Definite Integration

$I = \mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,.....\,\, + \,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right)$

Let ${2^n} = t$ and if $n \to \infty $ then $t \to \infty $

$I = \mathop {\lim }\limits_{n \to \infty } {1 \over t}\left( {\sum\limits_{r = 1}^{t - 1} {{1 \over {\sqrt {1 - {r \over t}} }}} } \right)$

$I = \int\limits_0^1 {{{dx} \over {\sqrt {1 - x} }} = \int\limits_0^1 {{{dx} \over {\sqrt x }}\,\,\,\,\,\,\,\,\,\left( {\int\limits_0^a {f(x)dx = \int\limits_0^a {f(a - x)dx} } } \right.} } $

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

$I = \int_{ - 3}^{101} {\left( {\left[ {\sin (\pi x)} \right] + {e^{[\cos (2\pi x)]}}} \right)dx} $

So,

$I = 52\int_0^2 {\left( {\left[ {\sin \pi x} \right] + {e^{[\cos 2\pi x]}}} \right)dx} $

$ = 52\left\{ {\int_1^2 { - 1} \,dx + \int_{{1 \over 4}}^{{3 \over 4}} {{e^{ - 1}}\,dx + \int_{{5 \over 4}}^{{7 \over 4}} {{e^{ - 1}}\,dx + \int_0^{{1 \over 4}} {{e^0}\,dx + \int_{{3 \over 4}}^{{5 \over 4}} {{e^0}\,dx + \int_{{7 \over 4}}^2 {{e^0}\,dx} } } } } } \right\}$

$ = {{52} \over e}$

Case 1 :

Let $0 \le x < 1$

then $\left[ x \right] = 0$, which is even

$\therefore$ $f(x) = 1 + \left[ x \right] - x$

$ = 1 + 0 - x$

$ = 1 - x$

Case 2 :

Let $1 \le x < 2$

then $\left[ x \right] = 1$, which is odd

$\therefore$ $f(x) = x - \left[ x \right]$

$ = x - 1$

Case 3 :

Let $2 \le x < 3$

then $\left[ x \right] = 2$, which is even

$\therefore$ $f(x) = 1 + \left[ x \right] - x$

$ = 1 + 2 - x$

$ = 3 - x$

Case 4 :

Let $3 \le x < 4$

then $\left[ x \right] = 3$, which is odd

$\therefore$ $f(x) = x - \left[ x \right]$

$ = x - 3$

$\therefore$ $f(x) = \left\{ {\matrix{ {1 - x} & ; & {0 \le x < 1} \cr {x - 1} & ; & {1 \le x < 2} \cr {3 - x} & ; & {2 \le x < 3} \cr {x - 3} & ; & {3 \le x < 4} \cr } } \right.$

$\therefore$ $f(x)$ is periodic and period of $f(x) = 2$

And period of $\cos \pi x = {{2\pi } \over \pi } = 2$

$\therefore$ Period of $f(x)\cos \pi x = 2$

Now,

$I = {{{\pi ^2}} \over {10}}\int_{ - 10}^{10} {f(x)\cos \pi x\,dx} $

$ = {{{\pi ^2}} \over {10}}\int_{ - 10}^{ - 10 + 10 \times 2} {f(x)\cos \pi x\,dx} $

$ = {{{\pi ^2}} \over {10}}\int_0^{10 \times 2} {f(x)\cos \pi x\,dx} $

$ = {{{\pi ^2}} \over {10}} \times 10\int_0^2 {f(x)\cos \pi x\,dx} $

$ = {\pi ^2}\int_0^2 {f(x)\cos \pi x\,dx} $

$\therefore$ $I = {\pi ^2}\left[ {\int_0^1 {f(x)\cos \pi x\,dx + \int_1^2 {f(x)\cos \pi x\,dx} } } \right]$

$ = {\pi ^2}\left[ {\int_0^1 {(1 - x)\cos \pi x\,dx + \int_1^2 {(x - 1)\cos \pi x\,dx} } } \right]$

$ = {\pi ^2}\left[ {\int_0^1 {\cos \pi x\,dx - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - \int_1^2 {\cos \pi x\,dx} } } } } \right]$

$ = {\pi ^2}\left[ {{1 \over \pi }\left[ {\sin \pi x} \right]_0^1 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - {1 \over \pi }\left[ {\sin \pi x} \right]_1^2} } } \right]$

$ = {\pi ^2}\left[ {0 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - 0} } } \right]$

$ = {\pi ^2}\left[ { - \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_0^1 + \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_1^2} \right]$

