ^{-1} \frac{\left(x+\frac{1}{2}\right)}{(\sqrt{3} / 2)}\right]_0^1=\frac{2}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)\right]_0^1\)
= \(\frac{2}{\sqrt{3}}\left[\tan ^{-1}(\sqrt{3})-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\right]=\frac{2}{\sqrt{3}} \cdot\left[\frac{\pi}{3}-\frac{\pi}{6}\right]=\frac{\pi}{3 \sqrt{3}} .\)
“Definite Integral Properties Explained Simply”
Example 5 Evaluate:
- \(\int_0^a \frac{d x}{\sqrt{a x-x^2}}\)
- \(\int_0^{\sqrt{2}} \sqrt{2-x^2} d x\)
Solution
(1) \(\int_0^a \frac{d x}{\sqrt{a x-x^2}}=\int_0^a \frac{d x}{\sqrt{-\left(x^2-a x+\frac{a^2}{4}\right)+\frac{a^2}{4}}}\)
= \(\int_0^a \frac{d x}{\sqrt{\left(\frac{a}{2}\right)^2-\left(x-\frac{a}{2}\right)^2}}\)
= \(\left[\sin ^{-1} \frac{\left(x-\frac{a}{2}\right)}{\left(\frac{a}{2}\right)}\right]_0^a=\left[\sin ^{-1}\left(\frac{2 x-a}{a}\right)\right]_0^a\)
= [sin-1 (1) – sin-1 (-1)]
= \(2 \sin ^{-1}(1)=\left(2 \times \frac{\pi}{2}\right)=\pi\)
(2) \(\int_0^{\sqrt{2}} \sqrt{2-x^2} d x=\int_0^{\sqrt{2}} \sqrt{(\sqrt{2})^2-x^2} d x\)
= \(\left[\frac{x}{2} \sqrt{2-x^2}+\frac{(\sqrt{2})^2}{2} \cdot \sin ^{-1} \frac{x}{\sqrt{2}}\right]_0^{\sqrt{2}}\)
= \(\left[0+\sin ^{-1}(1)\right]-\left[0+\sin ^{-1} 0\right]=\frac{\pi}{2}\).
Example 6 Evaluate:
- \(\int_0^{\pi / 2} x \cos x d x\)
- \(\int_0^\pi \cos 2 x \log \sin x d x\)
- \(\int_1^2 \frac{\log x}{x^2} d x\)
- \(\int_0^{\pi / 6}\left(2+3 x^2\right) \cos 3 x d x\)
Solution
(1) Integrating by parts, we get
\(\int_0^{\pi / 2} x \cos x d x=[x \sin x]_0^{\pi / 2}-\int_0^{\pi / 2} 1 \cdot \sin x d x\)= \(\frac{\pi}{2}+[\cos x]_0^{\pi / 2}=\left(\frac{\pi}{2}-1\right) .\)
(2) Integrating by parts, taking log(sin x) as the first function, we get
\(\int_0^\pi \cos 2 x \log \sin x d x = \left[(\log \sin x) \cdot \frac{\sin 2 x}{2}\right]_0^\pi-\int_0^\pi\left(\cot x \cdot \frac{\sin 2 x}{2}\right) d x\)= \(0-\int_0^\pi \frac{\cos x}{\sin x} \cdot \frac{2 \sin x \cos x}{2} d x=-\int_0^\pi \cos ^2 x d x\)
= \(-\frac{1}{2} \int_0^\pi 2 \cos ^2 x d x=-\frac{1}{2} \int_0^\pi(1+\cos 2 x) d x\)
= \(-\frac{1}{2} \cdot\left[x+\frac{\sin 2 x}{2}\right]_0^\pi=-\frac{\pi}{2}\).
(3) Integrating by parts, taking (log x) as the first function, we get
\(\int_1^2 \frac{\log x}{x^2} d x=\int_1^2(\log x) \cdot x^{-2} d x\)= \(\left[(\log x)\left(-\frac{1}{x}\right)\right]_1^2-\int_1^2 \frac{1}{x} \cdot\left(-\frac{1}{x}\right) d x\)
= \(\left[-\frac{\log 2}{2}+\frac{\log 1}{1}\right]+\int_1^2 \frac{d x}{x^2}\)
= \(\frac{-\log 2}{2}-\left[\frac{1}{x}\right]_1^2=\frac{-\log 2}{2}-\left\{\frac{1}{2}-1\right\}=\left(\frac{1-\log 2}{2}\right)\).
(4) \(\int_0^{\pi / 6}\left(2+3 x^2\right) \cos 3 x d x\)
= \(2 \int_0^{\pi / 6} \cos 3 x d x+3 \int_0^{\pi / 6} x^2 \cos 3 x d x\)
= \(\left.2\left[\frac{\sin 3 x}{3}\right]_0^{\pi / 6}+3 \cdot\left[x^2\left(\frac{\sin 3 x}{3}\right)\right]_0^{\pi / 6}-\int_0^{\pi / 6} 2 x\left(\frac{\sin 3 x}{3}\right) d x\right\}\) [integrating by parts]
= \(\frac{2}{3}+\frac{\pi^2}{36}-2 \int_0^{\pi / 6} x \sin 3 x d x\)
= \(\frac{2}{3}+\frac{\pi^2}{36}-2\left\{\left[x\left(\frac{-\cos 3 x}{3}\right)\right]_0^{\pi / 6}-\int_0^{\pi / 6} 1 \cdot\left(\frac{-\cos 3 x}{3}\right) d x\right\}\) [integrating by parts]
= \(\frac{2}{3}+\frac{\pi^2}{36}+\frac{2}{3}[x \cos 3 x]_0^{\pi / 6}-\frac{2}{3} \cdot\left[\frac{\sin 3 x}{3}\right]_0^{\pi / 6}\)
= \(\frac{2}{3}+\frac{\pi^2}{36}-\frac{2}{9}=\left(\frac{\pi^2}{36}+\frac{4}{9}\right)=\frac{1}{36}\left(\pi^2+16\right)\).
Example 7 Evaluate \(\int_1^2 \frac{d x}{x\left(1+x^2\right)}\)
Solution
Given:
⇒ \(\int_1^2 \frac{d x}{x\left(1+x^2\right)}\)
Let \(\frac{1}{x\left(1+x^2\right)}=\frac{A}{x}+\frac{B x+C}{\left(1+x^2\right)} .\)
Then, 1 ≡ A(1+x2)+(Bx+C)x. Putting x = 0, we get A = 1.
Comparing the coefficients of x2, we get A + B = 0 or B = -1.
Comparing coefficients of x, we get C = 0.
∴ \(\frac{1}{x\left(1+x^2\right)}=\left[\frac{1}{x}-\frac{x}{1+x^2}\right]\)
So, \(\int_1^2 \frac{d x}{x\left(1+x^2\right)}=\int_1^2 \frac{d x}{x}-\frac{1}{2} \int_1^2 \frac{2 x}{1+x^2} d x\)
⇒ \([\log x]_1^2-\frac{1}{2}\left[\log \left(1+x^2\right)\right]_1^2\)
⇒ \(\left[\frac{3}{2}(\log 2)-\frac{1}{2}(\log 5)\right]\)
⇒ \(\int_1^2 \frac{d x}{x\left(1+x^2\right)}\) = \(\left[\frac{3}{2}(\log 2)-\frac{1}{2}(\log 5)\right]\)
Example 8 Evaluate \(\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x\)
Solution
Given:
⇒ \(\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x\)
Integrating by parts, taking (x2+x) as the first function and \(\frac{1}{\sqrt{2 x+1}}\) as the second function, we get
⇒ \(\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x=\left[\left(x^2+x\right) \cdot \sqrt{2 x+1}\right]_2^4-\int_2^4(2 x+1) \cdot \sqrt{2 x+1} d x\)
⇒ \((60-6 \sqrt{5})-\int_2^4(2 x+1)^{3 / 2} d x\)
⇒ \((60-6 \sqrt{5})-\frac{1}{5} \cdot\left[(2 x+1)^{5 / 2}\right]_2^4\)
⇒ \((60-6 \sqrt{5})-\left(\frac{243}{5}-5 \sqrt{5}\right)\)
⇒ \(\left(\frac{57}{5}-\sqrt{5}\right)=\left(\frac{57-5 \sqrt{5}}{5}\right)\).
⇒ \(\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x\) = \(\left(\frac{57}{5}-\sqrt{5}\right)=\left(\frac{57-5 \sqrt{5}}{5}\right)\).
“Definite Integrals In Physics Applications”
Example 9 Evaluate:
- \(\int_0^{1 / 2} \frac{d x}{\sqrt{1-x}}\)
- \(\int_0^1\left(\frac{1-x}{1+x}\right) d x\)
Solution
- \(\int_0^{1 / 2} \frac{d x}{\sqrt{1-x}}=\int_0^{1 / 2}(1-x)^{-1 / 2} d x=\left[\frac{2 \sqrt{1-x}}{-1}\right]_0^{1 / 2}=(2-\sqrt{2}) .\)
- \(\int_0^1\left(\frac{1-x}{1+x}\right) d x=\int_0^1\left(-1+\frac{2}{x+1}\right) d x\) [on dividing (-x+1) by (x+1)]
= \([-x+2 \log |x+1|]_0^1=[(2 \log 2)-1] .\)
To Evaluate a Definite Integral by Substitution
In \(\int_a^b f(x) d x\), when the variable x is converted into a new variable t by some relation then we put x = a and x = b in that relation to btain the corresponding values of t, giving the lower limit and the upper limit respectively of the new integrand in t.
Solved Examples
Example 1 Evaluate:
- \(\int_0^2 e^{x / 2} d x\)
- \(\int_2^4 \frac{x}{\left(x^2+1\right)} d x\)
- \(\int_0^1 \cos ^{-1} x d x\)
- \(\int_0^1 \frac{(2 x+3)}{\left(5 x^2+1\right)} d x\)
Solution
(1) Put \(\frac{x}{2}\) = t so that dx = 2 dt.
Also, (x = 0 ⇒ t = 0) and (x = 2 ⇒ t = 1).
∴ \(\int_0^2 e^{x / 2} d x=2 \int_0^1 e^t d t=2\left[e^t\right]_0^1=2(e-1) .\)
(2) Put (x2+1) = t so that x dx = \(\frac{1}{2}\) dt.
Also, (x = 2 ⇒ t = 5) and (x = 4 ⇒ t = 17).
∴ \(\int_2^4 \frac{x}{\left(x^2+1\right)} d x=\frac{1}{2} \int_5^{17} \frac{d t}{t}=\frac{1}{2}[\log |t|]_5^{17}=\frac{1}{2}(\log 17-\log 5)\)
(3) Put x = cos t so that dx = -sin t dt.
Also, (x = 0 ⇒ t = \(\frac{\pi}{2}\)) and (x = 1 ⇒ t = 0).
∴ \(\int_0^1 \cos ^{-1} x d x=-\int_{\pi / 2}^0 \cos ^{-1}(\cos t) \sin t d t=\int_0^{\pi / 2} t \sin t d t\)
= \([t(-\cos t)]_0^{\pi / 2}-\int_0^{\pi / 2} 1 \cdot(-\cos t) d t\) [integrating by parts]
= \([\sin t]_0^{\pi / 2}=1\)
(4) Let (2x+3) ≡ A \(\cdot \frac{d}{d x}\left(5 x^2+1\right)\)+B
Then, (2x+3) ≡ (10x)A + B.
