Definite Integration
The value of the integral
$\int\limits_{ - \pi /2}^{\pi /2} {{{dx} \over {(1 + {e^x})({{\sin }^6}x + {{\cos }^6}x)}}} $ is equal to
$\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + {{{n^2}} \over {({n^2} + 9)(n + 3)}} + \,\,....\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)$ is equal to :
The value of the integral $\int\limits_{0}^{\frac{\pi}{2}} 60 \frac{\sin (6 x)}{\sin x} d x$ is equal to _________.
Explanation:
$I = \int\limits_0^{{\pi \over 2}} {60\,.\,{{\sin 6x} \over {\sin x}}dx} $
$ = 60\,.\,2\int\limits_0^{{\pi \over 2}} {(3 - 4{{\sin }^2}x)(4{{\cos }^2}x - 3)\cos x\,dx} $
$ = 120\int\limits_0^{{\pi \over 2}} {(3 - 4{{\sin }^2}x)(1 - 4{{\sin }^2}x)\cos x\,dx} $
Let $\sin x = t \Rightarrow \cos xdx = dt$
$ = 120\int\limits_0^1 {(3 - 4{t^2})(1 - 4{t^2})dt} $
$ = 120\int\limits_0^1 {(3 - 16{t^2} + 16{t^4})dt} $
$ = 120\left[ {3t - {{16{t^3}} \over 3} + {{16{t^5}} \over 5}} \right]_0^1$
$ = 104$
If $\int\limits_{0}^{\sqrt{3}} \frac{15 x^{3}}{\sqrt{1+x^{2}+\sqrt{\left(1+x^{2}\right)^{3}}}} \mathrm{~d} x=\alpha \sqrt{2}+\beta \sqrt{3}$, where $\alpha, \beta$ are integers, then $\alpha+\beta$ is equal to __________.
Explanation:
Put $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta $
$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15{{\tan }^3}\theta \,.\,{{\sec }^2}\theta \,d\theta } \over {\sqrt {1 + {{\tan }^2}\theta + \sqrt {{{\sec }^6}\theta } } }}} $
$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15{{\tan }^2}\theta {{\sec }^2}\theta \,d\theta } \over {\sec \theta \sqrt {1 + \sec \theta } }}} $
$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15({{\sec }^2}\theta - 1)\sec \theta \tan \theta \,d\theta } \over {\left( {\sqrt {1 + \sec \theta } } \right)}}} $
Now put $1 + \sec \theta = {t^2}$
$ \Rightarrow \sec \theta \tan \theta \,d\theta = 2tdt$
$ \Rightarrow I = \int\limits_{\sqrt 2 }^{\sqrt 3 } {{{15\left( {{{({t^2} - 1)}^2} - 1} \right)2t\,dt} \over t}} $
$ \Rightarrow I = 30\int\limits_{\sqrt 2 }^{\sqrt 3 } {({t^4} - 2{t^2} + 1 - 1)\,dt} $
$ \Rightarrow I = 30\int\limits_{\sqrt 2 }^{\sqrt 3 } {({t^4} - 2{t^2})\,dt} $
$ \Rightarrow I = \left. {30\left( {{{{t^5}} \over 5} - {{2{t^3}} \over 3}} \right)} \right|_{\sqrt 2 }^{\sqrt 3 }$
$ = 30\left[ {\left( {{9 \over 5}\sqrt 3 - 2\sqrt 3 } \right) - \left( {{{4\sqrt 2 } \over 5} - {{4\sqrt 2 } \over 3}} \right)} \right]$
$ = \left( {54\sqrt 3 - 60\sqrt 3 } \right) - \left( {24\sqrt 2 - 40\sqrt 2 } \right)$
$ = 16\sqrt 2 - 6\sqrt 3 $
$\therefore$ $\alpha = 16$ and $\beta = - 6$
$\alpha + \beta = 10.$
Let $f(x)=\min \{[x-1],[x-2], \ldots,[x-10]\}$ where [t] denotes the greatest integer $\leq \mathrm{t}$. Then $\int\limits_{0}^{10} f(x) \mathrm{d} x+\int\limits_{0}^{10}(f(x))^{2} \mathrm{~d} x+\int\limits_{0}^{10}|f(x)| \mathrm{d} x$ is equal to ________________.
