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 1:
\[\color{red}{Evaluate: \int_1^2 x^2\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int_1^2 x^2\ dx = \frac{x^3}{3} \Biggr]_{1}^{2}\]
\[= [\frac{2^3}{3} – \frac{1^3}{3}]\]
\[= \frac{8}{3} – \frac{1}{3}\]
\[= \frac{8 – 1}{3}\]
\[= \frac{7}{3}\]
\[\boxed{\int_1^2 x^2\ dx = \frac{7}{3}}\]
Example 2:
\[\color{red}{Evaluate: \int_0^1 (x^2\ -\ 2)\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int_0^1 (x^2\ -\ 2)\ dx = \frac{x^3}{3}\ -\ 2x \Biggr]_{0}^{1}\]
\[= [(\frac{1^3}{3}\ -\ 2(1))\ -\ 0]\]
\[= \frac{1}{3}\ -\ 2\]
\[= \frac{1 – 6}{3}\]
\[= \frac{-5}{3}\]
\[\boxed{\int_0^1 (x^2\ -\ 2)dx\ = \frac{-5}{3}}\]
Example 3:
\[\color{red}{Evaluate: \int_1^2 ( x – x^2)\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int_1^2 ( x – x^2)\ dx = (\frac{x^2}{2} -\frac{x^3}{3}) \Biggr]_{1}^{2}\]
\[= [(\frac{2^2 }{2}-\frac{2^3}{3}) – (\frac{1^2}{2} -\frac{1^3}{3})]\]
\[= [(\frac{4}{2} – \frac{8}{3}) – (\frac{1}{2} -\frac{1}{3})]\]
\[= [\frac{4}{2} – \frac{8}{3} – \frac{1}{2} +\frac{1}{3}]\]
\[ =\frac{ 12 – 16 -3 + 2}{6}\]
\[ =\frac{-5}{6}\]
\[\boxed{\int_1^2 ( x – x^2)\ dx = \frac{-5}{6}}\]
Example 4:
\[\color{red}{Evaluate: \int_1^2 \frac{1}{x}\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int_1^2 \frac{1}{x}\ dx = log\ x \Biggr]_{1}^{2}\]
\[=[log\ 2 – log\ 1]\]
\[= log\ 2 – 0\]
\[= log\ 2\]
\[\boxed{\int_1^2 \frac{1}{x}\ dx = log\ 2}\]
Example 5:
\[\color{red}{Evaluate: \int_0^1 (x^2+ 2x + 3)(x + 1)\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int_0^1 (x^2+ 2x + 3)(x + 1)\ dx = \int_0^1 (x^3+ x^2 + 2x^2 + 2x + 3x + 3)\ dx\]
\[=\int_0^1 (x^3+ 3x^2 + 5x + 3)\ dx\]
\[= \frac{x^4}{4} + 3 \frac{x^3}{3} + 5 \frac{x^2}{2} + 3x \Biggr]_{0}^{1}\]
\[= [(\frac{1^4}{4} + 1^3 + 5 \frac{1^2}{2} + 3(1))- 0)]\]
\[ = \frac{1}{4} + 1 + \frac{5}{2} + 3\]
\[ = \frac{1 + 4 + 10 + 12}{4}\]
\[= \frac{27}{4}\]
\[\boxed{\int_0^1 (x^2+ 2x + 3)(x + 1)\ dx =\frac{27}{4}}\]
Example 6:
\[\color{red}{Evaluate: \int_0^\frac{\pi}{2} sin^2 x\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[W.K.T\ sin^2 x = \frac{1 – cos\ 2x}{2}\]
\[\int_0^\frac{\pi}{2} sin^2 x\ dx = \int_0^\frac{\pi}{2} [ \frac{1 – cos\ 2x}{2}]\ dx\]
\[= \frac{1}{2}[x – \frac{sin\ 2x}{2}] \Biggr]_{0}^{\frac{\pi}{2}}\]
\[=\frac{1}{2}[(\frac{\pi}{2} – 0) – ( 0 – 0)]\]