Comparing the coefficients of like powers of x, we get
10 A = 2 or A = \(\frac{1}{5}\) and B = 3.
∴ (2x+3) = \(\frac{1}{5}\) (10x) + 3.
So, \(\int_0^1 \frac{(2 x+3)}{\left(5 x^2+1\right)} d x=\int_0^1 \frac{\frac{1}{5}(10 x)+3}{\left(5 x^2+1\right)} d x\)
= \(\frac{1}{5} \int_0^1 \frac{10 x}{\left(5 x^2+1\right)} d x+3 \int_0^1 \frac{d x}{\left(5 x^2+1\right)}\)
= \(\frac{1}{5}\left[\log \left|5 x^2+1\right|\right]_0^1+\frac{3}{5} \int_0^1 \frac{d x}{x^2+\left(\frac{1}{\sqrt{5}}\right)^2}\)
= \(\frac{1}{5} \log 6+\frac{3}{\sqrt{5}}\left(\tan ^{-1} \sqrt{5}\right)\).
Example 2 Evaluate:
- \(\int_1^3 \frac{\cos (\log x)}{x} d x\)
- \(\int_0^{\pi / 2} \sqrt{\cos \theta} \sin ^3 \theta d \theta\)
- \(\int_0^{\pi / 2} \frac{\cos x}{(1+\sin x)(2+\sin x)} d x\)
- \(\int_0^{\pi / 2} \frac{d x}{(1-2 \sin x)}\)
- \(\int_0^{\pi / 2} \frac{d x}{(3+2 \cos x)}\)
- \(\int_0^{\pi / 2} \frac{d x}{\left(4 \sin ^2 x+5 \cos ^2 x\right)}\)
Solution
(1) Put log x = t so that \(\frac{1}{x}\) dx = dt.
Also, (x = 1 ⇒ t = log 1 = 0) and (x = 3 ⇒ t = log 3).
∴ \(\int_1^3 \frac{\cos (\log x)}{x} d x=\int_0^{\log 3} \cos t d t=[\sin t]_0^{\log 3}=\sin (\log 3) .\)
(2) Put cos θ = t so that sin θ dθ = -dt.
Also, (θ = 0 ⇒ t = 1) and (θ = \(\frac{\pi}{2}\) ⇒ t = 0).
∴ \(\int_0^{\pi / 2} \sqrt{\cos \theta} \sin ^3 \theta d \theta=\int_0^{\pi / 2} \sqrt{\cos \theta} \cdot\left(1-\cos ^2 \theta\right) \sin \theta d \theta\)
= \(-\int_1^0 \sqrt{t}\left(1-t^2\right) d t=\int_0^1\left(t^{1 / 2}-t^{5 / 2}\right) d t\)
= \(\left[\frac{2}{3} t^{3 / 2}-\frac{2}{7} t^{7 / 2}\right]_0^1=\left(\frac{2}{3}-\frac{2}{7}\right)=\frac{8}{21} \text {. }\)
(3) Put sin x = t so that cos x dx = dt.
Also, (x = 0 ⇒ t = 0) and (x = \(\frac{\pi}{2}\) ⇒ t = 1).
∴ \(\int_0^{\pi / 2} \frac{\cos x}{(1+\sin x)(2+\sin x)} d x\)
= \(\int_0^1 \frac{d t}{(1+t)(2+t)}\)
= \(\int_0^1\left[\frac{1}{(1+t)}-\frac{1}{(2+t)}\right] d t\) [by partial fractions]
= \(\int_0^1 \frac{d t}{(1+t)}-\int_0^1 \frac{d t}{(2+t)}\)
= \([\log |1+t|]_0^1-[\log |2+t|]_0^1\)
= \([(log 2 – log 1) – (log 3 – log 2)] = (2 log 2) – (log 3)\).
(4) \(\int_0^{\pi / 2} \frac{d x}{(1-2 \sin x)}=\int_0^{\pi / 2} \frac{d x}{1-2\left\{\frac{2 \tan (x / 2)}{1+\tan ^2(x / 2)}\right\}}\)
= \(\int_0^{\pi / 2} \frac{\sec ^2(x / 2)}{\left[1+\tan ^2(x / 2)-4 \tan (x / 2)\right]} d x\)
= \(2 \int_0^1 \frac{d t}{\left(1+t^2-4 t\right)}\), where tan \(\frac{x}{2}\) = t
[x = 0 ⇒ t = 0 and x = \(\frac{\pi}{2}\) ⇒ t = 1]
= \(2 \int_0^1 \frac{d t}{(t-2)^2-(\sqrt{3})^2}=2 \cdot \frac{1}{2 \sqrt{3}}\left[\log \left|\frac{t-2-\sqrt{3}}{t-2+\sqrt{3}}\right|\right]_0^1\)
= \(\frac{1}{\sqrt{3}}\left[\log \left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)-\log \left(\frac{\sqrt{3}+2}{\sqrt{3}-2}\right)\right]\)
(5) \(\int_0^{\pi / 2} \frac{d x}{(3+2 \cos x)}=\int_0^{\pi / 2} \frac{d x}{3+2 \cdot\left[\frac{1-\tan ^2(x / 2)}{1+\tan ^2(x / 2)}\right]}\)
= \(\int_0^{\pi / 2} \frac{\sec ^2(x / 2)}{\tan ^2(x / 2)+5} d x\)
= \(2 \int_0^1 \frac{d t}{t^2+(\sqrt{5})^2}\), where tan \(\frac{x}{2}\) = t
[x = 0 ⇒ t = 0 and x = \(\frac{\pi}{2}\) ⇒ t = 1]
= \(2 \cdot \frac{1}{\sqrt{5}}\left[\tan ^{-1} \frac{t}{\sqrt{5}}\right]_0^1=\frac{2}{\sqrt{5}} \tan ^{-1} \frac{1}{\sqrt{5}} \text {. }\)
(6) \(\int_0^{\pi / 2} \frac{d x}{\left(4 \sin ^2 x+5 \cos ^2 x\right)}=\int_0^{\pi / 2} \frac{\sec ^2 x}{\left(4 \tan ^2 x+5\right)} d x\)
[dividing num. and denom. by cos2x]
= \(\int_0^{\infty} \frac{d t}{\left(4 t^2+5\right)}\), where tan x = t
[x = 0 ⇒ t = 0 and x = \(\frac{\pi}{2}\) ⇒ t = ∞]
= \(\frac{1}{4} \int_0^{\infty} \frac{d t}{t^2+\left(\frac{\sqrt{5}}{2}\right)^2}=\frac{1}{4} \cdot \frac{2}{\sqrt{5}}\left[\tan ^{-1} \frac{2 t}{\sqrt{5}}\right]_0^{\infty}\)
= \(\frac{1}{2 \sqrt{5}}\left[\tan ^{-1}(\infty)-\tan ^{-1}(0)\right]\)
= \(\frac{1}{2 \sqrt{5}}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4 \sqrt{5}}\).
“Fundamental Theorem Of Calculus And Definite Integrals”
Example 3 Evaluate:
- \(\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x\)
- \(\int_0^{1 / \sqrt{2}} \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}} d x\)
Solution
(1) Put x = tan θ so that dx = sec2θ dθ.
Clearly, x = 0 ⇒ θ = 0 and x = 1 ⇒ θ = (π/4).
∴ \(\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x=\int_0^{\pi / 4} \frac{\theta \tan \theta}{\sec ^3 \theta} \cdot \sec ^2 \theta d \theta=\int_0^{\pi / 4} \theta \sin \theta d \theta\)
⇒ \([-\theta \cos \theta]_0^{\pi / 4}-\int_0^{\pi / 4}(-\cos \theta) d \theta\) [integrating by parts]
⇒ \([-\theta \cos \theta]_0^{\pi / 4}+[\sin \theta]_0^{\pi / 4}=-\frac{\pi}{4} \cos \frac{\pi}{4}+\sin \frac{\pi}{4}\)
⇒ \(\left(\frac{-\pi}{4 \sqrt{2}}+\frac{1}{\sqrt{2}}\right)=\frac{(4-\pi)}{4 \sqrt{2}}=\frac{\sqrt{2}(4-\pi)}{8} .\)
(2) Put x = sin θ so that dx = cos θ dθ.
Clearly, (x = 0 ⇒ θ = 0) and (x = \(\frac{1}{\sqrt{2}}\) ⇒ θ = \(\frac{\pi}{4}\)).
∴ \(\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x=\int_0^{\pi / 4} \frac{\theta \tan \theta}{\sec ^3 \theta} \cdot \sec ^2 \theta d \theta=\int_0^{\pi / 4} \theta \sin \theta d \theta\)
⇒ \([-\theta \cos \theta]_0^{\pi / 4}-\int_0^{\pi / 4}(-\cos \theta) d \theta\) [integrating by parts]
⇒ \([-\theta \cos \theta]_0^{\pi / 4}+[\sin \theta]_0^{\pi / 4}=-\frac{\pi}{4} \cos \frac{\pi}{4}+\sin \frac{\pi}{4}\)
⇒ \(\frac{\pi}{4}+\log \left(\frac{1}{\sqrt{2}}\right)=\left(\frac{\pi}{4}-\frac{1}{2} \log 2\right) .\)
Example 4 Evaluate:
- \(\int_0^{\pi / 2} \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x\)
- \(\int_0^{\pi / 2} \frac{\cos x}{(1+\cos x+\sin x)} d x\)
Solution
(1) \(\int_0^{\pi / 2} \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x=\int_0^{\pi / 2} \frac{\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x\)
= \(\int_0^{\pi / 2} \frac{2\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2} d x=2 \cdot \int_1^{\sqrt{2}} \frac{d t}{t^2}=\left[\frac{-2}{t}\right]_1^{\sqrt{2}}=\sqrt{2}(\sqrt{2}-1) .\)
[putting cos \(\frac{x}{2}\) + sin \(\frac{x}{2}\) = t and \(\frac{1}{2}\)(cos \(\frac{x}{2}\) – sin \(\frac{x}{2}\))dx = dt; also, x = 0 ⇒ t = 1 and x = (π/2) ⇒ t = √2]
(2) \(\int_0^{\pi / 2} \frac{\cos x}{(1+\cos x+\sin x)} d x=\int_0^{\pi / 2} \frac{\cos x}{(1+\cos x)+\sin x} d x\)
= \(\int_0^{\pi / 2} \frac{\cos ^2(x / 2)-\sin ^2(x / 2)}{\left[2 \cos ^2(x / 2)+2 \sin (x / 2) \cos (x / 2)\right]} d x\)
= \(\frac{1}{2} \int_0^{\pi / 2} \frac{1-\tan ^2(x / 2)}{1+\tan (x / 2)} d x\)
[dividing num. and denom. by cos2(x/2)]
= \(\frac{1}{2} \int_0^{\pi / 2}[1-\tan (x / 2)] d x=\frac{1}{2} \int_0^{\pi / 2} d x-\frac{1}{2} \int_0^{\pi / 2} \frac{\sin (x / 2)}{\cos (x / 2)} d x\)
= \(\frac{1}{2} \cdot[x]_0^{\pi / 2}+[\log \cos (x / 2)]_0^{\pi / 2}\)
= \(\frac{\pi}{4}+\log \cos \frac{\pi}{4}=\frac{\pi}{4}+\log \left(\frac{1}{\sqrt{2}}\right)=\left(\frac{\pi}{4}-\frac{1}{2} \log 2\right) .\)
Example 5 Evaluate \(\int_{-a}^a \sqrt{\frac{a-x}{a+x}} d x\)
Solution
Given:
⇒ \(\int_{-a}^a \sqrt{\frac{a-x}{a+x}} d x\)
Put x = a cos θ so that dx = – a sin θ dθ.