Explanation:
$\because$ $f(x) = \min \,\{ [x - 1],[x - 2],\,......,\,[x - 10]\} = [x - 10]$
Also $|f(x)| = \left\{ {\matrix{ { - f(x),} & {if\,x \le 10} \cr {f(x),} & {if\,x \ge 10} \cr } } \right.$
$\therefore$ $\int\limits_0^{10} {f(x)dx + \int\limits_0^{10} {{{(f(x))}^2}dx + \int\limits_0^{10} {( - f(x))dx} } } $
$ = \int\limits_0^{10} {{{(f(x))}^2}dx} $
$ = {10^2} + {9^2} + {8^2}\, + \,.....\, + \,{1^2}$
$ = {{10 \times 11 \times 21} \over 6} = 385$
Let f be a differentiable function satisfying $f(x)=\frac{2}{\sqrt{3}} \int\limits_{0}^{\sqrt{3}} f\left(\frac{\lambda^{2} x}{3}\right) \mathrm{d} \lambda, x>0$ and $f(1)=\sqrt{3}$. If $y=f(x)$ passes through the point $(\alpha, 6)$, then $\alpha$ is equal to _____________.
Explanation:
$\because$ $f(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {f\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda ,\,x > 0} $
On differentiating both sides w.r.t., x, we get
$f'(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {{{{\lambda ^2}} \over 3}f'\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda } $
$f'(x) = {1 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {\lambda \,.\,{{2\lambda } \over 3}f'\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda } $
$\therefore$ $\sqrt 3 f'(x) = \left[ {{\lambda \over x}\,.\,f\left( {{{{\lambda ^2}x} \over 3}} \right)} \right]_0^{\sqrt 3 } - \int\limits_0^{\sqrt 3 } {{1 \over x}f\left( {{{{\lambda ^2}x} \over 3}} \right)dx} $
$\sqrt 3 x\,f'(x) = \sqrt 3 f(x) - {{\sqrt 3 } \over 2}f(x)$
$xf'(x) = {{f(x)} \over 2}$
On integrating we get : $\ln y = {1 \over 2}\ln x + \ln c$
$\because$ $f(1) = \sqrt 3 $ then $c = \sqrt 3 $
$\therefore$ ($\alpha$, 6) lies on
$\therefore$ $y = \sqrt {3x} $
$\therefore$ $6 = \sqrt {3\alpha } \Rightarrow \alpha = 12$.
If $\mathrm{n}(2 \mathrm{n}+1) \int_{0}^{1}\left(1-x^{\mathrm{n}}\right)^{2 \mathrm{n}} \mathrm{d} x=1177 \int_{0}^{1}\left(1-x^{\mathrm{n}}\right)^{2 \mathrm{n}+1} \mathrm{~d} x$, then $\mathrm{n} \in \mathbf{N}$ is equal to ______________.
Explanation:
$\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = \int_0^1 {1\,.\,{{(1 - {x^n})}^{2n + 1}}dx} } $
$ = \left[ {{{(1 - {x^n})}^{2n + 1}}\,.\,x} \right]_0^1 - \int_0^1 {x\,.\,(2n + 1){{(1 - {x^n})}^{2n}}\,.\, - n{x^{n - 1}}dx} $
$ = n(2n + 1)\int_0^1 {(1 - (1 - {x^n})){{(1 - {x^n})}^{2n}}dx} $
$ = n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n}}dx - n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx} } $
$(1 + n(2n + 1))\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n}}dx} } $
$(2{n^2} + n + 1)\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = 1177\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx} } $
$\therefore$ $2{n^2} + n + 1 = 1177$
$2{n^2} + n - 1176 = 0$
$\therefore$ $n = 24$ or $ - {{49} \over 2}$
$\therefore$ $n = 24$
Let $f$ be a twice differentiable function on $\mathbb{R}$. If $f^{\prime}(0)=4$ and $f(x) + \int\limits_0^x {(x - t)f'(t)dt = \left( {{e^{2x}} + {e^{ - 2x}}} \right)\cos 2x + {2 \over a}x} $, then $(2 a+1)^{5}\, a^{2}$ is equal to _______________.
