Definite Integration

$I = \int\limits_0^{{\pi \over 2}} {{{{{\left( {\sin \,x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}} dx$

We know $\left[ {{{\left( {\sin x + \cos x} \right)}^2} = 1 + \sin 2x} \right],\,$ So

$I = \int\limits_0^{{\pi \over 2}} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\left( {\sin x + \cos x} \right)}}} dx$

$ = \int\limits_0^{{\pi \over 2}} {\left( {\sin x + \cos x} \right)dx} $

$\left[ {} \right.$ $\sin x + \cos x > 0$ $\,\,if\,\,0 < x < {\pi \over 2}$ $\left. {} \right]$

or $I = \left[ { - \cos x + \sin x} \right]_0^{{\pi \over 2}} $

= - cos ${\pi \over 2}$ + sin ${\pi \over 2}$ + cos 0 - sin 0

= - 0 + 1 + 1 - 0

= 2

$\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|} dx = \int\limits_{ - 2}^3 {\left| {{x^2} - 1} \right|} dx$

Now $\left| {{x^2} - 1} \right| = \left\{ {\matrix{ {{x^2} - 1} & {if} & {x \le - 1} \cr {1 - {x^2}} & {if} & { - 1 \le x \le 1} \cr {{x^2} - 1} & {if} & {x \ge 1} \cr } } \right.$

$\therefore$ Integral is

$\int\limits_{ - 2}^{ - 1} {\left( {{x^2} - 1} \right)dx + \int\limits_{ - 1}^1 {\left( {1 - {x^2}} \right)} } dx + \int\limits_1^3 {\left( {{x^2} - 1} \right)dx} $

$\left[ {{{{x^3}} \over 3} - x} \right]_{ - 2}^{ - 1} + \left[ {x - {{{x^3}} \over 3}} \right]_{ - 1}^1 + \left[ {{{{x^3}} \over 3} - x} \right]_1^3$

$ = \left( { - {1 \over 3} + 1} \right) - \left( { - {8 \over 3} + 2} \right) + \left( {2 - {2 \over 3}} \right) + \left( {{{27} \over 3} - 3} \right) - \left( {{1 \over 3} - 1} \right)$

$ = {2 \over 3} + {2 \over 3} + {4 \over 3} + 6 + {2 \over 3} = {{28} \over 3}$

$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}}$

$ \Rightarrow f\left( { - x} \right) = {{{e^{ - x}}} \over {1 + {e^{ - x}}}} = {1 \over {{e^x} + 1}}$

$\therefore$ $f\left( x \right) + f\left( { - x} \right) = 1\forall x$

Now ${I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}} dx$

$ = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {\left( {1 - x} \right)} g\left\{ {x\left( {1 - x} \right)} \right\}dx$

$\left[ {} \right.$ using $\int\limits_a^b {f\left( x \right)} dx\,a$

$ = \int\limits_a^b {f\left( {a + b - x} \right)dx} $ $\left. \, \right]$

$ = {I_2} - {I_1} \Rightarrow 2{I_1} = {I_2}$

$\therefore$ ${{{I_2}} \over {{I_1}}} = 2$

$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)} dy$

$ = \int\limits_0^t {{e^{t - y}}} ydy = {e^t}\int\limits_0^t {{e^{ - y}}} \,ydy$

$ = {e^t}\left[ { - y{e^{ - y}} - {e^{ - y}}} \right]_0^t$

$ = - {e^t}\left[ {y{e^{ - y}} + {e^{ - y}}} \right]_0^t$

$ = - {e^t}\left[ {t\,{e^{ - t}} + {e^{ - t}} - 0 - 1} \right]$

$ = - {e^t}\left[ {{{t + 1 - {e^t}} \over {{e^t}}}} \right]$

$ = {e^t} - \left( {1 + t} \right)$

Given $f'\left( x \right) = f\left( x \right) \Rightarrow {{f'\left( x \right)} \over {f\left( x \right)}} = 1$

Integrating log

$f\left( x \right) = x + c \Rightarrow f\left( x \right) = {e^{x + c}}$

$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = {e^x}$

$\therefore$ $\int\limits_0^1 {f\left( x \right)g\left( x \right)} dx$

$ = \int\limits_0^1 {{e^x}} \left( {{x^2} - {e^x}} \right)dx$

$ = \int\limits_0^1 {{x^2}} {e^x}dx - \int\limits_0^1 {{e^{2x}}} dx$

$ = \left[ {{x^2}{e^x}} \right]_0^1 - 2\left[ {x{e^x} - {e^x}} \right]_0^1 - {1 \over {20}}\left[ {{e^{2x}}} \right]_0^1$