Also, (x = -a ⇒ θ = π) and (x = a ⇒ θ = 0).
∴ \(\int_{-a}^a \sqrt{\frac{a-x}{a+x}} d x=\int_\pi^0 \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \cdot(-a \sin \theta) d \theta\)
= \(a \int_0^\pi \sqrt{\frac{2 \sin ^2(\theta / 2)}{2 \cos ^2(\theta / 2)}} \cdot 2 \sin (\theta / 2) \cos (\theta / 2) d \theta\)
= \(a \int_0^\pi 2 \sin ^2(\theta / 2) d \theta=a \int_0^\pi(1-\cos \theta) d \theta\)
= \(a \int_0^\pi d \theta-a \int_0^\pi \cos \theta d \theta\)
= \(a \cdot[\theta]_0^\pi-a[\sin \theta]_0^\pi=a \pi .\)
⇒ \(\int_{-a}^a \sqrt{\frac{a-x}{a+x}} d x\) = \(a \cdot[\theta]_0^\pi-a[\sin \theta]_0^\pi=a \pi .\)
Example 6 Evaluate \(\int_0^1 x \cdot \sqrt{\frac{1-x^2}{1+x^2}} d x\)
Solution
Given:
⇒ \(\int_0^1 x \cdot \sqrt{\frac{1-x^2}{1+x^2}} d x\)
Put x2 = t and x dx = \(\frac{1}{2}\) dt. Then,
[x = 0 ⇒ t = 0] and [x = 1 ⇒ t = 1].
∴ I = \(\frac{1}{2} \cdot \int_0^1 \sqrt{\frac{1-t}{1+t}} d t\)
= \(\frac{1}{2} \cdot \int_0^1\left\{\frac{\sqrt{1-t}}{\sqrt{1+t}} \times \frac{\sqrt{1-t}}{\sqrt{1-t}}\right\} d t=\frac{1}{2} \cdot \int_0^1 \frac{(1-t)}{\sqrt{1-t^2}} d t\)
= \(\frac{1}{2} \cdot \int_0^1 \frac{d t}{\sqrt{1-t^2}}-\frac{1}{2} \cdot \int_0^1 \frac{t}{\sqrt{1-t^2}} d t\)
= \(\frac{1}{2} \cdot\left[\sin ^{-1} t\right]_0^1+\frac{1}{4} \cdot \int_0^1 \frac{(-2 t)}{\sqrt{1-t^2}} d t\)
= \(\frac{1}{2} \cdot\left[\sin ^{-1} 1-\sin ^{-1} 0\right]+\frac{1}{4} \cdot \int_1^0 \frac{1}{\sqrt{u}} d u\), where (1-t2) = u
= \(\frac{1}{2}\left(\frac{\pi}{2}-0\right)-\frac{1}{4} \cdot \int_0^1 \frac{d u}{\sqrt{u}}=\frac{\pi}{4}-\frac{1}{4}[2 \sqrt{u}]_0^1\)
= \(\frac{\pi}{4}-\frac{1}{2}[\sqrt{1}-\sqrt{0}]=\left(\frac{\pi}{4}-\frac{1}{2}\right)\)
⇒ \(\int_0^1 x \cdot \sqrt{\frac{1-x^2}{1+x^2}} d x\) = \(\frac{\pi}{4}-\frac{1}{2}[\sqrt{1}-\sqrt{0}]=\left(\frac{\pi}{4}-\frac{1}{2}\right)\)
Example 7 Evaluate \(\int_0^{\pi / 2} \frac{\cos x}{(3 \cos x+\sin x)} d x \text {. }\)
Solution
Given:
⇒ \(\int_0^{\pi / 2} \frac{\cos x}{(3 \cos x+\sin x)} d x \text {. }\)
Let cos x = \(A(3 \cos x+\sin x)+B \cdot \frac{d}{d x}(3 \cos x+\sin x)\)
Then, cos x = A(3 cos x + sin x) + B.(-3 sin x + cos x)
Comparing the coefficients of cos x, we get 3A + B = 1.
Comparing the coefficients of sin x, we get A – 3B = 0.
Solving 3A + B = 1 and A – 3B = 0, we get A = \(\frac{3}{10}\) and B = \(\frac{1}{10}\).
∴ cos x = \(\frac{3}{10}\)(3 cos x + sin x) + \(\frac{1}{10}\) (-3 sin x + cos x).
So, \(\int_0^{\pi / 2} \frac{\cos x}{(3 \cos x+\sin x)} d x\)
= \(\frac{3}{10} \int_0^{\pi / 2} \frac{(3 \cos x+\sin x)}{(3 \cos x+\sin x)} d x+\frac{1}{10} \int_0^{\pi / 2} \frac{(-3 \sin x+\cos x)}{(3 \cos x+\sin x)} d x\)
= \(\frac{3}{10} \int_0^{\pi / 2} d x+\frac{1}{10} \int_0^{\pi / 2} \frac{(-3 \sin x+\cos x)}{(3 \cos x+\sin x)} d x\)
= \(\frac{3}{10} \cdot[x]_0^{\pi / 2}+\frac{1}{10} \cdot[\log |3 \cos x+\sin x|]_0^{\pi / 2}=\left(\frac{3 \pi}{20}-\frac{\log 3}{10}\right)\)
⇒ \(\int_0^{\pi / 2} \frac{\cos x}{(3 \cos x+\sin x)} d x \text {. }\) = \(\frac{3}{10} \cdot[x]_0^{\pi / 2}+\frac{1}{10} \cdot[\log |3 \cos x+\sin x|]_0^{\pi / 2}=\left(\frac{3 \pi}{20}-\frac{\log 3}{10}\right)\)
“Numerical Methods For Definite Integrals”
Example 8 Evaluate \(\int_0^{\pi / 2}\{\sqrt{\tan x}+\sqrt{\cot x}\} d x\)
Solution
Given:
⇒ \(\int_0^{\pi / 2}\{\sqrt{\tan x}+\sqrt{\cot x}\} d x\)
We have
I = \(\int_0^{\pi / 2}\left\{\frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}}\right\} d x=\int_0^{\pi / 2} \frac{(\sin x+\cos x)}{\sqrt{\sin x \cos x}} d x\)
= \(\sqrt{2} \cdot \int_0^{\pi / 2} \frac{(\sin x+\cos x)}{\sqrt{2 \sin x \cos x}} d x=\sqrt{2} \cdot \int_0^{\pi / 2} \frac{(\sin x+\cos x)}{\sqrt{1-(\sin x-\cos x)^2}} d x\)
Put (sin x – cos x) = t and (cos x + sin x)dx = dt.
Also, [x = 0 ⇒ t = -1] and [x = \(\frac{\pi}{2}\) ⇒ t = 1].
∴ I = \(\sqrt{2} \cdot \int_{-1}^1 \frac{d t}{\sqrt{1-t^2}}=\sqrt{2}\left[\sin ^{-1} t\right]_{-1}^1\)
= \(\sqrt{2}\left\{\sin ^{-1}(1)-\sin ^{-1}(-1)\right\}=\sqrt{2}\left\{2 \sin ^{-1}(1)\right\}\)
= \(\left(\sqrt{2} \times 2 \times \frac{\pi}{2}\right)=\sqrt{2} \pi\)
⇒ \(\int_0^{\pi / 2}\{\sqrt{\tan x}+\sqrt{\cot x}\} d x\) = \(\left(\sqrt{2} \times 2 \times \frac{\pi}{2}\right)=\sqrt{2} \pi\)
Properties of Definite Integrals
Theorem 1 \(\int_a^b f(x) d x=\int_a^b f(t) d t\)
Proof
Let \(\int f(x) d x=F(x) .\) Then, \(\int f(t) d t=F(t)\)
∴ \(\int_a^b f(x) d x=[F(x)]_a^b=F(b)-F(a) .\)
And, \(\int_a^b f(t) d t=[F(t)]_a^b=F(b)-F(a)\)
Hence, \(\int_a^b f(x) d x=\int_a^b f(t) d t\)
Theorem 2 \(\int_a^b f(x) d x=-\int_b^a f(x) d x .\)
Proof
Let \(\int f(x) d x=F(x) .\)
Then, \(\int_a^b f(x) d x=[F(x)]_a^b=F(b)-F(a) .\)
And, \(-\int_b^a f(x) d x=-[F(x)]_b^a=-[F(a)-F(b)]=F(b)-F(a) .\)
Hence, \(\int_a^b f(x) d x=-\int_b^a f(x) d x .\)
Theorem 3 \(\int^b f(x) d x=\int^c f(x) d x+\int^b f(x) d x\), where a < c < b.
Proof
Let \(\int f(x) d x=F(x) .\) Then, \(\int_a f(x) d x=[F(x)]_a^b=F(b)-F(a)\)
⇒ \(\int^c f(x) d x+\int^b f(x) d x=[F(x)]_a^c+[F(x)]_c^b=\{F(c)-F(a)\}+\{F(b)-F(c)\}\)
= \(F(b)-F(a)=[F(x)]_a^b=\int_a^b f(x) d x\).
Hence, \(\int_a^c f(x) d x+\int_c^b f(x) d x=\int_a^b f(x) d x .\)
Remark If a < c1 < c2 < …. < cn < b then
⇒ \(\int_a^b f(x) d x=\int_a^{c_1} f(x) d x+\int_{c_1}^{c_2} f(x) d x+\ldots+\int_{c_n}^b f(x) d x .\)
Theorem 4 \(\int_0^a f(x) d x=\int_0^a f(a-x) d x .\)
Proof
In the RHS integral, put (a-x) = t so that dx = -dt.
Now, when x = 0 we have t = a.
And, when x = a we have t = 0.
∴ \(\int_0^a f(a-x) d x=-\int_a^0 f(t) d t=\int_0^a f(t) d t=\int_0^a f(x) d x .\)
Hence, \(\int_0^a f(x) d x=\int_0^a f(a-x) d x .\)
Theorem 5 \(\int_a^b f(a+b-x) d x=\int_a^b f(x) d x .\)
Proof
Putting a + b – x =t, we get dx = -dt.
Now, x = a ⇒ t = b.
And, x = b ⇒ t = a.