Explanation:
Here $f(0)=2 \hspace{0.5cm} ...(ii)$
On differentiating equation (i) w.r.t. $x$ we get :
$ \begin{aligned} & f^{\prime}(x)+\int_0^x f^{\prime}(t) d t+x f^{\prime}(x)-x f^{\prime}(x) \\\\ & = 2\left(e^{2 x}-e^{-2 x}\right) \cos 2 x-2\left(e^{2 x}+e^{-2 x}\right) \sin 2 x+\frac{2}{a} \\\\ & \Rightarrow \quad f(x)+f(x)-f(0)\\\\ &=2\left(e^{2 x}-e^{-2 x}\right) \cos 2 x-2\left(e^{2 x}+e^{-2 x}\right) \sin 2 x+\frac{2}{a} \\\\ & \quad \text { Replace } x \text { by } 0 \text { we get }: \\\\ & \Rightarrow \quad 4=\frac{2}{a} \Rightarrow a=\frac{1}{2} \cdot \\\\ & \therefore \quad(2 a+1)^5 \cdot a^2=2^5 \cdot \frac{1}{2^2}=2^3=8 \end{aligned} $
Let ${a_n} = \int\limits_{ - 1}^n {\left( {1 + {x \over 2} + {{{x^2}} \over 3} + \,\,.....\,\, + \,\,{{{x^{n - 1}}} \over n}} \right)dx} $ for every n $\in$ N. Then the sum of all the elements of the set {n $\in$ N : an $\in$ (2, 30)} is ____________.
Explanation:
$\because$ ${a_n} = \int\limits_{ - 1}^n {\left( {1 + {x \over 2} + {{{x^2}} \over 3}\, + \,....\, + \,{{{x^{n - 1}}} \over n}} \right)dx} $
$ = \left[ {x + {{{x^2}} \over {{2^2}}} + {{{x^3}} \over {{3^2}}}\, + \,......\, + \,{{{x^n}} \over {{n^2}}}} \right]_{ - 1}^n$
${a_n} = {{n + 1} \over {{1^2}}} + {{{n^2} - 1} \over {{2^2}}} + {{{n^3} + 1} \over {{3^2}}} + {{{n^4} - 1} \over {{4^2}}}\, + \,...\, + \,{{{n^n} + {{( - 1)}^{n + 1}}} \over {{n^2}}}$
Here, ${a_1} = 2,\,{a_2} = {{2 + 1} \over 1} + {{{2^2} - 1} \over 2} = 3 + {3 \over 2} = {9 \over 2}$
${a_3} = 4 + 2 + {{28} \over 9} = {{100} \over 9}$
${a_4} = 5 + {{15} \over 4} + {{65} \over 9} + {{255} \over {16}} > 31$.
$\therefore$ The required set is $\{ 2,3\} $. $\because$ ${a_n} \in (2,30)$
$\therefore$ Sum of elements = 5.
$ \begin{aligned} &\text { If } \lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(n k+1)+(n k+2)+\ldots+(n k+n)] \\ &=33 \cdot \lim _{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot\left[1^{k}+2^{k}+3^{k}+\ldots+n^{k}\right] \end{aligned}$, then the integral value of $\mathrm{k}$ is equal to _____________
Explanation:
$ \begin{aligned} &\Rightarrow \int_{0}^{1}(k+x) d x=33 \int_{0}^{1} x^{k} d x \\\\ &\Rightarrow \quad \frac{2 k+1}{2}=\frac{33}{k+1} \\\\ &\Rightarrow \quad k=5 \end{aligned} $
Let $f(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)dx} $. If $f(1) + f'(1) = \alpha e - {1 \over 6}$, then the value of 150$\alpha$ is equal to ___________.