$ = e - \left[ {{{{e^2}} \over 2} - {1 \over 2}} \right] - 2\left[ {e - e + 1} \right]$

$ = e - {{{e^2}} \over 2} - {3 \over 2}$

$I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}} dx$

$ = \int\limits_0^1 {\left( {1 - x} \right){{\left( {1 - 1 + x} \right)}^n}dx} $

$ = \int\limits_0^1 {\left( {1 - x} \right)} {x^n}dx$

$ = \left[ {{{{x^{n + 1}}} \over {n + 1}} - {{{x^{n + 2}}} \over {n + 2}}} \right]_0^1$

$ = {1 \over {n + 1}} - {1 \over {n + 2}}$

The given expression can be written as

$\mathop {\lim }\limits_{n \to \infty } {1 \over n}{\sum\limits_{r = 1}^n {\left( {{r \over n}} \right)} ^4} - \mathop {\lim }\limits_{n \to \infty } {1 \over n}.\mathop {\lim }\limits_{n \to \infty } {1 \over n}{\left( {{r \over n}} \right)^3}$

$ = \int\limits_0^1 {{x^4}} \,\,dx - \mathop {\lim }\limits_{n \to \infty } {1 \over n} \times \int\limits_0^1 {{x^3}} \,\,dx$

$ = \left[ {{{{x^5}} \over 5}} \right]_0^1 - 0 = {1 \over 5}$

$I = \int\limits_a^b {xf\left( x \right)} dx$

$ = \int\limits_a^b {\left( {a + b - x} \right)} f\left( {a + b - x} \right)dx$

$ = \left( {a + b} \right)\int\limits_a^b {f\left( {a + b - x} \right)} dx - \int\limits_a^b {xf} \left( {a + b - x} \right)dx$

$ = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)dx} - \int\limits_a^b {xf\left( x \right)dx} $

$\left[ {} \right.$ As given that $f\left( {a + b - x} \right) = f\left( x \right)$ $\left. {} \right]$

$2I = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)} dx$

$ \Rightarrow I = {{\left( {a + b} \right)} \over 2}\int\limits_a^b {f\left( x \right)} dx$

$\mathop {\lim }\limits_{x \to 0} {{{d \over {dx}}\int\limits_0^{{x^2}} {{{\sec }^2}tdt} } \over {{d \over x}\left( {x\sin x} \right)}}$

$ = \mathop {\lim }\limits_{x \to 0} {{{{\sec }^2}{x^2}.2x} \over {\sin \,x + x\,\cos \,x}}$ (by $L'$ Hospital rule)

$\mathop {\lim }\limits_{x \to 0} {{2{{\sec }^2}{x^2}} \over {\left( {{{\sin x} \over x} + \cos \,x} \right)}} = {{2 \times 1} \over {1 + 1}} = 1$

We have $\int\limits_0^2 {f\left( x \right)} dx = {3 \over 4};Now,$

$\int\limits_0^2 {xf'\left( x \right)} dx$

$ = x\int\limits_0^2 {f'\left( x \right)dx} - \int\limits_0^2 {f\left( x \right)} dx$

$ = \left[ {x\,f\left( x \right)} \right]_0^2 - {3 \over 4}$

$ = 2f\left( 2 \right) - {3 \over 4}$

$ = 0 - {3 \over 4}$

$\left( {} \right.$ As $f\left( 2 \right) = 0$ $\left. {} \right)$ $ = - {3 \over 4}.$

$\int\limits_0^2 {\left[ {{x^2}} \right]} dx = \int\limits_0^1 {\left[ {{x^2}} \right]dx} + \int\limits_1^{\sqrt 2 } {\left[ {{x^2}} \right]} dx + $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$\int\limits_{\sqrt 2 }^{\sqrt 3 } {\left[ {{x^2}} \right]} + \int\limits_{\sqrt 3 }^2 {\left[ {{x^2}} \right]} dx$

$ = \int\limits_0^1 {0dx} + \int\limits_1^{\sqrt 2 } {1dx} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {2dx} + \int\limits_{\sqrt 3 }^2 {3dx} $