∴ \(\int_a^b f(a+b-x) d x=-\int_b^a f(t) d t=\int_a^b f(t) d t=\int_a^b f(x) d x .\)
Hence, \(\int_a^b f(a+b-x) d x=\int_a^b f(x) d x\)
Theorem 6 \(\int_a^b\{f(x)+g(x)\} d x=\int_a^b f(x) d x+\int_a^b g(x) d x .\)
Proof
Let \(\int f(x) d x=F(x) \text { and } \int g(x) d x=G(x) \text {. }\)
Then, \(\int[f(x)+g(x)] d x=\int f(x) d x+\int g(x) d x=F(x)+G(x) .\)
∴ \(\int_a^b[f(x)+g(x)] d x=[F(x)+G(x)]_a^b\)
= [F(b) + G(b)] – [F(a) + G(a)]
= [F(b) – F(a)] + [G(b) – G(a)]
= \(\int_a^b f(x) d x+\int_a^b g(x) d x\)
Hence, \(\int_a^b\{f(x)+g(x)\} d x=\int_a^b f(x) d x+\int_a^b g(x) d x\)
“Difference Between Definite And Indefinite Integrals”
Theorem 7 \(\int_{-a}^a f(x) d x=\left\{\begin{array}{l}
0, \text { when } f(x) \text { is an odd function } \\
2 \int_0^a f(x) d x \text {, when } f(x) \text { is an even function. }
\end{array}\right.\)
Proof
We have \(\int_{-a}^a f(x) d x=\int_{-a}^0 f(x) d x+\int_0^a f(x) d x\) …(1)
In the first integral on the RHS of (1), put x = -t so that dx = -dt.
When x = -a, we have t = a. And, when x = 0, we have t = 0.
∴ \(\int_{-a}^0 f(x) d x=-\int_a^0 f(-t) d t=\int_0^a f(-t) d t=\int_0^a f(-x) d x\)
Thus, \(\int_{-a}^0 f(x) d x=\int_0^a f(-x) d x\) …(2)
Using (2) in (1), we have
⇒ \(\int_{-a}^a f(x) d x=\int_0^a f(-x) d x+\int_0^a f(x) d x=\int_0^a[f(-x)+f(x)] d x\)
= \(\left\{\begin{array}{l}
0, \text { when } f(x) \text { is odd } \\
2 \int_0^a f(x) d x \text {, when } f(x) \text { is even. }
\end{array}\right.\)
[∵ f(x) is odd ⇒ f(-x) = -f(x) and f(x) is even ⇒ f(-x) = f(x)]
Theorem 8 \(\int_0^{2 a} f(x) d x=\left\{\begin{array}{l}
0, \text { if } f(2 a-x)=-f(x) \\
2 \int_0^a f(x) d x, \text { if } f(2 a-x)=f(x)
\end{array}\right.\)
Proof
We have \(\int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_a^{2 \pi} f(x) d x\) …(1)
Now, in the second integral on the RHS of (1), put x = 2a – t so that dx = -dt.
When x = a, we have t = a. When x = 2a, we have t = 0.
∴ \(\int_a^{2 a} f(x) d x=-\int_a^0 f(2 a-t) d t=\int_a^a f(2 a-t) d t\)
= \(\int_0^a f(2 a-x) d x .\)
Thus, \(\int_a^{2 \pi} f(x) d x=\int_0^a f(2 a-x) d x\) …(2)
Using (2) in (1), we get
⇒ \(\int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_0^a f(2 a-x) d x\)
= \(\int_0^a\{f(x)+f(2 a-x)\} d x\)
= \(\left\{\begin{array}{l}
0, \text { when } f(2 a-x)=-f(x) \\
2 \int^a f(x) d x \text {, when } f(2 a-x)=f(x) .
\end{array}\right.\)
Hence, \(\int_0^{2 a} f(x) d x=\left\{\begin{array}{l}
0, \text { when } f(2 a-x)=-f(x) \\
2 \int_0^a f(x) d x, \text { when } f(2 a-x)=f(x) .
\end{array}\right.\)
Solved Examples
Example 1 Prove that \(\int_0^{\pi / 2} \frac{\sin x}{(\sin x+\cos x)} d x=\frac{\pi}{4} .\)
Solution
Given:
⇒ \(\int_0^{\pi / 2} \frac{\sin x}{(\sin x+\cos x)} d x=\frac{\pi}{4} .\)
Let I = \(\int_0^{\pi / 2} \frac{\sin x}{(\sin x+\cos x)} d x\) …(1)
Using the result \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\) in (1), we get
I = \(\int_0^{\pi / 2} \frac{\sin [(\pi / 2)-x]}{\sin [(\pi / 2)-x]+\cos [(\pi / 2)-x]} d x\)
or I = \(\int_0^{\pi / 2} \frac{\cos x}{(\cos x+\sin x)} d x=\int_0^{\pi / 2} \frac{\cos x}{(\sin x+\cos x)} d x\) …(2)
Adding (1) and (2), we get
2I = \(\int_0^{\pi / 2} \frac{\sin x}{(\sin x+\cos x)} d x+\int_0^{\pi / 2} \frac{\cos x}{(\sin x+\cos x)} d x\)
= \(\int_0^{\pi / 2} \frac{(\sin x+\cos x)}{(\sin x+\cos x)} d x=\int_0^{\pi / 2} d x=[x]_0^{\pi / 2}=\frac{\pi}{2}\)
∴ I = \(\frac{\pi}{4}, i.e., \int_0^{\pi / 2} \frac{\sin x}{(\sin x+\cos x)} d x=\frac{\pi}{4} \text {. }\)
Example 2 Prove that \(\int_0^{\pi / 2} \frac{\sqrt{\cos x}}{(\sqrt{\sin x}+\sqrt{\cos x})} d x=\frac{\pi}{4} \text {. }\)
Solution
Given:
⇒ \(\int_0^{\pi / 2} \frac{\sqrt{\cos x}}{(\sqrt{\sin x}+\sqrt{\cos x})} d x=\frac{\pi}{4} \text {. }\)
Let I = \(\int_0^{\pi / 2} \frac{\sqrt{\cos x}}{(\sqrt{\sin x}+\sqrt{\cos x})} d x\) …(1)
Using the result \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\) in (1), we get
I = \(\int_0^{\pi / 2} \frac{\sqrt{\cos [(\pi / 2)-x]}}{\sqrt{\sin [(\pi / 2)-x]}+\sqrt{\cos [(\pi / 2)-x]}} d x\)
or I = \(\int_0^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x=\int_0^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\) …(2)
Adding (1) and (2), we get
2I = \(\int_0^{\pi / 2} \frac{\sqrt{\cos x}}{(\sqrt{\sin x}+\sqrt{\cos x})} d x+\int_0^{\pi / 2} \frac{\sqrt{\sin x}}{(\sqrt{\sin x}+\sqrt{\cos x})} d x\)
= \(\int_0^{\pi / 2} \frac{(\sqrt{\sin x}+\sqrt{\cos x})}{(\sqrt{\sin x}+\sqrt{\cos x})} d x=\int_0^{\pi / 2} d x=[x]_0^{\pi / 2}=\frac{\pi}{2} .\)
∴ I = \(\frac{\pi}{4}, i.e., \int_0^{\pi / 2} \frac{\sqrt{\cos x}}{(\sqrt{\sin x}+\sqrt{\cos x})} d x=\frac{\pi}{4} .\)
Example 3 Evaluate
- \(\int_0^1 x(1-x)^n d x\)
- \(\int_0^1 x(1-x)^{3 / 2} d x\)
Solution
(1) We have
\(\int^1 x(1-x)^n d x=\int^1(1-x)[1-(1-x)]^n d x\)[∵ \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)]
= \(\int_0^1(1-x) x^n d x=\int_0^1 x^n d x-\int_0^1 x^{n+1} d x\)
= \(\left[\frac{x^{n+1}}{(n+1)}\right]_0^1-\left[\frac{x^{n+2}}{n+2}\right]_0^1=\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)
= \(\frac{1}{(n+1)(n+2)}\).
(2) We have \(\int_0^1 x(1-x)^{3 / 2} d x=\int_0^1(1-x)[1-(1-x)]^{3 / 2} d x\)
[∵ \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)]
= \(\int_0^1(1-x) x^{3 / 2} d x=\int_0^1 x^{3 / 2} d x-\int_0^1 x^{5 / 2} d x\)
= \(\left[\frac{2}{5} x^{5 / 2}\right]_0^1-\left[\frac{2}{7} x^{7 / 2}\right]_0^1=\left(\frac{2}{5}-\frac{2}{7}\right)=\frac{4}{35} .\)
“Definite Integral Vs Area Under Curve”
Example 4 Show that \(\int_0^{\pi / 2} \log (\tan x) d x=0 .\)
Solution
Given:
⇒ \(\int_0^{\pi / 2} \log (\tan x) d x=0 .\)
Let I = \(\int_0^{\pi / 2} \log (\tan x) d x\)
Then, I = \(\int_0^{\pi / 2} \log \left[\tan \left(\frac{\pi}{2}-x\right)\right] d x\) [∵ \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)]
or I = \(\int_0^{\pi / 2} \log (\cot x) d x=\int_0^{\pi / 2} \log \left(\frac{1}{\tan x}\right) d x=-\int_0^{\pi / 2} \log \tan x d x=-I\)
∴ I = -I or 2I = 0 or I = 0.
Hence, \(\int_0^{\pi / 2} \log (\tan x) d x=0\)
Example 5 Prove that \(\int_0^{\pi / 4} \log (1+\tan x) d x=\frac{\pi}{8}(\log 2)\)
Solution
Given:
⇒ \(\int_0^{\pi / 4} \log (1+\tan x) d x=\frac{\pi}{8}(\log 2)\)
Let I = \(\int_0^{\pi / 4} \log (1+\tan x) d x\) …(1)
Then, I = \(\int_0^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x\) [∵ \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)]
or I = \(\int_0^{\pi / 4} \log \left[1+\frac{1-\tan x}{1+\tan x}\right] d x\)
or I = \(\int_0^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x\)
or I = \(\int_0^{\pi / 4}[\log 2-\log (1+\tan x)] d x\)
or I = \((\log 2) \cdot \int_0^{\pi / 4} d x-\int_0^{\pi / 4} \log (1+\tan x) d x\) …(2)
Adding (1) and (2), we get
2I = \((\log 2) \int_0^{\pi / 4} d x=(\log 2) \cdot[x]=\frac{\pi}{4}(\log 2)\)
∴ I = \(\frac{\pi}{8}(\log 2), i.e., \int_0^{\pi / 4} \log (1+\tan x) d x=\frac{\pi}{8}(\log 2)\).
