Definite Integration

$I = \int\limits_0^1 {{{8\log \left( {1 + x} \right)} \over {1 + {x^2}}}} dx$

put $x = \tan \,\theta ,$

$\therefore$ ${{dx} \over {d\theta }} = {\sec ^2}\theta \Rightarrow dx = {\sec ^2}\theta d\theta $

$\therefore$ $I = 8\int\limits_0^{\pi /4} {{{\log \left( {1 + \tan \theta } \right)} \over {1 + {{\tan }^2}\theta }}} .{\sec ^2}\theta d\theta $

$I = 8\int\limits_0^{\pi /4} {\log \left( {1 + \tan \theta } \right)} d\theta \,...\left( i \right)$

$ = 8\int\limits_0^{\pi /4} {\log \left[ {1 + \tan \left( {{\pi \over 4} - \theta } \right)} \right]} d\theta $

$ = 8\int\limits_0^{\pi /4} {\log \left[ {1 + {{1 - \tan \theta } \over {1 + \tan \theta }}} \right]} d\theta $

$ = 8\int\limits_0^{\pi /4} {\log \left[ {{2 \over {1 + \tan \theta }}} \right]} d\theta $

$ = 8\int\limits_0^{\pi /4} {\left[ {\log 2 - \log \left( {1 + \tan \theta } \right)} \right]d\theta } $

$I = 8.\left( {\log 2} \right)\left[ x \right]_0^{\pi /4} - 8$

$\,\,\,\,\,\,\,\,\,\int\limits_0^{\pi /4} {\log \left( {1 + \tan \theta } \right)} d\theta $

$I = 8.{\pi \over 4}.\log 2 - I$ $\left[ {} \right.$ From equation $(i)$ $\left. {} \right]$

$ \Rightarrow 2I = 2\pi \log 2,$

$\therefore$ $I = \pi \log 2$

$p'\left( x \right) = p'\left( {1 - x} \right)$

$ \Rightarrow p\left( x \right) = - p\left( {1 - x} \right) + c$

at $x=0$

$p\left( 0 \right) = - p\left( 1 \right) + c \Rightarrow 42 = c$

Now, $p\left( x \right) = - p\left( {1 - x} \right) + 42$

$ \Rightarrow p\left( x \right) + p\left( {1 - x} \right) = 42$

$ \Rightarrow I = \int\limits_0^1 {p\left( x \right)dx\,\,\,\,\,\,\,\,\,\,...\left( i \right)} $

$ \Rightarrow I = \int\limits_0^1 {p\left( {1 - x} \right)} dx\,\,\,\,\,\,...\left( {ii} \right)$

on adding $(i)$ and $(ii),$

$\,\,\,\,\,\,\,\,\,\,\,$$2I = \int\limits_0^1 {\left( {42} \right)dx \Rightarrow I = 21} $

Let $I = \int_0^\pi {\left[ {\cot x} \right]dx\,\,\,\,\,\,...\left( 1 \right)} $

$ = \int_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]} dx$

$ = \int_0^\pi {\left[ { - \cot x} \right]dx\,\,\,\,\,\,...\left( 2 \right)} $

Adding two values of $I$ in $e{q^n}s\left( 1 \right)\,\,\& \,\,\left( 2 \right),$

We get $2I = \int_0^\pi {\left( {\left[ {\cot x} \right] + \left[ { - \cot x} \right]} \right)dx} $

$ = \int_0^\pi {\left( { - 1} \right)dx} $

$\left[ {} \right.$ As $\left[ x \right] + \left[ { - x} \right] = - 1,\,\,if\,\,$ $x \notin z$

and $\left[ x \right] + \left[ { - x} \right] = 0,\,\,if\,\,x \in z$ $\left. {} \right]$

$ = \left[ { - x} \right]_0^\pi = - \pi \Rightarrow I = - {\pi \over 2}$

We know that ${{\sin x} \over x} < 1,$ for $x \in \left( {0,1} \right)$

$ \Rightarrow {{\sin x} \over {\sqrt x }} < \sqrt x $ on $x \in \left( {0,1} \right)$

$ \Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx < \int\limits_0^1 {\sqrt x dx} = \left[ {{{2{x^{3/2}}} \over 3}} \right]} _0^1$

$ \Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}} dx < {2 \over 3} \Rightarrow I < {2 \over 3}$

Also ${{\cos x} \over {\sqrt x }} < {1 \over {\sqrt x }}$ for $x \in \left( {0,1} \right)$

$ \Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < \int\limits_0^1 {{x^{ - 1/2}}dx} } $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {2\sqrt x } \right]_0^1 = 2$

$ \Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < 2} $

$ \Rightarrow J < 2$

Given $f\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),$

where $f\left( x \right) = \int_1^x {{{\log \,t} \over {1 + t}}} \,dt$

$\therefore$ $F\left( e \right) = f\left( e \right) + f\left( {{1 \over e}} \right)$

$ \Rightarrow F\left( e \right)$

$ = \int_1^e {{{\log \,t} \over {1 + t}}dt + \int_1^{1/e} {{{\log t} \over {1 + t}}} } dt\,\,\,....\left( A \right)$

Now for solving, $I = \int_1^{1/e} {{{\log t} \over {1 + t}}dt} $

$\therefore$ Put ${1 \over t} = z \Rightarrow - {1 \over {{t^2}}}dt = dz$

$ \Rightarrow dt = - {{dz} \over {{z^2}}}$

and limit for $t = 1 \Rightarrow z = 1$ and for

$t = 1/e \Rightarrow z = e$

$\therefore$ $I = \int_1^e {{{\log \left( {{1 \over z}} \right)} \over {1 + {1 \over z}}}} \left( { - {{dz} \over {{z^2}}}} \right)$

$ = \int_1^e {{{\left( {\log 1 - \log z} \right).z} \over {z + 1}}\left( { - {{dz} \over {{z^2}}}} \right)} $

$ = \int_1^e { - {{\log z} \over {\left( {z + 1} \right)}}\left( { - {{dz} \over z}} \right)} $

[ as $\log 1 = 0$ ]

$ = \int_1^e {{{\log z} \over {z\left( {z + 1} \right)}}} dz$

$\therefore$ $I = \int_1^e {{{\log \,t} \over {t\left( {t + 1} \right)}}dt} $

$\left[ {} \right.$ By property $\int_a^b {f\left( t \right)dt} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = \int_a^b {f\left( x \right)dx} $ $\left. {} \right]$

Equation $(A)$ becomes

$F\left( e \right) = \int_1^e {{{\log t} \over {1 + t}}dt + \int_1^e {{{\log t} \over {t\left( {1 + t} \right)}}} } dt$

$ = \int_1^e {{{t.\log t + \log t} \over {t\left( {1 + t} \right)}}} dt$

$ = \int_1^e {{{\left( {\log t} \right)\left( {t + 1} \right)} \over {t\left( {1 + t} \right)}}} $

$ \Rightarrow F\left( e \right) = \int_1^e {{{\log t} \over t}dt} $

Let $\log t = x$

$\therefore$ ${1 \over t}dt = dx$

$\left[ {} \right.$ for limit $t = 1,x = 0$

and $t = e,x = \log \,e = 1$ $\left. {} \right]$

$\therefore$ $F\left( e \right) = \int_0^1 {x\,dx} $

$ \Rightarrow F\left( e \right) = \left[ {{{{x^2}} \over 2}} \right]_0^1$

$ \Rightarrow F\left( e \right) = {1 \over 2}$

$\int_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }}} = {\pi \over 2}$

$\therefore$ $\left[ {{{\sec }^{ - 1}}t} \right]_{\sqrt 2 }^x = {\pi \over 2}$

$\left[ {} \right.$ As $\int {{{dx} \over {x\sqrt {{x^2} - 1} }}} = {\sec ^{ - 1}}x$ $\left. {} \right]$

$ \Rightarrow {\sec ^{ - 1}}x - {\sec ^{ - 1}}\sqrt 2 = {\pi \over 2}$

$ \Rightarrow {\sec ^{ - 1}}x - {\pi \over 4} = {\pi \over 2}$

$ \Rightarrow {\sec ^{ - 1}}x = {\pi \over 2} + {\pi \over 4}$

$ \Rightarrow {\sec ^{ - 1}}x = {{3\pi } \over 4}$

$ \Rightarrow x = \sec {{3\pi } \over 4}$

$ \Rightarrow x = - \sqrt 2 $

Let $a = k + h$ where $k$ is an integer such that

$\left[ a \right] = k$ and $0 \le h < 1$

$\therefore$ $\int\limits_1^a {\left[ x \right]f'\left( x \right)dx = \int\limits_1^2 {1f'\left( x \right)} } \,dx$

$\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_2^3 {2f'\left( x \right)dx + } ....$

$\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_{k - 1}^k {\left( {k - 1} \right)dx + \int\limits_k^{k + h} {kf'\left( x \right)} } dx$

$\left\{ {f\left( 2 \right) - f\left( 1 \right)} \right\} + 2\left\{ {f\left( 3 \right) - f\left( 2 \right)} \right\}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ + 3\left\{ {f\left( 4 \right) - f\left( 3 \right)} \right\} + \,........\,$

$\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ + \left( {k - 1} \right)\left\{ {f\left( k \right) - f\left( {k - 1} \right)} \right\}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ + k\left\{ {f\left( {k + h} \right) - f\left( k \right)} \right\}$

$ = - f\left( 1 \right) - f\left( 2 \right) - f\left( 3 \right).......$

$\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ - f\left( k \right) + kf\left( {k + h} \right)$

$ = \left[ a \right]f\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right)} \right.$

$\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\left. { + f\left( 3 \right) + ........f\left( {\left[ a \right]} \right)} \right\}$

$I = \int\limits_0^\pi {xf\left( {\sin \,x} \right)dx} $

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

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

$ \Rightarrow 2I = \pi {\pi \over 0}f\left( {\sin x} \right)dx$

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

$ = \pi \int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx} $

$ = \pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx} $

$I = \int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} \,dx$

Put $x + \pi = t$

$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{t^3} + {{\cos }^2}t} \right)dt} $

$ = 2\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}} tdt$

$\left[ {} \right.$ using the property of even and odd function $\left. {} \right]$

$ = \int\limits_0^{{\pi \over 2}} {\left( {1 + \cos 2t} \right)} dt = {\pi \over 2} + 0$

${I_1} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,{I_2} = \int\limits_0^1 {{2^{{x^3}}}} dx,$

$ = {I_3} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,$

${I_4} = \int\limits_0^1 {{2^{{x^3}}}} dx\,\,$

$\forall 0 < x < 1,\,{x^2} > {x^3}$

$ \Rightarrow \int\limits_0^1 {{2^{{x^2}}}} \,dx > \int\limits_0^1 {{2^{{x^3}}}} dx$

and $\int\limits_1^2 {{2^{{x^3}}}dx} > \int\limits_1^2 {{2^{{x^2}}}dx} $

$ \Rightarrow {I_1} > {I_2}$ and ${I_4} > {I_3}$

$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}} + {3 \over {{n^2}}}se{c^2}{9 \over {{n^2}}} + ... + {1 \over n}{{\sec }^2}1} \right]$

$ = \mathop {\lim }\limits_{n \to \infty } {r \over {{n^2}}}{\sec ^2}{{{r^2}} \over {{n^2}}} = \mathop {\lim }\limits_{n \to \infty } {1 \over n}.{\pi \over n}{\sec ^2}{{{r^2}} \over {{n^2}}}$

$ \Rightarrow $ Given limit is equal to value of integral

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_0^1 {x{{\sec }^2}} \,{x^2}\,dx$

or ${1 \over 2}\int\limits_0^1 {2x\,\sec } \,\,{x^2}dx = {1 \over 2}\int\limits_0^1 {{{\sec }^2}} tdl$

$\left[ {} \right.$ put $\left. {{x^2} = t} \right]$

$ = {1 \over 2}\left( {\tan t} \right)_0^1 = {1 \over 2}\tan \,1.$

$I = \int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$I = \int\limits_3^6 {{{\sqrt {9 - x} } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

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

Adding equation $(1)$ and $(2)$

$2I = \int\limits_3^6 {dx} = 3 \Rightarrow I = {3 \over 2}$

$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}} dt$

$ = \mathop {\lim }\limits_{x \to 0} {{\int\limits_6^{f\left( x \right)} {4{t^3}dt} } \over {x - 2}}$

This limit resembles a derivative because the fraction has the form $0/0$ as $x \to 2$ since both the numerator (integral from $6$ to $f(x)$) and the denominator ($x - 2$) are zero when $x = 2$.

Applying $L'$ Hospital rule

$\mathop {\lim }\limits_{x \to 2} {{\left[ {4f{{\left( x \right)}^3}f'\left( x \right)} \right]} \over 1}$

$ = 4{\left( {f\left( 2 \right)} \right)^3}f'\left( 2 \right)$

$ = 4 \times {6^3} \times {1 \over {48}} = 18$

Let $I = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {a^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$ = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}\left( { - x} \right)} \over {1 + {a^{ - x}}}}} dx$

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

$ = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {\alpha ^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

Adding equations $(1)$ and $(2)$ we get

$2I = \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\left( {{{1 + {a^x}} \over {1 + {a^x}}}} \right)dx$

$ = \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\,dx$

$ = 2\int\limits_0^\pi {{{\cos }^2}} x\,dx$

$ = 2 \times 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} x\,dx$

$ = 4\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} x\,dx$

$ \Rightarrow I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} \,x\,dx$

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

$ \Rightarrow I\,\, = 2\int\limits_0^{{\pi \over 2}} {dx - 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} } \,x\,dx$

$ \Rightarrow I + I = 2\left( {{\pi \over 2}} \right) = \pi $

$ \Rightarrow I = {\pi \over 2}$

$\mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}} {e^{{r \over n}}}\,\,$

$\left[ {} \right.$ Using definite integrals as limit of sum $\left. {} \right]$

$ = \int\limits_\theta ^1 {{e^x}} dx = e - 1$

Let $I = \int\limits_0^\pi {xf\left( {\sin x} \right)} dx$

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

$\therefore$ $2I = \pi \int\limits_2^\pi {f\left( {\sin x} \right)} dx$

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

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

$ \Rightarrow A = \pi $

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

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}}$