$ = \left[ x \right]_1^{\sqrt n } + \left[ {2x} \right]_{\sqrt 2 }^{\sqrt 3 } + \left[ {3x} \right]_{\sqrt 3 }^2$

$ = \sqrt 2 - 1 + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3 $

$ = 5 - \sqrt 3 - \sqrt 2 $

${I_n} + {I_{n + 2}}$

$ = \int\limits_0^{\pi /4} {{{\tan }^n}} \,x\left( {1 + {{\tan }^2}x} \right)dx$

$ = \int\limits_0^{\pi /4} {{{\tan }^n}} x\,{\sec ^2}x\,dx$

$ = \left[ {{{{{\tan }^{n + 1}}x} \over {n + 1}}} \right]_0^{\pi /4}$

$ = {{1 - 0} \over {n + 1}} = {1 \over {n + 1}}$

$\therefore$ ${I_n} + {I_{n + 2}} = {1 \over {n + 1}}$

$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } n\left[ {{I_n} + {I_{n + 2}}} \right]$

$ = \mathop {\lim }\limits_{n \to \infty } \,n.{1 \over {n + 1}} = \mathop {\lim }\limits_{n \to \infty } {n \over {n + 1}}$

$ = \mathop {\lim }\limits_{n \to \infty } {n \over {n\left( {1 + {1 \over n}} \right)}} = 1$

$I = \int\limits_0^{10\pi } {\left| {\sin x} \right|} dx$

$ = 10\int\limits_0^\pi {\left| {\sin x\,} \right|} \,dx$

$ = 10\int\limits_0^\pi {\sin \,x\,dx} $

$\left[ \, \right.$ as $\left| {\sin x} \right|$ is periodic with period $\pi $

and $sin$ $x > 0$ if $0 < x$ $ < \pi $ $\left. \, \right]$

$I = 20\int\limits_0^{\pi /2} {\sin x} \,dx$

$ = 20\left[ { - \cos \,x_0^{\pi /2}} \right] = 20$

$\int_{ - \pi }^\pi {{{2x\left( {1 + \sin \,x} \right)} \over {1 + {{\cos }^2}x}}} dx$

$ = \int_{ - \pi }^\pi {{{2x\,dx} \over {1 + {{\cos }^2}x}} + 2\int_{ - \pi }^\pi {{{x\,\sin x} \over {1 + {{\cos }^2}x}}} } dx$

$ = 0 + 4\int_0^\pi {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} ;$

$\left[ \, \right.$ as $\int\limits_{ - a}^a {f\left( x \right)} dx = 0$ $\left. \, \right]$

if $f(x)$ is odd

$ = 2\int\limits_0^a {f\left( x \right)} dx$ if $f(x)$ is even.

$I = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \left( {\pi - x} \right)} \over {1 + {{\cos }^2}\left( {\pi - x} \right)}}} dx$

$I = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \,x} \over {1 + {{\cos }^2}x}}} $

$ \Rightarrow I = 4\pi \int_0^\pi {{{\sin x\,dx} \over {1 + {{\cos }^2}x}}} - 4\int {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} $

$ \Rightarrow 2I = 4\pi \int_0^\pi {{{\sin x} \over {1 + {{\cos }^2}x}}} dx$

put $\cos x = t \Rightarrow - \sin xdx = dt$

$\therefore$ $I = - 2\pi \int\limits_1^{ - 1} {{1 \over {1 + {t^2}}}} dt$

$ = 2\pi \int\limits_{ - 1}^1 {{1 \over {1 + {t^2}}}} dt$

$ = 2\pi \left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1$

$ = 2\pi \left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}\left( { - 1} \right)} \right]$

$ = 2\pi \left[ {{\pi \over 4} - \left( {{{ - \pi } \over 4}} \right)} \right] = 2\pi {\pi \over 2} = {\pi ^2}$

We have $\mathop {\lim }\limits_{x \to \infty } {{{1^p} + {2^p} + .... + {n^p}} \over {{n^{p + 1}}}};$

$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^n {{{{r^p}} \over {{n^p}.n}}} = \int\limits_0^1 {{x^p}} dx$

$ = {\left[ {{{{x^{p + 1}}} \over {p + 1}}} \right]_0} = {1 \over {p + 1}}$