DEFINITE INTEGRALS
21 Apr 2021
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Definition of Definite Integrals:
\[\int_{a}^{b} f(x)\ dx = F(b) – F(a)\]
\[Note:\ \int_{0}^{a} f(x)\ dx = \int_{0}^{a} f(a - x)\ dx\]
Example :
\[Evaluate: \int_1^2 x^2\ dx\]
Soln:
\[\int_1^2 x^2\ dx = \frac{x^3}{3} \Biggr]_{1}^{2}\]
\[= [\frac{2^3}{3} - \frac{1^3}{3}]\]
\[= \frac{8}{3} - \frac{1}{3}\]
\[= \frac{8 - 1}{3}\]
\[= \frac{7}{3}\]
Example :
\[Evaluate: \int_1^2 \frac{1}{x}\ dx\]
Soln:
\[\int \frac{1}{x}\ dx = log\ x \Biggr]_{1}^{2}\]
\[=[log\ 2 - log\ 1]\]
\[= log\ 2 - 0\]
\[= log\ 2\]
Example :
\[Evaluate: \int_0^\frac{\pi}{2} sin^2 x\ dx\]
Soln:
\[W.K.T\ sin^2 x = \frac{1 - cos\ 2x}{2}\]
\[\int_0^\frac{\pi}{2} sin^2 x\ dx = \int_0^\frac{\pi}{2} \frac{1}{2}[ \frac{1 - cos\ 2x}{2}]\ dx\]
\[= \frac{1}{2}[x - \frac{sin\ 2x}{2}] \Biggr]_{0}^{\frac{\pi}{2}}\]
\[=\frac{1}{2}[(\frac{\pi}{2} - 0) - ( 0 - 0)]\]
\[=\frac{1}{2}[\frac{\pi}{2}]\]
\[=\frac{\pi}{4}\]
Example :
\[Evaluate: \int_0^\frac{\pi}{2} sin^3 x\ dx\]
Soln:
\[W.K.T\ sin\ 3x = 3 sin\ x - 4sin^3\ x\]
\[ sin^3 x = \frac{1}{4}[3sin\ x - 4 sin^3\ x]\]
\[\int_0^\frac{\pi}{2} sin^3 x\ dx = \int_0^\frac{\pi}{2} \frac{1}{4}[3sin\ x - 4 sin^3\ x]\ dx\]
\[= \frac{1}{4}[3sin\ x - 4 sin^3\ x]\Biggr]_{0}^{\frac{\pi}{2}}\]
\[=\frac{1}{4}[(-3 cos \frac{\pi}{2} + \frac{(cos\ 3\frac{\pi}{2}}{3})\ - ( -3 cos\ 0 + \frac{cos3(0){3})]\]
Example :
\[Evaluate: \int_0^2 x^2\ \sqrt{1 + x^3}\ dx\]
Soln:
\[put\ u =1 + x^3\]
\[\frac{du}{dx}= 3x^2\]
\[x^2\ dx = \frac{1}{3}\ du\]
\[\int_0^2 x^2\ \sqrt{1 + x^3}\ dx = \frac{1}{3}\ \int_0^2 u^\frac{1}{2}\ du\]
\[= \frac{1}{3}[x - \frac{u^\frac{3}{2}}{\frac{3}{2}}] \Biggr]_{0}^{2}\]
\[=\frac{1}{3} × \frac{2}{3} \Biggr[u^\frac{3}{2} \Biggr]_{0}^{2}\]
\[=\frac{2}{9} \Biggr[(1 + x^3)^\frac{3}{2} \Biggr]_{0}^{2}\]
\[=\frac{2}{9}[(1 + 2^3)^\frac{3}{2} - (1 + 0^3)^\frac{3}{2}\]
\[=\frac{2}{9}[9^\frac{3}{2} - 1^\frac{3}{2}]\]
\[=\frac{2}{9}[27 - 1]\]
\[\int_0^2 x^2\ \sqrt{1 + x^3}\ dx\ = \frac{52}{9}\]]