Example 6 Prove that \(\int_0^{\pi / 2} \log (\sin x) d x=\int_0^{\pi / 2} \log (\cos x) d x=-\frac{\pi}{2}(\log 2) .\)
Solution
Given:
⇒ \(\int_0^{\pi / 2} \log (\sin x) d x=\int_0^{\pi / 2} \log (\cos x) d x=-\frac{\pi}{2}(\log 2) .\)
Let I = \(\int_0^{\pi / 2} \log (\sin x) d x\) …(1)
Then, I = \(\int_0^{\pi / 2} \log \left[\sin \left(\frac{\pi}{2}-x\right)\right] d x\) [∵ \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)]
or I = \(\int_0^{\pi / 2} \log (\cos x) d x\) …(2)
Adding (1) and (2), we get
2I = \(\int_0^{\pi / 2}[\log (\sin x)+\log (\cos x)] d x\)
= \(\int_0^{\pi / 2} \log (\sin x \cos x) d x=\int_0^{\pi / 2} \log \left(\frac{\sin 2 x}{2}\right) d x\)
= \(\int_0^{\pi / 2} \log (\sin 2 x) d x-\int_0^{\pi / 2}(\log 2) d x=\frac{1}{2} \int_0^\pi \log \sin t d t-(\log 2) \cdot \int_0^{\pi / 2} d x\)
[putting 2x = t in the 1st integral]
= \(\frac{1}{2} \int_0^\pi \log \sin t d t-(\log 2) \cdot[x]_0^{\pi / 2}\)
= \(\left(\frac{1}{2} \times 2\right) \cdot \int_0^{\pi / 2} \log \sin t d t-\frac{\pi}{2}(\log 2)\)
= \(\int_0^{\pi / 2} \log (\sin x) d x-\frac{\pi}{2}(\log 2) .\)
∴ 2I = \(I-\frac{\pi}{2}(\log 2) \text { or } I=-\frac{\pi}{2}(\log 2)\)
∴ \(\int_0^{\pi / 2} \log (\sin x) d x=\int_0^{\pi / 2} \log (\cos x) d x=-\frac{\pi}{2}(\log 2) .\)
Example 7 Prove that:
- \(\int_0^{\pi / 2} \sin 2 x \log (\tan x) d x=0\)
- \(\int_0^1 \log \left(\frac{1}{x}-1\right) d x=0\)
Solution
(1) Let I = \(\int_0^{\pi / 2} \sin 2 x \log (\tan x) d x\) …(1)
Then, I = \(\int_0^{\pi / 2} \sin 2\left(\frac{\pi}{2}-x\right) \log \tan \left(\frac{\pi}{2}-x\right) d x\)
[∵ \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)]
or I = \(\int_0^{\pi / 2} \sin 2 x \log (\cot x) d x\) …(2)
Adding (1) and (2), we get
2I = \(\int_0^{\pi / 2}[\sin 2 x \log (\tan x)+\sin 2 x \log (\cot x)] d x\)
= \(\int_0^{\pi / 2} \sin 2 x\{\log (\tan x)+\log (\cot x)\} d x\)
= \(\int_0^{\pi / 2} \sin 2 x \cdot \log (\tan x \cdot \cot x) d x=\int_0^{\pi / 2} \sin 2 x \cdot \log (1) d x=0\)
[∴ log 1 = 0]
∴ I = 0 ⇒ \(\int_0^{\pi / 2} \sin 2 x \log (\tan x) d x=0 .\)
(b) Put x = cos2t so that dx = -sin 2t dt
= \(\int_0^{\pi / 2} \sin 2 t \cdot \log (\tan t) d t=0\) [from (a)].
Example 8 Evaluate:
- \(\int_0^{\pi / 2} \frac{\sin ^2 x}{(\sin x+\cos x)} d x\)
- \(\int_0^{\pi / 2} \frac{\sin ^2 x}{(1+\sin x \cos x)} d x\)
Solution
(1) Let I = \(\int_0^{\pi / 2} \frac{\sin ^2 x}{(\sin x+\cos x)} d x\) …(1)
Then, I = \(\int_0^{\pi / 2} \frac{\sin ^2\left(\frac{\pi}{2}-x\right)}{\left[\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)\right]} d x\)
or I = \(\int_0^{\pi / 2} \frac{\cos ^2 x}{(\cos x+\sin x)} d x=\int_0^{\pi / 2} \frac{\cos ^2 x}{(\sin x+\cos x)} d x\) …(2)
Adding (1) and (2), we get
2I = \(\int_0^{\pi / 2} \frac{\left(\sin ^2 x+\cos ^2 x\right)}{(\sin x+\cos x)} d x=\int_0^{\pi / 2} \frac{d x}{(\sin x+\cos x)}\)
= \(\int_0^{\pi / 2} \frac{d x}{\left[\frac{2 \tan (x / 2)}{1+\tan ^2(x / 2)}+\frac{1-\tan ^2(x / 2)}{1+\tan ^2(x / 2)}\right]}\)
= \(\int_0^{\pi / 2} \frac{\sec ^2(x / 2)}{\left[1-\tan ^2(x / 2)+2 \tan (x / 2)\right]} d x\)
= \(2 \int_0^1 \frac{d t}{\left(1-t^2+2 t\right)}\), where tan \(\frac{x}{2}\) = t
[x = 0 ⇒ t = 0 and x = \(\frac{\pi}{2}\) ⇒ t = 1]
= \(2 \int_0^1 \frac{d t}{2-(t-1)^2}=2 \int_0^1 \frac{d t}{(\sqrt{2})^2-(t-1)^2}\)
= \(2 \cdot \frac{1}{2 \sqrt{2}}\left\{\log \left|\frac{\sqrt{2}+t-1}{\sqrt{2}-t+1}\right|\right\}_0^1=\frac{1}{\sqrt{2}}\left[0-\log \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)\right]\)
= \(-\frac{1}{\sqrt{2}} \log \left[\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)} \times \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}\right]=-\frac{1}{\sqrt{2}} \log (\sqrt{2}-1)^2\)
= -√2 log (√2 – 1).
(2) Let I = \(\int_0^{\pi / 2} \frac{\sin ^2 x}{(1+\sin x \cos x)} d x\) …(1)
Then, I = \(\int_0^{\pi / 2} \frac{\sin ^2[(\pi / 2)-x]}{1+\sin [(\pi / 2)-x] \cos [(\pi / 2)-x]} d x\)
or I = \(\int_0^{\pi / 2} \frac{\cos ^2 x}{(1+\sin x \cos x)} d x\) …(2)
Adding (1) and (2), we get
2I = \(\int_0^{\pi / 2} \frac{\left(\sin ^2 x+\cos ^2 x\right)}{(1+\sin x \cos x)} d x=\int_0^{\pi / 2} \frac{d x}{(1+\sin x \cos x)}\)
[∵ \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)]
= \(\int_0^{\pi / 2} \frac{\sec ^2 x}{\left(\sec ^2 x+\tan x\right)} d x\) [dividing num. and denom. by cos2x]
= \(\int_0^{\pi / 2} \frac{\sec ^2 x}{\left(1+\tan ^2 x+\tan x\right)} d x=\int_0^\pi \frac{d t}{\left(t^2+t+1\right)}\), where tan x = t
[x = 0 ⇒ t = 0 and x = \(\frac{\pi}{2}\) ⇒ t = ∞]
= \(\int_0^{\infty} \frac{d t}{\left(t+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\left[\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t+1}{\sqrt{3}}\right)\right]_0^{\infty}\)
= \(\frac{2}{\sqrt{3}}\left[\tan ^{-1}(\infty)-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\right]=\frac{2}{\sqrt{3}} \cdot\left(\frac{\pi}{2}-\frac{\pi}{6}\right)=\frac{2 \pi}{3 \sqrt{3}} .\)
“Practice Problems For Definite Integrals With Answers”
Example 9 Prove that \(\int_0^\pi \frac{x \tan x}{(\sec x+\tan x)} d x=\pi\left(\frac{\pi}{2}-1\right) .\)
Solution
Given:
⇒ \(\int_0^\pi \frac{x \tan x}{(\sec x+\tan x)} d x=\pi\left(\frac{\pi}{2}-1\right) .\)
Let I = \(\int_0^\pi \frac{x \tan x}{(\sec x+\tan x)} d x\) …(1)
Then, I = \(\int_0^\pi \frac{(\pi-x) \tan (\pi-x)}{[\sec (\pi-x)+\tan (\pi-x)]} d x\) [∵ \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)]
or I = \(\int_0^\pi \frac{(\pi-x) \tan x}{(\sec x+\tan x)} d x\) …(2)
Adding (1) and (2), we get
2I = \(\pi \int_0^\pi \frac{\tan x}{(\sec x+\tan x)} d x=\pi \int_0^\pi \frac{\tan x(\sec x-\tan x)}{\left(\sec ^2 x-\tan ^2 x\right)} d x\)
= \(\pi \cdot\left[\int_0^\pi \sec x \tan x d x-\int_0^\pi \tan ^2 x d x\right]\)
= \(\pi \cdot\left\{[\sec x]_0^\pi-\int_0^\pi\left(\sec ^2 x-1\right) d x\right\}\)
= \(\pi \cdot\left\{-2-[\tan x]_0^\pi+[x]_0^\pi\right\}=\pi(\pi-2)\)
∴ I = \(\pi\left(\frac{\pi}{2}-1\right), i.e., \int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x=\pi\left(\frac{\pi}{2}-1\right) .\)
Example 10 Evaluate \(\int_0^\pi \frac{x}{(1+\sin x)} d x\)
Solution
Given:
⇒ \(\int_0^\pi \frac{x}{(1+\sin x)} d x\)
Let I = \(\int_0^\pi \frac{x}{(1+\sin x)} d x\) …(1)
Then, I = \(\int_0^\pi \frac{(\pi-x)}{1+\sin (\pi-x)} d x=\int_0^\pi \frac{(\pi-x)}{(1+\sin x)} d x\) …(2)
Adding (1) and (2), we get
2I = \(\pi \int_0^\pi \frac{d x}{(1+\sin x)}=\pi \cdot \int_0^\pi \frac{1}{(1+\sin x)} \times \frac{(1-\sin x)}{(1-\sin x)} d x\)
or 2I = \(\int_0^\pi\left(\frac{1-\sin x}{\cos ^2 x}\right) d x=\pi \cdot\left[\int_0^\pi \sec ^2 x d x-\int_0^\pi \sec x \tan x d x\right]\)
= \(\pi \cdot\left\{[\tan x]_0^\pi-[\sec x]_0^\pi\right\}=2 \pi\)
∴ I = π, i.e., \(\int_0^\pi \frac{x}{(1+\sin x)} d x=\pi .\)
Example 11 Evaluate \(\int_0^\pi \frac{x \sin x}{\left(1+\cos ^2 x\right)} d x\)
Solution
Given:
⇒ \(\int_0^\pi \frac{x \sin x}{\left(1+\cos ^2 x\right)} d x\)
Let I = \(\int_0^\pi \frac{x \sin x}{\left(1+\cos ^2 x\right)} d x\) …(1)
Then, I = \(\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} d x\)
or I = \(\int_0^\pi \frac{(\pi-x) \sin x}{\left(1+\cos ^2 x\right)} d x\) …(2)
Adding (1) and (2), we get
2I = \(\pi \int_0^\pi \frac{\sin x}{\left(1+\cos ^2 x\right)} d x=-\pi \int_1^{-1} \frac{d t}{\left(1+t^2\right)}\), where cos x = t
[x = 0 ⇒ t = 1 and x = π ⇒ t = -1]
= \(\pi \int_{-1}^1 \frac{d t}{\left(1+t^2\right)}=\pi\left[\tan ^{-1} t\right]_{-1}^1\)
= \(\pi\left[\tan ^{-1} 1-\tan ^{-1}(-1)\right]=\pi\left[\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right]=\frac{\pi^2}{2} .\)
∴ I = \(\frac{\pi^2}{4} .\)
Example 12 Evaluate \(\int_0^{\pi / 2} \frac{x}{(\sin x+\cos x)} d x\)
Solution
Given
⇒ \(\int_0^{\pi / 2} \frac{x}{(\sin x+\cos x)} d x\)
Let I = \(\int_0^{\pi / 2} \frac{x}{(\sin x+\cos x)} d x\) …(1)
Then, I = \(\int_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right)}{\sin [(\pi / 2)-x]+\cos [(\pi / 2)-x]} d x\)
or I = \(\int_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right)}{(\cos x+\sin x)} d x=\int_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right)}{(\sin x+\cos x)} d x\) …(2)
Adding (1) and (2), we get
2I = \(\frac{\pi}{2} \int_0^{\pi / 2} \frac{d x}{(\sin x+\cos x)}\)
∴ I = \(\frac{\pi}{4} \int_0^{\pi / 2} \frac{d x}{(\sin x+\cos x)}=\frac{\pi}{4} \int_0^{\pi / 2} \frac{d x}{\left[\frac{2 \tan (x / 2)}{1+\tan ^2(x / 2)}+\frac{1-\tan ^2(x / 2)}{1+\tan ^2(x / 2)}\right]}\)
= \(\frac{\pi}{4} \int_0^{\pi / 2} \frac{\sec ^2(x / 2)}{1-\tan ^2(x / 2)+2 \tan (x / 2)} d x\)
= \(\frac{\pi}{4} \int_0^1 \frac{2 d t}{\left(1-t^2+2 t\right)}\), where t = tan \(\frac{x}{2}\)
[x = 0 ⇒ t = 0 and x = \(\frac{\pi}{2}\) ⇒ t = 1]
= \(\frac{\pi}{2} \int_0^1 \frac{d t}{\left[(\sqrt{2})^2-(t-1)^2\right]} d t\)
= \(\frac{\pi}{2} \cdot \frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2}+(t-1)}{\sqrt{2}-(t-1)}\right|_0^1=\frac{\pi}{4 \sqrt{2}} \log \left|\frac{\sqrt{2}+1}{\sqrt{2}-1}\right| .\)
⇒ \(\int_0^{\pi / 2} \frac{x}{(\sin x+\cos x)} d x\) = \(\frac{\pi}{2} \cdot \frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2}+(t-1)}{\sqrt{2}-(t-1)}\right|_0^1=\frac{\pi}{4 \sqrt{2}} \log \left|\frac{\sqrt{2}+1}{\sqrt{2}-1}\right| .\)
Example 13 Prove that \(\int_0^{\pi / 2}(2 \log \sin x-\log \sin 2 x) d x=-\frac{\pi}{2}(\log 2)\)
Solution
Given:
⇒ \(\int_0^{\pi / 2}(2 \log \sin x-\log \sin 2 x) d x=-\frac{\pi}{2}(\log 2)\)
Let I = \(\int_0^{\pi / 2}(2 \log \sin x-\log \sin 2 x) d x\) …(1)
Then, I = \(\int_0^{\pi / 2}\left[2 \log \sin \left(\frac{\pi}{2}-x\right)-\log \sin 2\left(\frac{\pi}{2}-x\right)\right] d x\)
or I = \(\int_0^{\pi / 2}(2 \log \cos x-\log \sin 2 x) d x\) …(2)
Adding (1) and (2), we get
2I = \(\int_0^{\pi / 2}[2(\log \sin x+\log \cos x)-2 \log \sin 2 x] d x\)
or I = \(\int_0^{\pi / 2}[\log (\sin x \cos x)-\log \sin 2 x] d x\)
= \(\int_0^{\pi / 2}\left[\log \left(\frac{\sin 2 x}{2}\right)-\log \sin 2 x\right] d x\)
= \(\int_0^{\pi / 2}(\log \sin 2 x-\log 2-\log \sin 2 x) d x\)
= \(-\log 2 \cdot \int_0^{\pi / 2} d x=(-\log 2) \cdot[x]_0^{\pi / 2}=-\frac{\pi}{2}(\log 2)\)
⇒ \(\int_0^{\pi / 2}(2 \log \sin x-\log \sin 2 x) d x=-\frac{\pi}{2}(\log 2)\) = \(-\log 2 \cdot \int_0^{\pi / 2} d x=(-\log 2) \cdot[x]_0^{\pi / 2}=-\frac{\pi}{2}(\log 2)\)
Example 14 Evaluate \(\int_0^\pi \frac{x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)} d x\)
Solution
Given
⇒ \(\int_0^\pi \frac{x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)} d x\)
Let I = \(\int_0^\pi \frac{x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)} d x\) …(1)
Then, I = \(\int_0^\pi \frac{(\pi-x)}{\left(a^2 \cos ^2(\pi-x)+b^2 \sin ^2(\pi-x)\right]} d x\)
or I = \(\int_0^\pi \frac{(\pi-x)}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)} d x\) …(2)
Adding (1) and (2), we get
2I = \(\int_0^\pi \frac{(x+\pi-x)}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)} d x=\pi \int_0^\pi \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)}\)
= \(2 \pi \int_0^{\pi / 2} \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)}=2 \pi \int_0^{\pi / 2} \frac{\sec ^2 x}{\left(a^2+b^2 \tan ^2 x\right)} d x\)
[dividing num. and denim. by cos2x]
= \(2 \pi \int_0^{\infty} \frac{d t}{\left(a^2+b^2 t^2\right)}\), where tan x = t
= \(\frac{2 \pi}{b^2} \int_0^{\infty} \frac{d t}{\left(\frac{a^2}{b^2}+t^2\right)}=\left[\frac{2 \pi}{b^2} \cdot \frac{b}{a} \tan ^{-1}\left(\frac{b t}{a}\right)\right]_0^{\infty}\)
= \(\frac{2 \pi}{a b}\left[\tan ^{-1}(\infty)-\tan ^{-1}(0)\right]=\frac{2 \pi}{a b}\left(\frac{\pi}{2}-0\right)=\left(\frac{2 \pi}{a b} \times \frac{\pi}{2}\right)=\frac{\pi^2}{a b} .\)
∴ I = \(\frac{\pi^2}{2 a b}\) ⇒ \(\int_0^\pi \frac{x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)} d x=\frac{\pi^2}{2 a b} \text {. }\)
Example 15 Prove that \(\int_0^\pi x \sin ^3 x d x=\frac{2 \pi}{3} .\)
Solution
Given:
⇒ \(\int_0^\pi x \sin ^3 x d x=\frac{2 \pi}{3} .\)
Let I = \(\int_0^\pi x \sin ^3 x d x\) …(1)
Then, I = \(\int_0^\pi(\pi-x) \sin ^3(\pi-x) d x\)
or I = \(\int_0^\pi(\pi-x) \sin ^3 x d x\) …(2)
Adding (1) and (2), we get
2I = \(\int_0^\pi \pi \sin ^3 x d x=\pi \int_0^\pi \sin ^2 x \cdot \sin x d x\)
= \(\pi \int_0^\pi\left(1-\cos ^2 x\right) \sin x d x\)
= \(-\pi \int_1^{-1}\left(1-t^2\right) d t\), where cos x = t
[x = 0 ⇒ t = 1 and x = π ⇒ t = -1]
= \(\pi \int_{-1}^1\left(1-t^2\right) d t=\pi\left[t-\frac{t^3}{3}\right]_{-1}^1=\frac{4 \pi}{3}\)
Hence, I = \(\frac{2 \pi}{3} ⇒ \int_0^\pi x \sin ^3 x d x=\frac{2 \pi}{3}\)
Example 16 Evaluate \(\int_0^1 \cot ^{-1}\left(1-x+x^2\right) d x .\)
Solution
Given
⇒ \(\int_0^1 \cot ^{-1}\left(1-x+x^2\right) d x .\)
We have
I = \(\int_0^1 \cot ^{-1}\left\{1-x+x^2\right\} d x\)
= \(\int_0^1 \tan ^{-1}\left(\frac{1}{1-x+x^2}\right) d x=\int_0^1 \tan ^{-1}\left\{\frac{x+(1-x)}{1-x+x^2}\right\} d x\)
= \(\int_0^1\left\{\tan ^{-1} x+\tan ^{-1}(1-x)\right\} d x\)
= \(\int_0^1 \tan ^{-1} x d x+\int_0^1 \tan ^{-1}(1-x) d x\)
= \(\int_0^1 \tan ^{-1} x d x+\int_0^1 \tan ^{-1}\{1-(1-x)\} d x\)
[∵ \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)]
= \(\int_0^1 \tan ^{-1} x d x+\int_0^1 \tan ^{-1} x d x=2 \int_0^1 \tan ^{-1} x d x\)
= \(2 \int_0^1\left(\tan ^{-1} x \cdot 1\right) d x\)
= \(2\left[\left(\tan ^{-1} x\right) x-\int_0^1 \frac{1}{\left(1+x^2\right)} \cdot x d x\right]_0^1\)
= \(2\left[\left(\tan ^{-1} x\right) \cdot x\right]_0^1-2 \int_0^1 \frac{x}{\left(1+x^2\right)} d x\)
= \(2\left\{\left(\tan ^{-1} 1\right) \cdot 1-0\right\}-\left[\log \left(1+x^2\right)\right]_0^1\)
= \(\left(2 \times \frac{\pi}{4}\right)-(\log 2-\log 1)=\left(\frac{\pi}{2}-\log 2\right)\)
⇒ \(\int_0^1 \cot ^{-1}\left(1-x+x^2\right) d x .\) = \(\left(2 \times \frac{\pi}{4}\right)-(\log 2-\log 1)=\left(\frac{\pi}{2}-\log 2\right)\)
Example 17 Evaluate \(\int_{\pi / 5}^{3 \pi / 10} \frac{\sin x}{(\sin x+\cos x)} d x .\)
Solution
Given
⇒ \(\int_{\pi / 5}^{3 \pi / 10} \frac{\sin x}{(\sin x+\cos x)} d x .\)
Let I = \(\int_{\pi / 5}^{3 \pi / 10} \frac{\sin x}{(\sin x+\cos x)} d x .\) …(1)
Here, a = \(\frac{\pi}{5}\) and b = \(\frac{3 \pi}{10}\) ⇒ (a+b) = \(\left(\frac{\pi}{5}+\frac{3 \pi}{10}\right)=\frac{\pi}{2}\).
Using \(\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\), we have
I = \(\int_{\pi / 5}^{3 \pi / 10} \frac{\sin \left(\frac{\pi}{2}-x\right) d x}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x\)
⇒ I = \(\int_{\pi / 5}^{3 \pi / 10} \frac{\cos x}{(\sin x+\cos x)} d x\) …(2)
Adding (1) and (2), we get
2I = \(\int_{\pi / 5}^{3 \pi / 10} d x=[x]_{\pi / 5}^{3 \pi / 10}=\left(\frac{3 \pi}{10}-\frac{\pi}{5}\right)=\frac{\pi}{10}\)
⇒ I = \(\frac{\pi}{20}\).
Integrals of the form \(\int_{-a}^a f(x) d x\), where f(x) is even or odd
We know that
- f(x) is odd, if f(-x) = -f(x);
- f(x) is even, if f(-x) = f(x);
- \(\int_{-\pi}^a f(x) d x=0\), when f(x) is odd;
- \(\int_{-a}^a f(x) d x=2 \int_0^a f(x) d x\), when f(x) is even.
Example 18 Show that \(\int_{-\pi / 2}^{\pi / 2} \sin ^7 x d x=0\)
Solution
⇒ \(\int_{-\pi / 2}^{\pi / 2} \sin ^7 x d x=0\)
Let f(x) = sin7x. Then,
f(-x) = [sin (-x)]7 = -sin7x = -f(x).
∴ f(x) is an odd function of x.
But, \(\int_{-a}^a f(x) d x\) = 0, when f(x) is odd.
∴ \(\int_{-\pi / 2}^{\pi / 2} \sin ^7 x d x=0\)
Example 19 Evaluate \(\int_{-\pi / 2}^{\pi / 2} \sin ^2 x d x\)
Solution
Given
⇒ \(\int_{-\pi / 2}^{\pi / 2} \sin ^2 x d x\)
Let f(x) = sin2x.
Then, f(-x) = [sin(-x)]2 = (-sin x)2 = sin2x = f(x).
∴ f(x) is an even function.
So, \(\int_{-\pi / 2}^{\pi / 2} \sin ^2 x d x=2 \int_0^{\pi / 2} \sin ^2 x d x=2 \int_0^{\pi / 2}\left(\frac{1-\cos 2 x}{2}\right) d x\)
= \(\int_0^{\pi / 2}(1-\cos 2 x) d x=\left[x-\frac{\sin 2 x}{2}\right]_0^{\pi / 2}=\frac{\pi}{2}\)
⇒ \(\int_{-\pi / 2}^{\pi / 2} \sin ^2 x d x\) = \(\int_0^{\pi / 2}(1-\cos 2 x) d x=\left[x-\frac{\sin 2 x}{2}\right]_0^{\pi / 2}=\frac{\pi}{2}\)
Example 20 Prove that \(\int_{-1}^1 \log \left(\frac{2-x}{2+x}\right) d x=0 .\)
Solution
Given
⇒ \(\int_{-1}^1 \log \left(\frac{2-x}{2+x}\right) d x=0 .\)
Let f(x) = \(\log \left(\frac{2-x}{2+x}\right)\)
Then, f(-x) = \(\log \left(\frac{2-x}{2+x}\right)\) = \(\log \left(\frac{2-x}{2+x}\right)^{-1}=-\log \left(\frac{2-x}{2+x}\right)=-f(x)\)
∴ f(x) is an odd function of x.
But, we know that \(\int_{-a}^a f(x) d x=0\), when f(x) is an odd function of x.
∴ \(\int_{-1}^1 \log \left(\frac{2-x}{2+x}\right) d x=0\)
Example 21 Evaluate \(\int_1^4 f(x) d x \text {, where } f(x)= \begin{cases}4 x+3, & \text { if } 1 \leq x \leq 2 \\ 3 x+5, & \text { if } 2 \leq x \leq 4 .\end{cases}\)
Solution
Given
⇒ \(\int_1^4 f(x) d x \text {, where } f(x)= \begin{cases}4 x+3, & \text { if } 1 \leq x \leq 2 \\ 3 x+5, & \text { if } 2 \leq x \leq 4 .\end{cases}\)
⇒ \(\int_1^4 f(x) d x=\int_1^2 f(x) d x+\int_2^4 f(x) d x\)
= \(\int_1^2(4 x+3) d x+\int_2^4(3 x+5) d x\)
= \(\left[2 x^2+3 x\right]_1^2+\left[\frac{3 x^2}{2}+5 x\right]_2^4=(9+28)=37\)
Integrals of Modulus Functions
Example 22 Evaluate:
- \(\int_{-1}^2|x| d x\)
- \(\int_0^1|5 x-3| d x\)
- \(\int_0^\pi|\cos x| d x\)
Solution
(1) Clearly, |x| = \(\left\{\begin{array}{r}
-x \text { when }-1 \leq x \leq 0 \\
x \text { when } 0 \leq x \leq 2
\end{array}\right.\)
∴ \(\int_{-1}^2|x| d x=\int_{-1}^0|x| d x+\int_0^2|x| d x\)
= \(\int_{-1}^0(-x) d x+\int_0^2 x d x=\left[\frac{-x^2}{2}\right]_{-1}^0+\left[\frac{x^2}{2}\right]_0^2=\left(\frac{1}{2}+2\right)=\frac{5}{2} .\)
(2) Clearly, |5x-3| = \(\left\{\begin{array}{r}
-(5 x-3) \text { when } 0 \leq x \leq \frac{3}{5} \\
(5 x-3) \text { when } \frac{3}{5} \leq x \leq 1 .
\end{array}\right.\)
∴ \(\int_0^1|5 x-3| d x=\int_0^{3 / 5}|5 x-3| d x+\int_{3 / 5}^1|5 x-3| d x\)
= \(\int_0^{3 / 5}-(5 x-3) d x+\int_{3 / 5}^1(5 x-3) d x\)
= \(\left[3 x-\frac{5 x^2}{2}\right]_0^{3 / 5}+\left[\frac{5 x^2}{2}-3 x\right]_{3 / 5}^1\)
= \(\left(\frac{9}{5}-\frac{9}{10}\right)+\left(-\frac{1}{2}+\frac{9}{10}\right)=\frac{13}{10}\)
(3) Clearly, |cos x| = \(\left\{\begin{array}{r}
\cos x \text { when } 0 \leq x \leq \frac{\pi}{2} \\
-\cos x \text { when } \frac{\pi}{2} \leq x \leq \pi
\end{array}\right.\)
∴ \(\int_0^\pi|\cos x| d x=\int_0^{\pi / 2}|\cos x| d x+\int_{\pi / 2}^\pi|\cos x| d x\)
= \(\int_0^{\pi / 2} \cos x d x+\int_{\pi / 2}^\pi-\cos x d x\)
= \([\sin x]_0^{\pi / 2}-[\sin x]_{\pi / 2}^\pi=(1+1)=2 .\)
Example 23 Evaluate \(\int_1^4 f(x) d x \text {, where } f(x)=|x-1|+|x-2|+|x-3|\)
Solution
Given:
⇒ \(\int_1^4 f(x) d x \text {, where } f(x)=|x-1|+|x-2|+|x-3|\)
⇒ \(\int_1^4 f(x) d x=\int_1^2 f(x) d x+\int_2^3 f(x) d x+\int_3^4 f(x) d x\)
= \(\int_1^2\{(x-1)-(x-2)-(x-3)\} d x+\int_2^3\{(x-1)+(x-2)-(x-3)\} d x\)+
⇒ \(\int_3^4\{(x-1)+(x-2)+(x-3)\} d x\)
= \(\int_1^2(-x+4) d x+\int_2^3 x d x+\int_3^4(3 x-6) d x\)
= \(\left[\frac{-x^2}{2}+4 x\right]_1^2+\left[\frac{x^2}{2}\right]_2^3+\left[\frac{3 x^2}{2}-6 x\right]_3^4=\left(\frac{5}{2}+\frac{5}{2}+\frac{9}{2}\right)=\frac{19}{2}\)
“Step-By-Step Guide To Definite Integrals”
Example 24 Evaluate:
- \(\int_{-\pi / 2}^{\pi / 2}|\sin x| d x\)
- \(\int_{-1}^1 e^{|x|} d x\)
- \(\int_{-2}^1|2 x+1| d x\)
Solution
(1) Clearly, |sin x| is an even function of x.
∴ \(\int_{-\pi / 2}^{\pi / 2}|\sin x| d x=2 \int_0^{\pi / 2}|\sin x| d x\)
= \(2 \int_0^{\pi / 2} \sin x d x\) [∵ sin x ≥ 0, when 0 ≤ x ≤ \(\frac{\pi}{2}\)
= \([-2 \cos x]_0^{\pi / 2}=2\)
(2) Clearly, \(e^{|x|}\) is an even function of x.
∴ \(\int_{-1}^1 e^{|x|} d x=2 \int_0^1 e^{|x|} d x\)
= \(2 \int_0^1 e^x d x\) [∵ |x| = x, when 0 ≤ x ≤ 1]
= \(\left[2 e^x\right]_0^1=(2 e-2)=2(e-1)\)
(3) \(\left[-2 \leq x<-\frac{1}{2} \Rightarrow 2 x+1<0\right] \text { and }\left[-\frac{1}{2} \leq x \leq 1 \Rightarrow 2 x+1 \geq 0\right]\)
∴ \(\int_{-2}^1|2 x+1| d x=\int_{-2}^{-1 / 2}|2 x+1| d x+\int_{-1 / 2}^1|2 x+1| d x\)
= \(\int_{-2}^{-1 / 2}-(2 x+1) d x+\int_{-1 / 2}^1(2 x+1) d x\)
= \(\left[-x^2-x\right]_{-2}^{-1 / 2}+\left[x^2+x\right]_{-1 / 2}^1\)
= \(\left(-\frac{1}{4}+\frac{1}{2}\right)-(-4+2)+\left[2-\left(\frac{1}{4}-\frac{1}{2}\right)\right]\)
= \(\frac{1}{4}+2+\frac{9}{4}=\frac{9}{2}\)
Example 25 Evaluate \(\int_0^{2 \pi}|\sin x| d x\)
Solution
Given
⇒ \(\int_0^{2 \pi}|\sin x| d x\)
We know that sin x is positive, when 0 ≤ x ≤ π and sin x is negative when π ≤ x ≤ 2π.
∴ \(|\sin x|=\left\{\begin{aligned}
\sin x, & \text { when } 0 \leq x \leq \pi \\
-\sin x, & \text { when } \pi \leq x \leq 2 \pi
\end{aligned}\right.\)
∴ \(\int_0^{2 \pi}|\sin x| d x=\int_0^\pi|\sin x| d x+\int_\pi^{2 \pi}|\sin x| d x\)
= \(\int_0^\pi \sin x d x+\int_\pi^{2 \pi}(-\sin x) d x\)
= \([-\cos x]_0^\pi+[\cos x]_\pi^{2 \pi}=(2+2)=4\)
⇒ \(\int_0^{2 \pi}|\sin x| d x\) = \([-\cos x]_0^\pi+[\cos x]_\pi^{2 \pi}=(2+2)=4\)
Example 26 Evaluate \(\int_{-\pi / 2}^{\pi / 2} f(x) d x, \text { where } f(x)=\sin |x|+\cos |x| \text {. }\)
Solution
Given
\(\int_{-\pi / 2}^{\pi / 2} f(x) d x, \text { where } f(x)=\sin |x|+\cos |x| \text {. }\)f(x) = sin|x| + cos|x|
⇒ f(-x) = sin |-x| + cos|-x| = sin|x| + cos|x| = f(x)
⇒ f(x) is an even function.
∴ I = \(\int_{-\pi / 2}^{\pi / 2} f(x) d x\)
= \(2 \int_0^{\pi / 2} f(x) d x=2 \int_0^{\pi / 2}\{\sin |x|+\cos |x|] d x\)
= \(2 \int_0^{\pi / 2}(\sin x+\cos x) d x [∵ |x| = x in 0 < x < \frac{\pi}{2}]\)
= \(2 \cdot[-\cos x+\sin x]_0^{\pi / 2}=2\left[\left(-\cos \frac{\pi}{2}+\sin \frac{\pi}{2}\right)-(-\cos 0+\sin 0)\right]\)
Definite Integral as the Limit of a Sum
Let f(x)be a continuous real-valued function, defined in the closed interval [a,b]. Then we define
\(\int_a^b f(x) d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+\ldots+f[a+(n-1) h)]\) where nh = (b-a).
The method of evaluating \(\int_a^b f(x) d x\) by using the above definition is called integration from first principles.
Some Useful Results for Direct Applications
(1) \(1+2+3+\ldots+(n-1)=\frac{1}{2} n(n-1) .\)
(2) \(1^2+2^2+3^2+\ldots+(n-1)^2=\frac{1}{6}(n-1) n(2 n-1)\)
(3) \(1^3+2^3+3^3+\ldots+(n-1)^3=\left\{\frac{n(n-1)}{2}\right\}^2 .\)
(4) \(a+a r+a r^2+\ldots+a r^{n-1}=\frac{a\left(r^n-1\right)}{(r-1)}\)
(5) \(\begin{aligned}
\sin a+\sin (a+h)+ & \sin (a+2 h)+\ldots+\sin [a+(n-1) h] \\
& =\frac{\sin \left\{a+\left(\frac{n-1}{2}\right) h\right\} \sin \left(\frac{n h}{2}\right)}{\sin (h / 2)}
\end{aligned}\)
(6) \(\begin{aligned}
\cos a+\cos (a+h)+ & \cos (a+2 h)+\ldots+\cos [a+(n-1) h] \\
& =\frac{\cos \left\{a+\frac{(n-1)}{2} h\right\} \sin \left(\frac{n h}{2}\right)}{\sin (h / 2)} .
\end{aligned}\)
Solved Examples
Example 1 Evaluate the following integrals as a limit of sums:
- \(\int_0^5(x+1) d x\)
- \(\int_1^3(2 x+3) d x\)
Solution
(1) Let f(x) = (x+1); a = 0; b = 5 and nh = (5-0) = 5. Then,
\(\int_0^5(x+1) d x=\lim _{h \rightarrow 0} h[f(0)+f(0+h)+f(0+2 h)+\ldots+f\{0+(n-1) h\}]\)= \(\lim _{h \rightarrow 0} h[1+(h+1)+(2 h+1)+\ldots+\{(n-1) h+1\}]\)
= \(\lim _{h \rightarrow 0} h[n+\{h+2 h+3 h+\ldots+(n-1) h\}]\)
= \(\lim _{h \rightarrow 0} h[n+\{1+2+3+\ldots+(n-1)\} h]\)
= \(\lim _{h \rightarrow 0} h\left[n+\frac{n(n-1)}{2} h\right]=\lim _{h \rightarrow 0}\left[n h+\frac{n h(n h-h)}{2}\right]\)
= \(\lim _{h \rightarrow 0}\left[5+\frac{5(5-h)}{2}\right]\) [∵ nh = 5]
= \(\frac{35}{2}\).
(2) Let f(x) = (2x+3)’ a = 1; b = 3 and nh = (3-1) = 2.
Then, \(\int_1^3(2 x+3) d x=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+\ldots+f\{1+(n-1) h\}]\)
= \(\lim _{h \rightarrow 0} h[5+(5+2 h)+(5+4 h)+\ldots+\{5+2(n-1) h\}]\)
[∵ f(1) = 5, f(1+h) = 5+2h, etc.]
= \(\lim _{h \rightarrow 0} h[5 n+(2 h+4 h+\ldots+2(n-1) h\}]\)
= \(\lim _{h \rightarrow 0} h[5 n+2\{1+2+3+\ldots+(n-1)\} h]\)
= \(\lim _{h \rightarrow 0} h\left[5 n+2 \cdot \frac{n(n-1)}{2} h\right]\)
= \(\lim _{h \rightarrow 0}[5 n h+n h(n h-h)]\)
= \(\lim _{h \rightarrow 0}[10+2(2-h)]\) [∵ nh = 2]
= 14.
Example 2 Evaluate the following integrals as limit of sums:
- \(\int_0^2\left(x^2+1\right) d x\)
- \(\int_1^3\left(x^2+x\right) d x\)
Solution
(1) Let f(x) = (x2+1); a = 0; b = 2 and nh = (2-0) =2.
∴ \(\int_0^2\left(x^2+1\right) d x=\lim _{h \rightarrow 0} h[f(0)+f(0+h)+\ldots+f\{0+(n-1) h\}]\)
= lim h [1 + (h2+1) + (4h2+1) + (9h2+1) + … + {(n-1)2h2 + 1}]
[∵ f(0) = 1, f(0+h) = (h2+1), f(0+2h) = (4h2+1), etc.]
= \(\lim _{h \rightarrow 0} h\left[n+\left\{1^2+2^2+3^2+\ldots+(n-1)^2\right\} h^2\right]\)
= \(\lim _{h \rightarrow 0} h\left[n+\frac{(n-1) n(2 n-1)}{6} \cdot h^2\right]\)
= \(\lim _{h \rightarrow 0}\left[n h+\frac{(n h-h) n h(2 n h-h)}{6}\right]\)
= \(\lim _{h \rightarrow 0}\left[2+\frac{(2-h) 2(4-h)}{6}\right]\) [∵ nh = 2]
= \(\frac{14}{3}\).
(2) Let f(x) = (x2+x); a = 1; b = 3 and nh = (3-1) = 2.
∴ \(\begin{aligned}
\int_1^3\left(x^2+x\right) d x=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+f(1+2 h) & +\ldots \\
& +f[1+(n-1) h]]
\end{aligned}\)
= \(\begin{aligned}
& \lim _{h \rightarrow 0} h\left[\left(1^2+1\right)+\left\{(1+h)^2+(1+h)\right\}+\left\{(1+2 h)^2\right.\right. \\
& \left.+(1+2 h)\}+\ldots+(1+(n-1) h\}^2+\{1+(n-1) h\}\right]
\end{aligned}\)
= \(\begin{aligned}
\lim _{h \rightarrow 0} h\left[2 n+h^2\left\{1^2+2^2\right.\right. & \left.+\ldots+(n-1)^2\right\} \\
& +3 h\{1+2+\ldots+(n-1)\}]
\end{aligned}\)
= \(\lim _{h \rightarrow 0} h\left[2 n+h^2 \cdot \frac{(n-1) n(2 n-1)}{6}+3 h \cdot \frac{n(n-1)}{2}\right]\)
= \(\lim _{h \rightarrow 0}\left[2 n h+\frac{(n h-h) n h(2 n h-h)}{6}+\frac{3}{2} \cdot n h(n h-h)\right]\)
= \(\lim _{h \rightarrow 0}\left[4+\frac{(2-h) 2(4-h)}{6}+\frac{3}{2} \cdot 2(2-h)\right]\) [∵ nh = 2]
= \(\frac{38}{3}\).
Example 3 Evaluate \(\int_0^1\left(3 x^2+2 x+1\right) d x\) as limit of sums.
Solution
Given
⇒ \(\int_0^1\left(3 x^2+2 x+1\right) d x\)
Let f(x) = (3x2 + 2x +1); a = 0, b = 1 and nh = (1-0) = 1.
Then, \(\int_0^1\left(3 x^2+2 x+1\right) d x\)
= \(\lim _{h \rightarrow 0} h[f(0)+f(0+h)+f(0+2 h)+\ldots+f\{0+(n-1) h\}]\)
= \(\begin{aligned}
\lim _{h \rightarrow 0} h\left[1+\left(3 h^2+2 h+1\right)\right. & +\left(3 \cdot 2^2 h^2+2 \cdot 2 h+1\right)+\ldots \\
& \left.+\left(3 \cdot(n-1)^2 h^2+2 \cdot(n-1) h+1\right)\right]
\end{aligned}\)
= \(\begin{aligned}
\lim _{h \rightarrow 0} h\left[n+3 h^2\left[1^2+2^2+3^2\right.\right. & \left.+\ldots+(n-1)^2\right] \\
& +2 h(1+2+3+\ldots+(n-1)]]
\end{aligned}\)
= \(\lim _{h \rightarrow 0} h\left[n+3 h^2 \cdot \frac{(n-1) n(2 n-1)}{6}+2 h \cdot \frac{n(n-1)}{2}\right]\)
= \(\lim _{h \rightarrow 0}\left[n h+\frac{1}{2}(n h-h)(n h)(2 n h-h)+(n h)(n h-h)\right]\)
= \(\lim _{h \rightarrow 0}\left[1+\frac{1}{2}(1-h) \cdot 1 \cdot(2-h)+1 \cdot(1-h)\right]\) [∵ nh =1]
= 3.
⇒\(\int_0^1\left(3 x^2+2 x+1\right) d x\) = 3.
Example 4 Evaluate \(\int_{-1}^1 e^x d x\) as limit of sums.
Solution
Given
⇒ \(\int_{-1}^1 e^x d x\)
Let f(x) = ex and nh = {1-(-1)} = 2. Then,
⇒ \(\int_{-1}^1 e^x d x=\lim _{h \rightarrow 0} h[f(-1)+f(-1+h)+\ldots+f[-1+(n-1) h]]\)
= \(\lim _{h \rightarrow 0} h\left[e^{-1}+e^{(-1+h)}+e^{(-1+2 h)}+\ldots+e^{(-1+(x-1) h)}\right]\)
[∵ f(-1) = e-1 , f(-1+h) = e(-1+h) , etc.]
= \(\frac{1}{e} \cdot \lim _{h \rightarrow 0} h\left[\frac{\left(e^h\right)^n-1}{\left(e^h-1\right)}\right]=\frac{1}{e} \cdot \lim _{h \rightarrow 0} \frac{h\left[e^{n h}-1\right]}{\left(e^h-1\right)}\)
= \(\frac{1}{e} \cdot \lim _{h \rightarrow 0} \frac{h\left(e^2-1\right)}{\left(e^h-1\right)}\) [∵ nh =2]
= \(\frac{1}{e} \cdot \lim _{h \rightarrow 0} \frac{h\left(e^2-1\right)}{\left\{1+h+\frac{h^2}{\lfloor 2}+\frac{h^3}{\lfloor 3}+\ldots\right\}-1}\)
= \(\frac{1}{e} \cdot \lim _{h \rightarrow 0} \frac{e^2-1}{\left(1+\frac{h}{[2}+\frac{h^2}{[3}+\ldots\right)}=\frac{1}{e}\left(e^2-1\right)=\left(e-\frac{1}{e}\right) .\)
Example 5 Evaluate \(\int_a^b \sin x d x\) from first principles.
Solution
Given
⇒ \(\int_a^b \sin x d x\)
Let f(x) = sin x and let nh = (b-a). Then,
⇒ \(\int_a^b \sin x d x=\lim _{h \rightarrow 0} h[f(a)+f(a+h)+\ldots+f(a+(n-1) h\}]\)
= \(\lim _{h \rightarrow 0} h[\sin a+\sin (a+h)+\ldots+\sin \{a+(n-1) h\}]\)
= \(\lim _{h \rightarrow 0} h \cdot \frac{\sin \left\{a+\left(\frac{n-1}{2}\right) h\right\} \sin \left(\frac{n h}{2}\right)}{\sin (h / 2)}\)
= \(\lim _{h \rightarrow 0}\left[2 \sin \left\{a+\frac{1}{2} n h-\frac{1}{2} h\right\} \sin \left(\frac{n h}{2}\right) \times \frac{(h / 2)}{\sin (h / 2)}\right]\)
= \(\lim _{h \rightarrow 0}\left[2 \sin \left\{a+\frac{1}{2}(b-a)-\frac{1}{2} h\right\} \sin \left(\frac{b-a}{2}\right) \times \frac{(h / 2)}{\sin (h / 2)}\right]\)
[∵ nh = (b-a)]
= \(\lim _{h \rightarrow 0} 2 \sin \left\{\left(\frac{b+a}{2}\right)-\frac{1}{2} h\right\} \sin \left(\frac{b-a}{2}\right) \cdot \lim _{h \rightarrow 0} \frac{(h / 2)}{\sin (h / 2)}\)
= \(2 \sin \left(\frac{b+a}{2}\right) \sin \left(\frac{b-a}{2}\right)\)
= (cos a – cos b)
[∵ 2sin A sin B = cos(A-B)-cos(A+B)].