Explanation:
Given,
$f(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)dx} $
$f'(t) = {e^{{t^3}}}\left( {{{{t^8}} \over {{{({t^6} + 2{t^3} + 2)}^2}}}} \right)$
$\therefore$ $f'(1) = {e^1}\left( {{1 \over {{{(1 + 2 + 2)}^2}}}} \right)$
$ = {e \over {{5^2}}}$
Now, $f(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^2} + 2)}^2}}}} \right)dx} $
Let ${x^3} = z \Rightarrow 3{x^2}dx = dz$
$ = \int\limits_0^{{t^3}} {{e^z}\left( {{{{x^6}\,.\,{x^2}dx} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)} $
$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{{{z^2}dz} \over {{{({z^2} + 2z + 2)}^2}}}} \right)} $
$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{{{z^2} + 2z + 2 - 2z - 2} \over {{{({z^2} + 2z + 2)}^2}}}} \right)} dz$
$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{{{z^2} + 2z + 2} \over {{{({z^2} + 2z + 2)}^2}}} - {{2z + 2} \over {{{({z^2} + 2z + 2)}^2}}}} \right)dz} $
$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{1 \over {{z^2} + 2z + 2}} - {{2z + 2} \over {{{({z^2} + 2z + 2)}^2}}}} \right)dz} $
$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {f(z) + f'(z)} \right)dz} $
$ = {1 \over 3}\left[ {{e^z}f(z)} \right]_0^{{t^3}}$
$ = {1 \over 3}\left[ {{e^z} \times {1 \over {{z^2} + 2z + 2}}} \right]_0^{{t^3}}$
$ = {1 \over 3}\left[ {{e^{{t^3}}} \times {1 \over {{t^6} + 2{t^3} + 2}} - {1 \over 2}} \right]$
$\therefore$ $f(t) = {1 \over 3}\left[ {{{{e^{{t^3}}}} \over {{t^6} + 2{t^3} + 2}} - {1 \over 2}} \right]$
So, $f(1) = {1 \over 3}\left[ {{e \over {1 + 2 + 2}} - {1 \over 2}} \right]$
$ = {1 \over 3}\left[ {{e \over 5} - {1 \over 2}} \right]$
Given, $f(1) + f'(1) = \alpha e - {1 \over 6}$
$ \Rightarrow {1 \over 3}\left[ {{e \over 5} - {1 \over 2}} \right] + {e \over {25}} = \alpha e - {1 \over 6}$
$ \Rightarrow {e \over {15}} - {1 \over 6} + {e \over {25}} = \alpha e - {1 \over 6}$
$ \Rightarrow {e \over {15}} + {e \over {25}} = \alpha e$
$ \Rightarrow {{10e + 6e} \over {150}} = \alpha e$
$ \Rightarrow {{16e} \over {150}} = \alpha e$
$ \Rightarrow \alpha = {{16} \over {150}}$
$\therefore$ $150\alpha = 150 \times {{16} \over {150}} = 16$
The integral ${{24} \over \pi }\int_0^{\sqrt 2 } {{{(2 - {x^2})dx} \over {(2 + {x^2})\sqrt {4 + {x^4}} }}} $ is equal to ____________.
Explanation:
$I = {{24} \over \pi }\int_0^{\sqrt 2 } {{{2 - {x^2}} \over {(2 + {x^2})\sqrt {4 + {x^4}} }}dx} $
Let $x = \sqrt 2 t \Rightarrow dx = \sqrt 2 dt$
$I = {{24} \over \pi }\int_0^1 {{{(2 - 2{t^2})\,.\,\sqrt 2 dt} \over {(2 + 2{t^2})\sqrt {4 + 4{t^4}} }}} $
$ = {{12\sqrt 2 } \over \pi }\int_0^1 {{{\left( {{1 \over {{t^2}}} - 1} \right)dt} \over {\left( {t + {1 \over t}} \right)\sqrt {{{\left( {t + {1 \over t}} \right)}^2} - 2} }}} $
Let $t + {1 \over t} = u$
$ \Rightarrow \left( {1 - {1 \over {{t^2}}}} \right)dt = du$
$ = {{12\sqrt 2 } \over \pi }\int_\infty ^2 {{{ - du} \over {u\sqrt {{4^2} - 2} }}} $
$ = {{12\sqrt 2 } \over \pi }\int_2^\infty {{{du} \over {{u^2}\sqrt { - {{\left( {{{\sqrt 2 } \over u}} \right)}^2}} }}} $
$ = {{12\sqrt 2 } \over \pi }\int_{{1 \over {\sqrt 2 }}}^0 {{{ - {1 \over {\sqrt 2 }}dp} \over {\sqrt {1 - {p^2}} }}} $
$ = {{12} \over \pi }\left[ {{{\sin }^{ - 1}}p} \right]_0^{{1 \over {\sqrt 2 }}}$
$ = {{12} \over \pi }\,.\,{\pi \over 4} = 3$
Let f(x) = max {|x + 1|, |x + 2|, ....., |x + 5|}. Then $\int\limits_{ - 6}^0 {f(x)dx} $ is equal to __________.
Explanation:
For $\left| {x + 1} \right|$ critical point, x + 1 = 0 $\Rightarrow$ x = $-$1
For $\left| {x + 2} \right|$ critical point, x + 2 = 0 $\Rightarrow$ x = $-$2
For $\left| {x + 3} \right|$ critical point, x + 3 = 0 $\Rightarrow$ x = $-$3
For $\left| {x + 4} \right|$ critical point, x + 4 = 0 $\Rightarrow$ x = $-$4
For $\left| {x + 5} \right|$ critical point, x + 5 = 0 $\Rightarrow$ x = $-$5
Here maximum function is represent by the dotted line.