Example :
\[Evaluate: \int_0^\frac{\pi}{2} \frac{cos^2 x}{ 1- sin x} \ dx\]
Soln:
\[\int_0^\frac{\pi}{2} \frac{cos^2 x}{ 1- sin x} \ dx = \int_0^\frac{\pi}{2} \frac{1 - sin^2 x}{ 1- sin x} \ dx\]
\[= \int_0^\frac{\pi}{2} \frac{(1+ sin x)(1 - sin x)}{1- sin x}\ dx\]
\[=\int_0^\frac{\pi}{2} ( 1 + sin x)\ dx\]
\[=\Biggr[(x + cos x)\Biggr]_{0}^{\frac{\pi}{2}}\]
\[=[(\frac{\pi}{2} + cos \frac{\pi}{2}) - ( 0 + cos\ 0)]\]
\[=[(\frac{\pi}{2} +0) - ( 0 + 1)]\]
\[=\frac{\pi}{2} - 1\]
Example :
\[Evaluate: \int_0^\frac{\pi}{2} (2 + sin x)^3 cos x \ dx\]
Soln:
\[put\ u =2 + sin x\]
\[\frac{du}{dx}= cos x\]
du = cos x dx
\[\int_0^\frac{\pi}{2} (2 + sin x)^3 cos x \ dx = \int_0^\frac{\pi}{2} u^3 \ du\]
\[=\Biggr[\frac{u^4}{4}\Biggr]_{0}^{\frac{\pi}{2}}\]
\[=\Biggr[\frac{(2 + sin x)^4}{4}\Biggr]_{0}^{\frac{\pi}{2}}\]
\[=[(\frac{(2 + sin \frac{\pi}{2})^4}{4} - \frac{( 2 + sin\ 0)^4}{4}]\]
\[=[(\frac{(2 + 1)^4}{4} - \frac{( 2 + 0)^4}{4}]\]
\[=[(\frac{(3)^4}{4} - \frac{(2)^4}{4}]\]
\[=[(\frac{81}{4} - \frac{16}{4})]\]
\[=[(\frac{81- 16}{4})]\]
\[=[\frac{65}{4}]\]
Example :
\[Prove\ that \int_0^\frac{\pi}{2} \frac{sin\ x}{sin\ x + cos\ x}\ dx = \frac{\pi}{4}\]
Soln:
\[Use\ the\ property\ \int_{0}^{a} f(x)\ dx = \int_{0}^{a} f(a - x)\ dx\]
\[Let\ I= \int_0^\frac{\pi}{2} \frac{sin\ x}{sin\ x + cos\ x}\ dx ------------------ (1)\]
\[= \int_0^\frac{\pi}{2} \frac{sin\ (\frac{\pi}{2} - x)}{sin\ (\frac{\pi}{2} - x) + cos\ (\frac{\pi}{2} - x)}\ dx\]
\[= \int_0^\frac{\pi}{2} \frac{cos\ x}{cos\ x + sin\ x}\ dx----------------------( 2 )\]
Adding ( 1 ) & ( 2 )
\[2\ I= \int_0^\frac{\pi}{2} \frac{sin\ x}{sin\ x + cos\ x}\ dx + \int_0^\frac{\pi}{2} \frac{cos\ x}{cos\ x + sin\ x}\ dx\]
\[= \int_0^\frac{\pi}{2} \frac{sin\ x + cos\ x}{sin\ x + cos\ x}\ dx \]
\[= \int_0^\frac{\pi}{2} 1\ dx \]
\[=\Biggr[x \Biggr]_{0}^{\frac{\pi}{2}}\]
\[=\frac{\pi}{2} - 0 \]
\[=\frac{\pi}{2}\]
\[I =\frac{\pi}{4}\]
\[\int_0^\frac{\pi}{2} \frac{sin\ x}{sin\ x + cos\ x}\ dx = \frac{\pi}{4}\]