$\left[ {\mathrm{As}\,\int {x\cos \pi x\,dx = x\,.\,\int {\cos \pi x - \int {\left( {1\,.\,{{\sin \pi x} \over \pi }} \right)dx = x\,.\,{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x + c} } } } \right]$

$ = {\pi ^2}\left[ { - \left[ {\left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\,.\,\cos \pi } \right) - \left( {0 + {1 \over {{\pi ^2}}}\,.\,\cos 0} \right)} \right] + \left[ {\left( {2\,.\,{{\sin 2\pi } \over \pi } + {1 \over {{\pi ^2}}}\cos 2\pi } \right) - \left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\cos \pi } \right)} \right]} \right]$

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

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

$ = {\pi ^2}\left[ {{2 \over {{\pi ^2}}} + {2 \over {{\pi ^2}}}} \right]$

$ = {\pi ^2} \times {4 \over {{\pi ^2}}}$

$ = 4$

$\mathop {\lim }\limits_{n \to \alpha } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} $

$ = \mathop {\lim }\limits_{n \to \alpha } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)} \over {2{{\left( {{r \over n}} \right)}^2} - 7\left( {{r \over n}} \right) + 6}}} $

$ = \int_a^b {f(x)dx} $

$a = \mathop {\lim }\limits_{n \to \alpha } \left( {{1 \over n}} \right) = 0$

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

and ${r \over n} \to x$

$ = \int_0^1 {{x \over {2{x^2} - 7x + 6}}dx} $

$ = \int_0^1 {{x \over {2{x^2} - 3x - 4x + 6}}dx} $

$ = \int_0^1 {{x \over {(2x - 3)(x - 2)}}dx} $

$ = \int_0^1 {\left[ {{A \over {(2x - 3)}} + {B \over {(x - 2)}}} \right]dx} $

$ = \int_0^1 {\left( {{{ - 3} \over {2x - 3}} + {2 \over {x - 2}}} \right)dx} $

$ = \left[ { - {{3\log |2x - 3|} \over 2} + 2\log |x - 2|} \right]_0^1$

$ = - {3 \over 2}\left[ {\log ( - 1) - \log ( - 3)} \right] + 2\left[ {\log ( - 1) - \log ( - 2)} \right]$

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

$ = + {3 \over 2}\log 3 - 2\log 2$

$ = \log \sqrt {{3^3}} - \log 4$

$ = \log {{\sqrt {{3^2} \times 3} } \over 4}$

$ = \log {{3\sqrt 3 } \over 4}$

Given,

$f(x) = x + \int_0^1 {(x - t)f(t)dt} $

$ = x + x\int_0^1 {f(t)dt - \int_0^1 {tf(t)dt} } $

$ = x\left( {1 + \int_0^1 {f(t)dt} } \right) - \int_0^1 {tf(t)dt} $

Now,

let $A = 1 + \int_0^1 {f(t)dt} $

and $B = \int_0^1 {tf(t)dt} $

$\therefore$ $f(x) = Ax - B$

$ \Rightarrow f(t) = At - B$

So, $A = 1 + \int_0^1 {f(t)dt} $

$ = 1 + \int_0^1 {(At - B)dt} $

$ = 1 + \left[ {{{A{t^2}} \over 2} - Bt} \right]_0^1$

$ = 1 + {A \over 2} - B$

$ \Rightarrow {A \over 2} = 1 - B$ ...... (1)

$B = \int_0^1 {tf(t)dt} $

$ = \int_0^1 {t(At - B)dt} $

$ = \int_0^1 {(A{t^2} - Bt)dt} $

$ = \left[ {{{A{t^3}} \over 3} - {{B{t^2}} \over 2}} \right]_0^1$

$ = {A \over 3} - {B \over 2}$

$ \Rightarrow {{3B} \over 2} = {A \over 3}$ ....... (2)

Solving equation (1) and (2) we get,

$A = {{18} \over {13}}$ and $B = {4 \over {13}}$

$\therefore$ $f(x) = {{18} \over {13}}x - {4 \over {13}}$

By checking all the options you can see when x = 6 we get

$y = f(x) = {{18} \over {13}} \times 6 - {4 \over {13}}$

$ = {{108 - 4} \over {13}} = 8$

$\therefore$ Point (6, 8) lies on the curve.

Given,

$\int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } } $

Now,

$L.H.S. = \int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx} $

$ = \sqrt 2 \int_0^2 {{x^{1/2}}dx - \int_0^2 {\sqrt { - ({x^2} - 2x)} dx} } $

$ = \sqrt 2 \left[ {{{{x^{3/2}}} \over {{3 \over 2}}}} \right]_0^2 - \int_0^2 {\sqrt { - \left[ {{{(x - 1)}^2} - 1} \right]} dx} $

$ = \sqrt 2 \times {2 \over 3}\left[ {{2^{{3 \over 2}}}} \right] - \int_0^2 {\sqrt {1 - {{(x - 1)}^2}} dx} $

[We know,

$\sqrt {{a^2} - {x^2}} = {x \over 2}\sqrt {{a^2} - {x^2}} + {{{a^2}} \over 2}{\sin ^{ - 1}}\left( {{x \over a}} \right) + C$]

$ = {{2\sqrt 2 \times 2\sqrt 2 } \over 3} - \left[ {{{x - 1} \over 2}\sqrt {1 - {{(x - 1)}^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{{x - 1} \over 1}} \right)} \right]_0^2$

$ = {8 \over 3} - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {\left( {{1 \over 2}} \right)\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}( - 1)} \right)} \right]$

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

$ = {8 \over 3} - {1 \over 2}\,.\,{\pi \over 2} + {1 \over 2}\,.\,{{3\pi } \over 2}$

$ = {8 \over 3} - {\pi \over 4} + {{3\pi } \over 4}$

$ = {8 \over 3} - {{2\pi } \over 4}$

$ = {8 \over 3} - {\pi \over 2}$

Now in R.H.S.,

$\int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy} $

$ = \int_0^1 {dy - \int_0^1 {\sqrt {1 - {y^2}} - \int_0^1 {{{{y^2}} \over 2}dy} } } $

$ = \left[ y \right]_0^1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1 - \left[ {{{{y^3}} \over 6}} \right]_0^1$

$ = 1 - \left[ {\left( {{1 \over 2}\sqrt {1 - 0} + {1 \over 2}{{\sin }^{ - 1}}(1)} \right) - \left( {0 + {{\sin }^{ - 1}}(0)} \right)} \right] - {1 \over 6}$

$ = 1 - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) - {1 \over 6}$

$ = 1 - {1 \over 2} \times {\pi \over 2} - {1 \over 6}$

$ = {5 \over 6} - {\pi \over 4}$

Now,

$\int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy} $

$ = \left[ {2y - {{{y^3}} \over 6}} \right]_1^2$

$ = \left[ {\left( {4 - {8 \over 6}} \right) - \left( {2 - {1 \over 6}} \right)} \right]$

$ = 4 - {8 \over 6} - 2 + {1 \over 6}$

$ = 2 - {7 \over 6}$

$ = {5 \over 6}$

$\therefore$ $R.H.S. = {5 \over 6} - {\pi \over 4} + {5 \over 6} + I$

As $L.H.S. = R.H.S.$

$ \Rightarrow {8 \over 3} - {\pi \over 2} = {{10} \over 6} - {\pi \over 4} + I$

$ \Rightarrow {{10} \over 6} - {8 \over 3} + I = - {\pi \over 2} + {\pi \over 4}$

$ \Rightarrow {{ - 6} \over 6} + I = - {\pi \over 4}$

$ \Rightarrow I = 1 - {\pi \over 4}$

From option (c)

$\int_0^1 {(1 - \sqrt {1 - {y^2}} )dy} $

$ = \left[ y \right]_0^1 - \int_0^1 {(\sqrt {1 - {y^2}} )dy} $

$ = 1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1$

$ = 1 - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {0 + {1 \over 2}{{\sin }^{ - 1}}(0)} \right)} \right]$

$ = 1 - \left[ {{1 \over 2} \times {\pi \over 2} - 0} \right]$

$ = 1 - {\pi \over 4} = I$

$\therefore$ Option (c) is the right answer.

Note :

There is no way to guess which option is correct. You have to check all the options to see which give value equal to I.

We know,

$\left[ {{x \over 2}} \right]$ is discontinuous at 1, 2, 3, 4 ........

$\therefore$ [ x ] is discontinuous at 2, 4, 6, 8 .....

In between 0 to 5 it is discontinuous at 2 and 4.

Break the integration into 3 parts

(1) 0 to 2

(2) 2 to 4

(3) 4 to 5

$\therefore$ $\int\limits_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx} $

$ = \int\limits_0^2 {\cos \left( {\pi (x - 0)} \right)dx + \int\limits_2^4 {\cos \left( {\pi (x - 1)} \right)dx + \int\limits_4^5 {\cos \left( {\pi (x - 2)} \right)dx} } } $

$ = \int\limits_0^2 {\cos \pi x\,dx + \int\limits_2^4 {\cos (\pi x - \pi )dx + \int\limits_4^5 {\cos (\pi x - 2\pi )dx} } } $

$ = \int\limits_0^2 {\cos \pi dx - \int\limits_2^4 {\cos \pi x\,dx + \int\limits_4^5 {\cos \pi x\,dx} } } $

$ = \left[ {{{\sin \pi x} \over \pi }} \right]_0^2 - \left[ {{{\sin \pi x} \over \pi }} \right]_2^4 + \left[ {{{\sin \pi x} \over \pi }} \right]_4^5$

$ = 0 - 0 + 0$

$ = 0$

$f(x) = \int_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt} $

$f'(x) = 2x\left( {{{{x^4} - 5{x^2} + 4} \over {2 + {e^{{x^2}}}}}} \right) = 0$

$x = 0$, or $({x^2} - 4)({x^2} - 1) = 0$

$x = 0,$ $x = \pm 2,\, \pm 1$

Now, $f'(x) = {{2x(x + 1)(x - 1)(x + 2)(x - 2)} \over {\left( {{e^{{x^2}}} + 2} \right)}}$

f'(x) changes sign from positive to negative at x = $-$1, 1 So, number of local maximum points = 2

f'(x) changes sign from negative to positive at x = $-$2, 0, 2 So, number of local minimum points = 3

$\therefore$ m = 2, n = 3

$\int\limits_{\cos x}^1 {{t^2}f(t)dt = {{\sin }^3}x + \cos x} $

$ \Rightarrow \sin x{\cos ^2}x\,f(\cos x) = 3{\sin ^2}x\cos x - \sin x$

$ \Rightarrow f(\cos x) = 3\tan x - {\sec ^2}x$

$ \Rightarrow f'(\cos x).\,( - \sin x) = 3{\sec ^2}x - 2{\sec ^2}x\tan x$

Put $\cos x = {1 \over {\sqrt 3 }}$,

$\therefore$ $f'\left( {{1 \over {\sqrt 3 }}} \right)\left( { - {{\sqrt 2 } \over {\sqrt 3 }}} \right) = 9 - 6\sqrt 2 $

${1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right) = 6 - {9 \over {\sqrt 2 }}$

$\int_0^1 {{1 \over {{7^{\left[ {{1 \over x}} \right]}}}}dx} $, let ${1 \over x} = t$

${{ - 1} \over {{x^2}}}dx = dt$

$ = \int_\infty ^1 {{1 \over { - {t^2}{7^{[t]}}}}dt = \int_1^\infty {{1 \over {{t^2}{7^{[t]}}}}dt} } $

$ = \int_1^2 {{1 \over {7{t^2}}}dt + \int_2^3 {{1 \over {{7^2}{t^2}}}dt + \,\,....} } $

$ = {1 \over 7}\left[ { - {1 \over t}} \right]_1^2 + {1 \over {{7^2}}}\left[ {{{ - 1} \over t}} \right]_2^3 + {1 \over {{7^3}}}\left[ { - {1 \over t}} \right]_2^3 + \,\,....$

$ = \sum\limits_{n = 1}^\infty {{1 \over {{7^n}}}\left( {{1 \over n} - {1 \over {n + 1}}} \right)} $

$ = \sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^n}} \over n} - 7\sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^{n + 1}}} \over {n + 1}}} } $

$ = - \log \left( {1 - {1 \over 7}} \right) + 7\log \left( {1 - {1 \over 7}} \right) + 1$

$ = 1 + 6\log {6 \over 7}$

$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{x|x|}} + 1}}dx} $ ..... (i)

$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{ - x|x|}} + 1}}dx} $ ..... (ii)

$2I = \int\limits_{ - 2}^2 {|{x^3} + x|dx} $

$2I =2 \int\limits_0^2 {({x^3} + x)dx} $

$I = \int\limits_0^2 {({x^3} + x)dx} $

$ = \left. {{{{x^4}} \over 4} + {{{x^2}} \over 2}} \right]_0^2$

$ = \left( {{{16} \over 4} + {4 \over 2}} \right) - 0$

$ = 4 + 2 = 6$

${b_n} - {b_{n - 1}} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx - {{\cos }^2}(n - 1)x} \over {\sin x}}dx} $

$ = \int_0^{{\pi \over 2}} {{{ - \sin (2n - 1)x\,.\,\sin x} \over {\sin x}}dx} $

$ = \left. {{{\cos (2n - 1)x} \over {2n - 1}}} \right|_0^{\pi /2} = - {1 \over {2n - 1}}$

So, ${b_3} - {b_2}$, ${b_4} - {b_3}$, ${b_5} - {b_4}$ are in H.P.

$ \Rightarrow {1 \over {{b_3} - {b_2}}},\,{1 \over {{b_4} - {b_3}}},\,{1 \over {{b_5} - {b_4}}}$ are in A.P. with common difference $-$2.

$\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {\left( {1 + {{\cos }^2}x} \right)\left( {{e^{\cos x}} + {e^{ - \cos x}}} \right)}}dx} $

Let $\cos x = t$

$\sin xdx = dt$

$ = \int\limits_1^{ - 1} {{{ - {e^t}dt} \over {\left( {1 + {t^2}} \right)\left( {{e^t} + {e^{ - t}}} \right)}}} $

$I = \int\limits_{ - 1}^1 {{{{e^t}} \over {\left( {1 + {t^2}} \right)\left( {{e^t} + {e^{ - t}}} \right)}}dt} $ ...... (i)

$I = \int\limits_{ - 1}^1 {{{{e^{ - t}}} \over {\left( {1 + {t^2}} \right)\left( {{e^{ - t}} + {e^t}} \right)}}dt} $ .... (ii)

Adding (i) and (ii)

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

$2I = \left. {{{\tan }^{ - t}}} \right|_{ - 1}^1$

$2I = {\pi \over 4} - \left( { - {\pi \over 4}} \right)$

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

$I = {\pi \over 4}$

$I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^x})({{\sin }^6}x + {{\cos }^6}x)}}} $ ...... (i)

$I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^{ - x}})(si{n^6}x + {{\cos }^6}x)}}} $ ..... (ii)

(i) and (ii)

From equation (i) & (ii)

$2I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {{{\sin }^6}x + {{\cos }^6}x}}} $

$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{dx} \over {{{\sin }^6}x + {{\cos }^6}x}} = \int\limits_0^{{\pi \over 2}} {{{dx} \over {1 - {3 \over 4}{{\sin }^2}2x}}} } $

$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{4{{\sec }^2}2xdx} \over {4 + {{\tan }^2}2x}} = 2\int\limits_0^{{\pi \over 4}} {{{4{{\sec }^2}2x} \over {4 + {{\tan }^2}2x}}dx} } $

when x = 0, t = 0

Now, $\tan 2x = t$

when $x = {\pi \over 4},\,\,t \to \infty $

$2{\sec ^2}2x\,dx = dt$

$\therefore$ $I = 2\int\limits_0^\infty {{{2dt} \over {4 + {t^2}}} = 2\left( {{{\tan }^{ - 1}}{t \over 2}} \right)_0^\infty } $

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

$\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + \,\,...\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)$

$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{n^2}} \over {({n^2} + {r^2})(n + r)}}} $

$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{1 \over {\left[ {1 + {{\left( {{r \over n}} \right)}^2}} \right]\left[ {1 + \left( {{r \over n}} \right)} \right]}}} $

$ = \int\limits_0^1 {{1 \over {(1 + {x^2})(1 + x)}}dx} $

$ = {1 \over 2}\int_0^1 {\left[ {{1 \over {1 + x}} - {{(x - 1)} \over {(1 + {x^2})}}} \right]dx} $

$ = {1 \over 2}\left[ {\ln (1 + x) - {1 \over 2}\ln \left( {1 + {x^2}} \right) + {{\tan }^{ - 1}}x} \right]_0^1$

$ = {1 \over 2}\left[ {{\pi \over 4} + {1 \over 2}\ln 2} \right] = {\pi \over 8} + {1 \over 4}\ln 2$

Given,

$\int_1^e {{{{{\left( {\log _e^x} \right)}^{1/2}}} \over {x{{\left( {a - {{\left( {\log _e^x} \right)}^{3/2}}} \right)}^2}}} = 1} $

Let $\log _e^x = t$

$ \Rightarrow x = {e^t}$

$ \Rightarrow dx = {e^t}dt$

When $x = 1$ then $t = \log _e^1 = 0$

and when $x = e$ then $t = \log _e^e = 1$

$ \Rightarrow \int_0^1 {{{\sqrt t \,.\,{e^t}dt} \over {{e^t}{{(a - t\sqrt t )}^2}}} = 1} $

$ \Rightarrow \int_0^1 {{{\sqrt t \,dt} \over {{{(a - t\sqrt t )}^2}}} = 1} $

Let ${t^{1/2}} = z$

$ \Rightarrow {1 \over {2\sqrt t }}dt = dz$

$ \Rightarrow dt = 2\sqrt t \,dz \Rightarrow dt = 2z\,dz$

When $t = 0$, then $z = 0$

When $t = 1$, then $z = {1^{1/2}} = 1$

$\therefore$ $\int_0^1 {{{z \times 2z\,dz} \over {{{(a - {z^3})}^2}}} = 1} $

$ \Rightarrow \int_0^1 {{{2{z^2}\,dz} \over {{{(a - {z^3})}^2}}} = 1} $

Now let $a - {z^3} = p$

$ \Rightarrow - 3{z^2}\,dz = dp$

When $z = 0$ then $p = a - 0 = a$

When $z = 1$ then $p = a - {1^3} = a - 1$

$\therefore$ $\int_a^{a - 1} {{{2\,.\,\left( { - {{dp} \over 3}} \right)} \over {{p^2}}} = 1} $

$ \Rightarrow - {2 \over 3}\int_a^{a - 1} {{{dp} \over {{p^2}}} = 1} $

$ \Rightarrow - {2 \over 3}\left[ { - {1 \over p}} \right]_a^{a - 1} = 1$

$ \Rightarrow - {2 \over 3}\left[ { - {1 \over {a - 1}} + {1 \over a}} \right] = 1$

$ \Rightarrow - {2 \over 3}\left[ {{{ - a + a - 1} \over {a(a - 1)}}} \right] = 1$

$ \Rightarrow {2 \over 3}\left( {{1 \over {a(a - 1)}}} \right) = 1$

$ \Rightarrow 3{a^2} - 3a = 2$

$ \Rightarrow 3{a^2} - 3a - 2 = 0$

$ \Rightarrow a = {{3\, \pm \,\sqrt {9 - 4(3)( - 2)} } \over {2\,.\,3}}$

$\therefore$ $a = {{3\, \pm \,\sqrt {33} } \over 6},\,{{3 - \sqrt {33} } \over 6}$

$ \text { } \begin{aligned} & \text { Let } I=\int_0^4 \| x-2|-x| d x \\ & I=\int_0^2|-(x-2)-x| d x+\int_2^4|x-2-x| d x \\ &=\int_0^2|2-2 x| d x+\int_2^4 2 d x \\ &=\int_0^1(2-2 x) d x+\int_1^2-(2-2 x) d x+\int_2^4 2 d x \\ &=\left[2 x-x^2\right]_0^1-\left[2 x-x^2\right]_1^2+2[x]_2^4 \\ &=[2-1-0]-[4-4-2+1]+2[4-2] \\ &=1+1+4 \\ &=6 \end{aligned} $

$\int_{-a}^a f(x) \cdot d x=\int_0^a f(x) d x+\int_0^a g(x) d x$

As we know,

$ \int_{-a}^a f(x) d x=\int_{-a}^0 f(x) d x+\int_0^a f(x) d x $

In 2nd term, on putting $x=-t$

$ \Rightarrow \quad d x=-d t $

Limits when $x=0, t=0$

When $x=-a, t=a$

$ \begin{aligned} \therefore \quad \int_{-a}^a f(x) d x & =\int_a^0 f(-t),(-d t)+\int_0^a f(x) d x \\ & =\int_0^a f(x) d x-\int_a^0 f(-t) d t \\ & =\int_0^a f(x)+\int_0^a f(-t) d t \\ & =\int_0^a f(x)+\int_0^a f(-x) d x \\ \therefore g(x)=f(-x) . & \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n p} f\left(\frac{r}{n}\right)=\int_0^p f(x) d x \\ & f: R \rightarrow R, f(x)=x^2+2 \\ & \text { Now, } \lim _{n \rightarrow \infty} \frac{3}{n}\left[\begin{array}{r} f\left(\frac{7}{n}\right)+f\left(\frac{14}{n}\right)+f\left(\frac{21}{n}\right) \\ +\ldots+f(7) \end{array}\right] \\ & =\lim _{n \rightarrow \infty} \frac{3}{n}\left[\begin{array}{r} \left(\frac{7}{n}\right)^2+2+\left(\frac{14}{n}\right)^2+2+\left(\frac{21}{n}\right)^2 \\ +2+\ldots+\left(\frac{7 n}{n}\right)^2+2 \end{array}\right] \\ & =\lim _{n \rightarrow \infty} \frac{3}{n}\left[\begin{array}{r} 2 n+\left(\frac{7}{n}\right)^2\left(1^2+2^2+3^2\right. \\ +\ldots+n^2 \end{array}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{n \rightarrow \infty} \frac{3}{n}\left[2 n+\frac{49}{n^2} \frac{n(n+1)(2 n+1)}{6}\right] \\ & =\lim _{n \rightarrow \infty} 3\left[2+\frac{49}{6} \frac{n^2\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{n^2}\right] \\ & =3\left[2+\frac{49}{6} \times 2\right] \\ & =55 \end{aligned} $

Given,

$ \begin{aligned} & f(x)=\left|\begin{array}{ccc} 2 \cos ^2 x & \sin 2 x & \sin x \\ \sin 2 x & 2 \sin ^2 x & -\cos x \\ \sin x & -\cos x & 0 \end{array}\right| \\ & \text { Applying } R_2 \rightarrow R_2-2 \cos x R_3 \\ & f(x)=\left|\begin{array}{ccc} 2 \cos ^2 x & \sin 2 x & \sin x \\ 0 & 2 & -\cos x \\ \sin x & -\cos x & 0 \end{array}\right| \end{aligned} $

Expanding along $R_2$,

$ \begin{aligned} & f(x)=\left[0+2\left(0-\sin ^2 x\right)\right. \\ &+ \cos x\left(-2 \cos ^3 x-\sin x \sin 2 x\right] \\ & \Rightarrow f(x)=-2 \sin ^2 x \\ &-2 \cos ^2 x\left(\sin ^2 x+\cos ^2 x\right)=-2 \\ & \Rightarrow f^{\prime}(x)=0 \\ & \therefore \int_0^{\pi / 4}\left[2|f(x)|+5 f^{\prime}(x)\right] d x \\ &=\int_0^{\pi / 4}(4+0) d x=\pi \end{aligned} $

$ \text { } \begin{aligned}Let \,\,\, I & =\int_0^3\left(\sin \left(\frac{\pi}{3} x\right)-\cos \left(\frac{\pi}{3} x\right)\right) d x \\ \quad I & =\left[\frac{-\cos \frac{\pi}{3} x}{\frac{\pi}{3}}-\frac{\sin \left(\frac{\pi}{3} x\right)}{\frac{\pi}{3}}\right]_0^3 \end{aligned} $

$ \begin{aligned} & =-\frac{3}{\pi}\left[\cos \frac{\pi}{3} x+\sin \left(\frac{\pi x}{3}\right)\right]_0^3 \\ & =-\frac{3}{\pi}[(\cos \pi+\sin \pi)-(\cos 0+\sin 0)] \\ & =-\frac{3}{\pi}[-1-1]=\frac{6}{\pi} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_0^{\pi / 2} \sin ^4 \theta \cos ^3 \theta d \theta \\ & =\int_0^{\pi / 2} \sin ^4 \theta+\left(1-\sin ^2 \theta\right) \cos \theta d \theta \end{aligned} $

Let $\sin \theta=t$, then $\cos \theta d \theta=d t$

When $\theta \rightarrow 0, t \rightarrow 0$

$ \begin{aligned} & \text { and } \theta \rightarrow \frac{\pi}{2}, t \rightarrow 1 \\ & \qquad \begin{aligned} I & =\int_0^1 t^4\left(1-t^2\right) d t \\ & =\int_0^1\left(t^4-t^6\right) d t \\ & =\left[\frac{t^5}{5}-\frac{t^7}{7}\right]_0^1=\frac{1}{5}-\frac{1}{7} \\ & =\frac{2}{35} \end{aligned} \end{aligned} $

Let $1+x^2=t$

$2 x d x=d t$

When $x=0$ and $t=1$, when $x=1, t=2$

$ \begin{aligned} \therefore \exp \left[\int_1^2 \log t\right] & =\exp \left[[t \log t-t]_1^2\right] \\ & =\exp [2 \log 2-2+1] \\ & =\exp [\log 4-1] \\ & =e^{\log 4} \times e^{-1}=\frac{4}{e} \end{aligned} $

$ \text { } \begin{aligned}Let\,\, I & =\int_0^{2 a} f(x) d x \\ & =\int_0^a f(x) d x+\int_a^{2 a} f(x) d x \ldots \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned}Let\,\, I_1 & =\int_a^{2 a} f(x) d x \\ Let\,\,\,x & =t+a \\ d x & =d t \end{aligned} \end{aligned} $

$\,\,\,\,\,\,\,\,\,\, I_1=\int_0^a f(t+a) d t=\int_0^a f(a+x) d x $

$ \therefore \quad I=\int_0^a f(x) d x+\int_0^a f(a+x) d x $

Let $I=\int_1^2 x \sqrt{4-x^2} d x$

On putting $4-x^2=1$,

$ \begin{aligned} & -2 x d x =d t \\ \Rightarrow &\,\,\,\, x d x =-\frac{d t}{2} \end{aligned} $

When $x \rightarrow 1, t \rightarrow 3$

$ \begin{aligned} &\\ &\begin{aligned} \text { } k= & \int_{\frac{-3}{2}}^{\frac{3}{2}}[2 x-3] d x=\int_{\frac{-3}{2}}^{\frac{3}{2}}([2 x]-3) d x \\ = & \int_{-3 / 2}^{-1}(-3) d x+\int_{-1}^{-1 / 2}(-2) d x+\int_{\frac{-1}{2}}^0(-1) d x+\int_0^{1 / 2}(0) d x +\int_{\frac{1}{2}}^1 d x+\int_1^{3 / 2} 2 d x-\int_{\frac{-3}{2}}^{3 / 2} 3 d x \\ = & \frac{-3}{2}-1-\frac{1}{2}+\frac{1}{2}+1-9=\frac{-3}{2}-9 \\ \therefore & \left|k+\frac{1}{2}\right|=\left|-\frac{3}{2}-9+\frac{1}{2}\right|=10 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Given, } \int_1^4\left(x+\sqrt{x}+\frac{1}{x}\right) d x-\int_1^{2 \log 2} 1 d x \\ & =\left[\frac{x^2}{2}+\frac{2}{3} x^{\frac{3}{2}}+\ln x\right]_1^4-[x]_1^{2 \log 2} \\ & =\left(8+\frac{16}{3}+\ln 4\right)-\left(\frac{1}{2}+\frac{2}{3}+0\right) \\ & -(2 \log 2-1)=\frac{17}{2}+\frac{14}{3}=\frac{79}{6} \end{aligned} $

Let

$ I=\int_{-\pi / 4}^{\pi / 4} \frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x $

Let

$ \begin{aligned} f(x) & =\frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) \\ f(-x) & =\frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4-\sin x}{4+\sin x}\right)\right) \\ & =\frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}-\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) \end{aligned} $

$ \begin{aligned} f(x) & +f(-x)=\frac{6}{\pi(2-\cos 2 x)} \\ I & =\int_{-\pi / 4}^{\pi / 4} f(x) d x \\ & =\int_0^{\pi / 4} f(x)+f(-x) d x \\ & =\frac{6}{\pi} \int_0^{\pi / 4} \frac{1}{2-\cos 2 x} d x \\ I & =\frac{6}{\pi} \int_0^{\pi / 4} \frac{1}{2-\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right)} d x \\ & =\frac{6}{\pi} \int_0^{\pi / 4} \frac{1+\tan ^2 x}{1+3 \tan ^2 x} d x \\ & =\frac{6}{\pi} \int_0^{\pi / 4} \frac{\sec ^2 x}{1+3 \tan ^2 x} d x \end{aligned} $

$ \begin{aligned} &\text { On putting } \tan x=t\\ &\begin{aligned} & \sec ^2 x d x=d t \\ & I=\frac{6}{\pi} \int_0^1 \frac{d t}{1+(\sqrt{3} t)^2}=\frac{6}{\pi} \cdot \frac{1}{\sqrt{3}}\left[\tan ^{-1} \sqrt{3} t\right]_0^1 \\ & =\frac{2 \sqrt{3}}{\pi}\left[\tan ^{-1} \sqrt{3}-0\right] \\ & \quad I=\frac{2 \sqrt{3}}{\pi} \times \frac{\pi}{3}=\frac{2}{\sqrt{3}} \\ & \Rightarrow \quad I^2=\frac{4}{3} \\ & \quad 3 I^2=4 \end{aligned} \end{aligned} $

Given, $I=\int_\limits0^T f(x) d x$

If $f(x+T)=f(x)$

Now, $\int_\limits0^{5 T} f(2 x) d x$

On putting $2 x=y$

$\Rightarrow \quad d x=\frac{1}{2} d y$

$\frac{1}{2} \int_\limits0^{10 T} f(y) d y=\frac{10 I}{2}=5 I$

$\begin{aligned} \text { Let } I & =\int_\limits1^3 x \sqrt[n]{x^2-1} d x=6 \\ & =\frac{1}{2} \int_\limits1^3 2 x\left(x^2-1\right)^{\frac{1}{n}} d x=6 \end{aligned}$

$\begin{aligned} & \Rightarrow \quad\left[\frac{\left(n^2-1\right)^{\frac{1}{n}+1}}{\frac{1}{n}+1}\right]_1^3=12 \\ & \Rightarrow \quad \frac{(8)^{\frac{1}{n}+1}}{\frac{1}{n}+1}=12 \\ & \Rightarrow \quad \frac{(8)^{\frac{1}{n}}}{(n+1)}(n)=\frac{3}{2} \end{aligned}$

Now, on putting $n=3$, L.H.S = R. H. S

$\therefore n=3$