$\therefore$ Point of intersection A of line y = $-$x $-$1 and y = x + 5 :
$ - x - 1 = x + 5$
$ \Rightarrow 2x = - 6$
$ \Rightarrow x = - 3$
$\therefore$ $y = - 3 + 5 = 2$
$\therefore$ Point $A = ( - 3,2)$
$\therefore$ $\int_{ - 6}^0 {f(x)dx} $
$ = \int_{ - 6}^{ - 3} {( - x - 1)dx + \int_{ - 3}^0 {(x + 5)dx} } $
$ = \left( { - {{{x^2}} \over 2} - x} \right)_{ - 6}^{ - 3} + \left[ {{{{x^2}} \over 2} + 5x} \right]_{ - 3}^0$
$ = \left[ {\left( { - {9 \over 2} + 3} \right) - \left( { - {{36} \over 2} + 6} \right)} \right] + \left[ {0 - \left( {{9 \over 2} - 15} \right)} \right]$
$ = \left( { - {3 \over 2} + 12} \right) + {{21} \over 2}$
$ = + {{21} \over 2} + {{21} \over 2}$
$ = 21$
The value of the integral
${{48} \over {{\pi ^4}}}\int\limits_0^\pi {\left( {{{3\pi {x^2}} \over 2} - {x^3}} \right){{\sin x} \over {1 + {{\cos }^2}x}}dx} $ is equal to __________.
Explanation:
$I = {{48} \over {{\pi ^4}}}\int_0^\pi {\left[ {{{\left( {{\pi \over 2} - x} \right)}^3} - {{3{\pi ^2}} \over 4}\left( {{\pi \over 2} - x} \right) + {{{\pi ^3}} \over 4}} \right]{{\sin xdx} \over {1 + {{\cos }^2}x}}} $
Using $\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx} } $
$I = {{48} \over {{\pi ^4}}}\int_0^\pi {\left[ { - {{\left( {{\pi \over 2} - x} \right)}^3} + {{3{\pi ^4}} \over 4}\left( {{\pi \over 2} - x} \right) + {{{\pi ^3}} \over 4}} \right]{{\sin xdx} \over {1 + {{\cos }^2}x}}} $
Adding these two equations, we get
$2I = {{48} \over {{\pi ^4}}}\int_0^\pi {{{{\pi ^3}} \over 2}\,.\,{{\sin xdx} \over {1 + {{\cos }^2}x}}} $
$ \Rightarrow I = {{12} \over \pi }\left[ { - {{\tan }^{ - 1}}(\cos x)} \right]_0^\pi = {{12} \over \pi }\,.\,{\pi \over 2} = 6$
The value of b > 3 for which $12\int\limits_3^b {{1 \over {({x^2} - 1)({x^2} - 4)}}dx = {{\log }_e}\left( {{{49} \over {40}}} \right)} $, is equal to ___________.
Explanation:
$I = \int {{1 \over {({x^2} - 1)({x^2} - 4)}}dx = {1 \over 3}\int {\left( {{1 \over {{x^2} - 4}} - {1 \over {{x^2} - 1}}} \right)dx} } $
$ = {1 \over 3}\left( {{1 \over 4}\ln \left| {{{x - 2} \over {x + 2}}} \right| - {1 \over 2}\ln \left| {{{x - 1} \over {x + 1}}} \right|} \right) + C$
$12I = \ln \left| {{{x - 2} \over {x + 2}}} \right| + 2\ln \left| {{{x - 1} \over {x + 1}}} \right| + C$
$12\int\limits_3^b {{{dx} \over {({x^2} - 4)({x^2} - 1)}}} $
$ = \ln \left( {{{b - 2} \over {b + 2}}} \right) - 2\ln \left( {{{b - 1} \over {b + 1}}} \right) - \left( {\ln \left( {{1 \over 5}} \right) - 2\ln \left( {{1 \over 2}} \right)} \right)$
$ = \ln \left( {\left( {{{b - 2} \over {b + 2}}} \right)\,.\,{{{{(b + 1)}^2}} \over {{{(b - 1)}^2}}}} \right) - \left( {\ln {4 \over 5}} \right)$
So, ${{49} \over {40}} = {{(b - 2)} \over {(b + 2)}}{{{{(b + 1)}^2}} \over {{{(b - 1)}^2}}}\,.\,{5 \over 4}$
$ \Rightarrow b = 6$
Let $f(\theta ) = \sin \theta + \int\limits_{ - \pi /2}^{\pi /2} {(\sin \theta + t\cos \theta )f(t)dt} $. Then the value of $\left| {\int_0^{\pi /2} {f(\theta )d\theta } } \right|$ is _____________.