\[=\frac{1}{2}[\frac{\pi}{2}]\]
\[=\frac{\pi}{4}\]
\[\boxed{\int_0^\frac{\pi}{2} sin^2 x\ dx =\frac{\pi}{4}}\]
Example 7:
\[\color{red}{Evaluate: \int_0^\frac{\pi}{2} sin^3 x\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[W.K.T\ sin\ 3x = 3 sin\ x – 4sin^3\ x\]
\[ sin^3 x = \frac{1}{4}[3sin\ x\ -\ sin\ 3x]\]
\[\int_0^\frac{\pi}{2} sin^3 x\ dx = \int_0^\frac{\pi}{2} \frac{1}{4}[3sin\ x\ -\ sin\ 3x]\ dx\]
\[= \frac{1}{4}[-\ 3\ cos\ x\ +\ \frac{cos\ 3x}{3}]\Biggr]_{0}^{\frac{\pi}{2}}\]
\[=\frac{1}{4}[(-3 cos \frac{\pi}{2} + \frac{cos\ 3\frac{\pi}{2}}{3})\ -\ ( -3 cos\ 0 + \frac{cos3(0)}{3})]\]
\[=\frac{1}{4}[0\ – (-3\ +\ \frac{1}{3})]\]
\[=\frac{1}{4}[3\ -\ \frac{1}{3}]\]
\[=\frac{1}{4}[\frac{9\ -\ 1}{3}]\]
\[=\frac{1}{4}[\frac{8}{3}]\]
\[=\frac{2}{3}\]
\[\boxed{\int_0^\frac{\pi}{2} sin^3 x\ dx =\frac{2}{3}}\]
Example 8:
\[\color{red}{Evaluate: \int_0^\frac{\pi}{2} cos\ 3 x\ cos\ x\ dx}\ \hspace{18cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[W.K.T\ cos A\ cos B = \frac{1}{2}[cos (A + B) + cos ( A – B)]\]
\[ cos\ 3x\ cos x = \frac{1}{2}[cos (3x\ +\ x)\ +\ cos (3x\ -\ x)]\]
\[ = \frac{1}{2}[cos\ 4x\ +\ cos\ 2x]\]
\[\int_0^\frac{\pi}{2} cos\ 3 x\ cos\ x\ dx = \int_0^\frac{\pi}{2} \frac{1}{2}[cos\ 4x\ +\ cos\ 2x]\ dx\]
\[= \frac{1}{2}[\frac{sin\ 4x}{4}\ +\ \frac{sin\ 2x}{2}]\Biggr]_{0}^{\frac{\pi}{2}}\]
\[=\frac{1}{2}[(\frac{sin\ 4\frac{\pi}{2}}{4} + \frac{sin\ 2\frac{\pi}{2}}{2})\ -\ (\frac{sin\ 4(0)}{4}\ + \frac{sin\ 2(0)}{2})]\]
\[=\frac{1}{2}[(\frac{0}{4}\ +\ \frac{0}{2})- (\frac{0}{4}\ +\ \frac{0}{2})]\]
\[=\frac{1}{2}[0\ -\ 0]\]
\[\boxed{\int_0^\frac{\pi}{2} cos\ 3 x\ cos\ x\ dx\ =\ 0}\]
Example 9:
\[\color{red}{Evaluate: \int_0^2 x^2\ \sqrt{1 + x^3}\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[put\ u =1 + x^3\]
\[\frac{du}{dx}= 3x^2\]
\[x^2\ dx = \frac{1}{3}\ du\]
\[\int_0^2 x^2\ \sqrt{1 + x^3}\ dx = \frac{1}{3}\ \int_0^2 u^\frac{1}{2}\ du\]
\[= \frac{1}{3}[\frac{u^\frac{3}{2}}{\frac{3}{2}}] \Biggr]_{0}^{2}\]
\[=\frac{1}{3} × \frac{2}{3} \Biggr[u^\frac{3}{2} \Biggr]_{0}^{2}\]
\[=\frac{2}{9} \Biggr[(1 + x^3)^\frac{3}{2} \Biggr]_{0}^{2}\]
\[=\frac{2}{9}[(1 + 2^3)^\frac{3}{2} – (1 + 0^3)^\frac{3}{2}\]
\[=\frac{2}{9}[9^\frac{3}{2} – 1^\frac{3}{2}]\]
\[=\frac{2}{9}[27 – 1]\]
\[\int_0^2 x^2\ \sqrt{1 + x^3}\ dx\ = \frac{52}{9}\]
\[\boxed{\int_0^2 x^2\ \sqrt{1 + x^3}\ dx =\frac{52}{9}}\]
Example 10:
\[\color{red}{Evaluate: \int_0^\frac{\pi}{2} \frac{cos^2 x}{ 1- sin x} \ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int_0^\frac{\pi}{2} \frac{cos^2 x}{ 1- sin x} \ dx = \int_0^\frac{\pi}{2} \frac{1 – sin^2 x}{ 1- sin x} \ dx\]
\[= \int_0^\frac{\pi}{2} \frac{(1+ sin x)(1 – sin x)}{1- sin x}\ dx\]
\[=\int_0^\frac{\pi}{2} ( 1 + sin x)\ dx\]
\[=\Biggr[(x – cos x)\Biggr]_{0}^{\frac{\pi}{2}}\]
\[=[(\frac{\pi}{2} – cos \frac{\pi}{2}) – ( 0 – cos\ 0)]\]
\[=[(\frac{\pi}{2} – 0) – ( 0 – 1)]\]
\[=\frac{\pi}{2} + 1\]
\[\boxed{\int_0^\frac{\pi}{2} \frac{cos^2 x}{ 1- sin x}\ dx =\frac{\pi}{2} + 1}\]
Example 11:
\[\color{red}{Evaluate: \int_0^\frac{\pi}{2} (2 + sin x)^3 cos x \ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[put\ u =2 + sin x\]
\[\frac{du}{dx}= cos x\]
du = cos x dx
\[\int_0^\frac{\pi}{2} (2 + sin x)^3 cos x \ dx = \int_0^\frac{\pi}{2} u^3 \ du\]
\[=\Biggr[\frac{u^4}{4}\Biggr]_{0}^{\frac{\pi}{2}}\]
\[=\Biggr[\frac{(2 + sin x)^4}{4}\Biggr]_{0}^{\frac{\pi}{2}}\]
\[=[\frac{(2 + sin \frac{\pi}{2})^4}{4} – \frac{( 2 + sin\ 0)^4}{4}]\]
\[=[\frac{(2 + 1)^4}{4} – \frac{( 2 + 0)^4}{4}]\]
\[=[\frac{(3)^4}{4} – \frac{(2)^4}{4}]\]
\[=\ \frac{81}{4} – \frac{16}{4}\]
\[=\ \frac{81- 16}{4}\]
\[=\ \frac{65}{4}\]
\[\boxed{\int_0^\frac{\pi}{2} (2 + sin x)^3 cos x\ dx =\ \frac{65}{4}}\]
Example 12:
\[\color{red}{Prove\ that \int_0^\frac{\pi}{2} \frac{sin\ x}{sin\ x + cos\ x}\ dx = \frac{\pi}{4}}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[Use\ the\ property\ \int_{0}^{a} f(x)\ dx = \int_{0}^{a} f(a – x)\ dx\]
\[Let\ I= \int_0^\frac{\pi}{2} \frac{sin\ x}{sin\ x + cos\ x}\ dx —————— (1)\]
\[= \int_0^\frac{\pi}{2} \frac{sin\ (\frac{\pi}{2} – x)}{sin\ (\frac{\pi}{2} – x) + cos\ (\frac{\pi}{2} – x)}\ dx\]
\[= \int_0^\frac{\pi}{2} \frac{cos\ x}{cos\ x + sin\ x}\ dx———————-( 2 )\]
Adding ( 1 ) & ( 2 )
\[2\ I= \int_0^\frac{\pi}{2} \frac{sin\ x}{sin\ x + cos\ x}\ dx + \int_0^\frac{\pi}{2} \frac{cos\ x}{cos\ x + sin\ x}\ dx\]
\[= \int_0^\frac{\pi}{2} \frac{sin\ x + cos\ x}{sin\ x + cos\ x}\ dx \]
\[= \int_0^\frac{\pi}{2} 1\ dx \]
\[=\Biggr[x \Biggr]_{0}^{\frac{\pi}{2}}\]
\[=\frac{\pi}{2} – 0 \]
\[=\frac{\pi}{2}\]
\[I =\frac{\pi}{4}\]
\[\boxed{\int_0^\frac{\pi}{2} \frac{sin\ x}{sin\ x + cos\ x}\ dx = \frac{\pi}{4}}\]
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