Explanation:
Clearly $f(\theta)=a \sin \theta+b \cos \theta$
Where $a=1+\int_{-\pi / 2}^{\pi / 2}(a \sin t+b \cos t) d t \Rightarrow a=1+2 b\quad\quad...(i)$
and $b=\int_{-\pi / 2}^{\pi / 2}(a t \sin t+b t \cos t) d t \Rightarrow b=2 a\quad\quad...(ii)$
from (i) and (ii) we get
$ a=-\frac{1}{3} \text { and } b=-\frac{2}{3} $
So $f(\theta)=-\frac{1}{3}(\sin \theta+2 \cos \theta)$
$ \Rightarrow\left|\int_{0}^{\pi / 2} f(\theta) d \theta\right|=\frac{1}{3}(1+2 \times 1)=1 $
Let $\mathop {Max}\limits_{0\, \le x\, \le 2} \left\{ {{{9 - {x^2}} \over {5 - x}}} \right\} = \alpha $ and $\mathop {Min}\limits_{0\, \le x\, \le 2} \left\{ {{{9 - {x^2}} \over {5 - x}}} \right\} = \beta $.
If $\int\limits_{\beta - {8 \over 3}}^{2\alpha - 1} {Max\left\{ {{{9 - {x^2}} \over {5 - x}},x} \right\}dx = {\alpha _1} + {\alpha _2}{{\log }_e}\left( {{8 \over {15}}} \right)} $ then ${\alpha _1} + {\alpha _2}$ is equal to _____________.
Explanation:
So, $\alpha=f(1)=2$ and $\beta=\min (f(0), f(2))=\frac{5}{3}$
Now, $\int_{-1}^{3} \max \left\{\frac{x^{2}-9}{x-5}, x\right\} d x=\int_{-1}^{9 / 5} \frac{x^{2}-9}{x-5} d x+\int_{9 / 5}^{3} x d x$
$ =\int_{-1}^{9 / 5}\left(x+5+\frac{16}{x-5}\right) d x+\left.\frac{x^{2}}{2}\right|_{9 / 5} ^{3} $
$ =\frac{28}{25}+14+16 \ln \left(\frac{8}{15}\right)+\frac{72}{25}=18+16 \ln \left(\frac{8}{15}\right) $
Clearly $\alpha_{1}=18$ and $\alpha_{2}=16$, so $\alpha_{1}+\alpha_{2}=34$.
$f(x) = x + \int\limits_0^{\pi /2} {\sin x.\cos y\,f(y)\,dy} $, is :
${\pi ^2}\int\limits_0^2 {\left( {\sin {{\pi x} \over 2}} \right)(x - [x]} {)^{[x]}}dx$ is equal to :
$\int\limits_0^5 {{{x + [x]} \over {{e^{x - [x]}}}}dx = \alpha {e^{ - 1}} + \beta } $, where $\alpha$, $\beta$ $\in$ R, 5$\alpha$ + 6$\beta$ = 0, and [x] denotes the greatest integer less than or equal to x; then the value of ($\alpha$ + $\beta$)2 is equal to :
$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{{{n^2}} \over {{n^2} + 4{r^2}}}} $ is :
$\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{dx} \over {(1 + {e^{x\cos x}})({{\sin }^4}x + {{\cos }^4}x)}}} $ is equal to :
integral $\int\limits_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx} $ is :
where [x] is the greatest integer less than or equal to x. Which of the following is true?
$\int_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}} dx = \alpha {e^{ - 1}} + \beta {e^{ - {1 \over 2}}} + \gamma $, where $\alpha$, $\beta$, $\gamma$ are integers and [x] denotes the greatest integer less than or equal to x, then the value of $\alpha$ + $\beta$ + $\gamma$ is equal to :
$g(\alpha ) = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{{{\sin }^\alpha }x} \over {{{\cos }^\alpha }x + {{\sin }^\alpha }x}}dx} $
$I = \int_0^{10} {{{[x]{e^{[x]}}} \over {{e^{x - 1}}}}dx} $,
where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :