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CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 10 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Definite Integrals
Quiz 10
Let $$\displaystyle u = \overset{\infty}{\underset{0}{\int}} \dfrac{dx}{x^4 + 7x^2 + 1} \& v = \overset{\infty}{\underset{0}{\int}} \dfrac{x^2 dx}{x^4 + 7x^2 + 1}$$ then:
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v > u
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6v = $$\displaystyle \pi $$
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$$\displaystyle 3u+2v=5\pi /6$$
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$$\displaystyle u+v=\pi /3$$
Explanation
$$\displaystyle v=\int_{0}^{\infty }\frac{x^{2}dx}{x^{4}+7x^{2}+1}$$ put $$\displaystyle x=\frac{1}{t}\Rightarrow dx=\frac{-1}{t^{2}}dt$$
$$\displaystyle v=-\int_{\infty }^{0}\frac{\frac{1}{t^{2}}\cdot \frac{1}{t^{2}}dt}{\frac{1}{t^{4}}+\frac{1}{t^{2}}+1}=\int_{0}^{\infty }\frac{dx}{x^{4}+7x^{2}+1}$$
v=u hence $$\displaystyle 2u=\int_{0}^{\infty }\left ( \frac{x^{2}+1}{x^{4}+7x^{2}+1} \right )dx$$
$$\displaystyle =\int_{0}^{\infty }\left ( \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+7} \right )dx=\int_{0}^{\infty }\frac{d\left ( x-\frac{1}{x} \right )}{\left ( x-\frac{1}{x} \right )^{2}+3^{2}}=\int_{0}^{\infty }\frac{dt}{t^{2}+9}$$
$$\displaystyle \frac{2}{3}\left [ \tan^{-1}\frac{t}{3} \right ]_{0}^{\infty }$$
$$\displaystyle 2u=\pi /3$$
If $$\displaystyle I=\int _{8}^{15} \frac{dx}{(x-3)\sqrt{x+1} }$$ then$$ I$$ equals
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$$ \displaystyle \frac{1}{2}\log \frac {5}{3} $$
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$$\displaystyle 2 \log \frac{1}{3}$$
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$$ \displaystyle \frac{1}{2}-\log \frac {1}{5} $$
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$$\displaystyle 2 \log \frac{5}{3}$$
Explanation
Put $$\sqrt{x+1} =t $$ or $$ x + 1 =t^{2} $$
$$\displaystyle
\therefore I=\int _{3}^{4} \frac{2t}{(t^{2}-4) t}
dt=\frac{2}{(2)(2)}\log \left. \left | \frac{t-2}{t+2}\right |\right
]_{3}^{4}$$
$$\displaystyle =\frac{1}{2} \left[ \log \frac{1}{3}-\log \frac{1}{5}\right] $$
$$\displaystyle =\frac{1}{2}\log \frac{5}{3}$$
37 If $$ n > 1,$$ and $$ \displaystyle I=\int _{0}^{\infty} \frac{dx}{(x+\sqrt{1+x^{2}})^{n}}$$ then $$ I$$ equals
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$$ \displaystyle \frac{n}{n^{2}-1}$$
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$$ \displaystyle \frac{2n}{n^{2}-1}$$
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$$ \displaystyle \frac{n}{2(n^{2}-1)}$$
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$$ \displaystyle \sqrt{ n^{2}-1}$$
Explanation
As $$x $$ and $$\sqrt{1+x^{2}}$$ are involved,
put $$ \sqrt{1-x^{2}} = t - x$$ or $$1 + x^{2}=t^{2} -2tx + x^{2}$$
$$\displaystyle
\Rightarrow x= \frac{1}{2}\left(t-\frac{1}{t}\right) $$
$$\displaystyle \Rightarrow dx = \frac{1}{2} \left
(1+\frac{1}{t^{2}}\right )$$
$$ \displaystyle \therefore I= \int_{0}^{\infty} \frac{1}{2} \left(1+\frac{1}{t^{2}}\right) \frac{dt}{t^{n}}$$
$$=\displaystyle \frac{1}{2} \left.\left( \frac{t^{-n+1}}{1-n}-\frac{t^{-n-1}}{n+1}\right) \right ]_{1}^{\infty} $$
$$=\displaystyle
\frac{1}{2}\left. \left ( \frac{1}{1-n}\frac{1}
{t^{n-1}}-\frac{1}{n+1}\frac{1}{t^{n+1}} \right) \right ]_{1}^{\infty}
$$
$$ =\displaystyle 0+\frac{1}{2} \left( \frac{1}{n-1}+\frac{1}{n+1}\right) =\frac{n}{n^{2}-1} $$
A function $$f $$ is defined by $$\displaystyle f(x)=\frac{1}{2^{r-1}},\frac{1}{2r}<x\leq \frac{1}{2^{r-1}},r=1,2,3,.....$$ then the value of $$ \displaystyle \int _{0}^{1}f(x)dx $$ is equal
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$$\cfrac 13$$
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$$\cfrac 14$$
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$$\cfrac 23$$
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$$\cfrac 12$$
Explanation
$$\displaystyle \int_{0}^{1}f(x) dx =\sum_{r=1}^{\infty} \int_{2^{-r}}^{2^{-(r-1)}} \frac{1}{2^{r-1}}dx $$
$$\displaystyle =\sum_{r=1}^{\infty}\frac{1}{2^{r-1}}[2^{-(r-1)}-2^{-r}]$$
$$\displaystyle =\sum_{1}^{\infty}2^{-2(r-1)}- \sum _{1}^{\infty} 2^{-2r+1}$$
$$\displaystyle =(2^{2}-2)\sum_{1}^{\infty}2^{-2r}=2\cdot \frac{1}{4}\cdot \frac{1}{1-1/4}=\frac{2}{3}\cdot $$
If $$\displaystyle \int_{0}^{\alpha}\frac{dx}{1-\cos \alpha \cos x}=\frac{A}{\sin \alpha}+B(a\neq 0)$$
Then possible values of $$ A$$ and $$B$$ are
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$$\displaystyle A=\frac{\pi}{2},B=0$$
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$$\displaystyle A=\frac{\pi}{4},B=\frac{\pi}{4\sin \alpha} $$
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$$\displaystyle A=\frac{\pi}{6},B=\frac{\pi}{\sin \alpha} $$
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$$\displaystyle A=\pi ,B=\frac{\pi}{\sin \alpha} $$
Explanation
Let $$ \displaystyle I=\int _{ 0 }^{ \alpha }{ \frac { dx }{ 1-\cos { \alpha } \cos { \alpha } } } $$
$$ \displaystyle =\int _{ 0 }^{ \alpha }{ \frac { dx }{ \left( \cos ^{ 2 }{ \frac { x }{ 2 } } +\sin ^{ 2 }{ \frac { x }{ 2 } } \right) -\cos { \alpha } \left( \cos ^{ 2 }{ \frac { x }{ 2 } } -\sin ^{ 2 }{ \frac { x }{ 2 } } \right) } } $$
$$ \displaystyle =\int _{ 0 }^{ \alpha }{ \frac { dx }{ \left( 1-\cos { \alpha } \right) \cos ^{ 2 }{ \frac { x }{ 2 } } +\left( 1+\cos { \alpha } \right) \sin ^{ 2 }{ \frac { x }{ 2 } } } } $$
$$ \displaystyle =\int _{ 0 }^{ \alpha }{ \frac { dx }{ 2\sin ^{ 2 }{ \frac { \alpha }{ 2 } } \cos ^{ 2 }{ \frac { x }{ 2 } } +2\cos ^{ 2 }{ \frac { \alpha }{ 2 } } \sin ^{ 2 }{ \frac { x }{ 2 } } } } $$
$$ \displaystyle =\frac { 1 }{ 2 } \int _{ 0 }^{ \alpha }{ \frac { \sec ^{ 2 }{ \frac { \alpha }{ 2 } } \sec ^{ 2 }{ \frac { x }{ 2 } } }{ \tan ^{ 2 }{ \frac { \alpha }{ 2 } } +\tan ^{ 2 }{ \frac { x }{ 2 } } } dx } $$
put $$ \displaystyle \tan { \frac { x }{ 2 } } =t\Rightarrow \frac { 1 }{ 2 } \sec ^{ 2 }{ \frac { x }{ 2 } } dx=dt$$
$$ \displaystyle \therefore I=\int _{ 0 }^{ \tan { \frac { \alpha }{ 2 } } }{ \frac { \sec ^{ 2 }{ \frac { \alpha }{ 2 } } }{ { t }^{ 2 }+\tan ^{ 2 }{ \frac { \alpha }{ 2 } } } dt } $$
$$ \displaystyle =\sec ^{ 2 }{ \frac { \alpha }{ 2 } } \cot { \frac { \alpha }{ 2 } } \left[ \tan ^{ -1 }{ \left( \frac { t }{ \tan { \frac { \alpha }{ 2 } } } \right) } \right] _{ 0 }^{ \tan { \frac { \alpha }{ 2 } } }$$
$$ \displaystyle =\frac { 2 }{ \sin { \alpha } } .\frac { \pi }{ 4 } =\frac { \pi }{ 2\sin { \alpha } } $$
$$ \displaystyle A=\frac { \pi }{ 2 } ,B=0$$ or $$ \displaystyle A=\frac { \pi }{ 4 } ,B=\frac { \pi }{ 4\sin { \alpha } } $$
$$\displaystyle \int_{e^{e^{e}}}^{e^{e^{e^{e}}}}\frac{dx}{xlnx\cdot ln\left ( lnx \right )\cdot ln\left ( ln\left ( lnx \right ) \right )}$$ equals
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1
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1/e
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e-1
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1+e
Explanation
Let $$\displaystyle I=\int _{ { e }_{ { e }_{ e } } }^{ { e }^{ { e }^{ { e }^{ e } } } }{ \frac { 1 }{ x\log { x } .\log { \left( \log { x } \right) . } \log { \left( \log { \left( \log { x } \right) } \right) } } } dx$$
Substitute
$$\displaystyle \log { \left( \log { x } \right) } =t\Rightarrow \frac { 1 }{ x\log { x } } dx=dt$$
$$\displaystyle I=\int _{ e }^{ { { e }^{ e } } }{ \frac { 1 }{ t\log { t } } dt } =\left[ \log { \log { t } } \right] _{ e }^{ { e }^{ e } }=\left[ 1-0 \right] =1$$
If $$f\left( x \right) =\int _{ 0 }^{ 1 }{ \left( xf\left( t \right) +1 \right) dt,then\int _{ 0 }^{ 3 }{ f\left( x \right) dx=12 } } $$
because
Statement-2: f(x) = 3x + 1
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Statements-1 is true, statements-2 is true and statements-2 is correct explanation for statement-1.
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Statements-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statements-1.
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Statements-1 is true, statements-2 is false.
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Statements-1 is false, statements-2 is true.
Value of $$\displaystyle \int_{0}^{\pi /2}\displaystyle \frac{\sin 4\Theta }{\sin \Theta }\: d\Theta $$ is
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$$1/3$$
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$$2/3$$
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$$1$$
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$$4/3$$
Explanation
$$\displaystyle I=\int _{ 0 }^{ \pi /2 }{ \frac { \sin { 4\Theta } }{ \sin { \Theta } } d\Theta } =\int _{ 0 }^{ \pi /2 }{ \sin { 4\Theta } \csc { \Theta } d\Theta } $$
$$=\int _{ 0 }^{ \pi /2 }{ \left( 2\cos ^{ 3 }{ \Theta } +2\cos { \Theta } -6\sin ^{ 2 }{ \Theta } \cos { \Theta } \right) d\Theta } $$
$$\displaystyle ={ \left[ -2\sin ^{ 3 }{ \Theta } +\frac { 10\sin { \Theta } }{ 3 } +\frac { 2 }{ 3 } \sin { \Theta } \cos ^{ 2 }{ \Theta } \right] }_{ 0 }^{ \pi /2 }=\frac { 4 }{ 3 } $$
Evaluate $$\displaystyle \int_{0}^{1}\left ( tx+1-x \right )^{n}dx,$$ where n is a positive integer and t is a parameter independent of x. Hence $$\displaystyle \int_{0}^{1}x^{k}\left ( 1-x \right )^{n-k}dx=\frac{P}{\left [ ^{n}C_{k}\left ( n+1 \right ) \right ]}for\:k=0,1,......n$$, then $$P=$$
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2
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1
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3
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None of these
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ 1 }{ { \left( tx+1-x \right) }^{ n } } dx=\int _{ 0 }^{ 1 }{ { \left\{ \left( t-1 \right) x+1 \right\} }^{ n } } dx=\left[ \frac { \left( t-1 \right) { x+1 }^{ n+1 } }{ \left( n+1 \right) \left( t-1 \right) } \right] _{ 0 }^{ 1 }$$
$$\displaystyle =\frac { 1 }{ n+1 } \left( \frac { { t }^{ n+1 }-1 }{ t-1 } \right) =\frac { 1 }{ n+1 } \left( 1+t+{ t }^{ 2 }+...{ t }^{ n } \right) $$ ...(1)
Again $$\displaystyle\int _{ 0 }^{ 1 }{ { \left( tx+1-x \right) }^{ n } } dx=\int _{ 0 }^{ 1 }{ { \left[ \left( 1-x \right) +tx \right] }^{ n } } dx$$
$$\displaystyle=\int _{ 0 }^{ 1 }{ \left[ _{ }^{ n }{ { C }_{ 0 } }{ \left( 1-x \right) }^{ n }+^{ n }{ { C }_{ 1 } }{ \left( 1-x \right) }^{ n-1 }\left( tx \right) +^{ n }{ { C }_{ 2 } }{ \left( 1-x \right) }^{ n-2 }{ \left( tx \right) }^{ 2 }+...+^{ n }{ { C }_{ n } }{ \left( tx \right) }^{ n } \right] } dx$$
$$\displaystyle=\int _{ 0 }^{ 1 }{ \sum _{ r=1 }^{ n }{ ^{ n }{ { C }_{ r } } } } { \left( 1-x \right) }^{ n-r }{ \left( tx \right) }^{ r }dx=\sum _{ r=1 }^{ n }{ ^{ n }{ { C }_{ r } } } \left[ \int _{ 0 }^{ 1 }{ { \left( 1-x \right) }^{ n-r }{ x }^{ r }dx } \right] { t }^{ r }$$ ...(2)
From (1) and (2)
$$\displaystyle \sum _{ r=1 }^{ n }{ ^{ n }{ { C }_{ r } } } \left[ \int _{ 0 }^{ 1 }{ { \left( 1-x \right) }^{ n-r }.{ x }^{ r }dx } \right] { t }^{ r }=\frac { 1 }{ n+1 } \left( 1+t+...{ t }^{ n } \right) $$
On equating coefficient of $${ t }^{ k }$$ on both sides , we get
$$\displaystyle ^{ n }{ { C }_{ k } }\left[ \int _{ 0 }^{ 1 }{ { \left( 1-x \right) }^{ n-r },{ x }^{ r } } dx \right] \Rightarrow \int _{ 0 }^{ 1 }{ { \left( 1-x \right) }^{ n-k } } { x }^{ h }dx=\frac { 1 }{ { \left( n+1 \right) }^{ n }{ { C }_{ k } } } $$
For x > 0, let f(x) = $$\displaystyle \int_{1}^{x}{ \frac{log t}{1 + t}} dt $$. Then f(x) + f$$\displaystyle \left( \frac{1}{x} \right)$$ is equal to:
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$$\displaystyle \frac{1}{4} log x^2$$
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log x
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$$\displaystyle \frac{1}{2} (log x)^2$$
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$$\displaystyle \frac{1}{4} (log x)^2$$
Explanation
$$f(x)=\int _{ 1 }^{ x }{ \cfrac { \log { t } }{ 1+{ t } } } $$ $$--(1)$$
$$f\left( \cfrac { 1 }{ x } \right) =\int _{ 1 }^{ 1/x }{ \cfrac { \log { t } }{ 1+{ t } } } $$
Put t=1/u
$$f\left( \cfrac { 1 }{ x } \right) =\displaystyle\int _{ 1 }^{ x }{ \cfrac { -u.\log { u } }{ (1+{ u) } } \cfrac { -du }{ { u }^{ 2 } } } $$
$$f\left( \cfrac { 1 }{ x } \right) =\displaystyle \int _{ 1 }^{ x }{ \cfrac { \log { u } }{ u(1+{ u) } } du } $$ $$--(2)$$
Add (1) and (2)
$$=\displaystyle \int _{ 1 }^{ x }{ \cfrac { \log { u } }{ u } du }$$
$$z=logu\\ dz=\cfrac { du }{ u } $$
$$\int _{ 1 }^{ x }{ zdz } $$
$$={ \cfrac { 1 }{ 2 } (\log { x } })^{ 2 }$$
The value of $$\displaystyle\int _{ 0 }^{ { 1 }/{ \sqrt { 2 } } }{ \dfrac { \sin ^{ -1 }{ x } }{ { \left( 1-{ x }^{ 2 } \right) }^{ { 3 }/{ 2 } } } dx } $$ is
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$$\dfrac { \pi }{ 2 } -\log { 2 } $$
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$$\dfrac { \pi }{ 4 } -\dfrac { 1 }{ 2 } \log { 2 } $$
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$$\dfrac { \pi }{ 2 } +\dfrac { 1 }{ 2 } \log { 2 } $$
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$$\pi -\dfrac { 1 }{ 2 } \log { 2 } $$
Explanation
Let $$l=\displaystyle\int _{ 0 }^{ { 1 }/{ \sqrt { 2 } } }{ \dfrac { \sin ^{ -1 }{ x } }{ { \left( 1-{ x }^{ 2 } \right) }^{ { 3 }/{ 2 } } } dx }$$
When $$x=0$$, then $$\sin { t } =0=\sin { 0 } \Rightarrow t=0$$
When $$x=\dfrac { 1 }{ \sqrt { 2 } }$$, then $$\sin { t } =\dfrac { 1 }{ \sqrt { 2 } } =\sin { \dfrac { \pi }{ 4 } }$$
$$ \Rightarrow t=\dfrac { \pi }{ 4 }$$
Put $$ x=\sin { t } dx=\cos { t } dt$$
$$\therefore l=\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ \dfrac { t }{ \cos ^{ 3 }{ t } } \cdot \cos { t } dt } =\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ t\sec ^{ 2 }{ t } dt }$$
$$ ={ \left[ t\cdot \tan { t } \right] }_{ 0 }^{ { \pi }/{ 4 } }-\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ \tan { t } dt }$$
$$ =\dfrac { \pi }{ 4 } -{ \left[ \log { \sec { t } } \right] }_{ 0 }^{ { \pi }/{ 4 } }$$
$$=\dfrac { \pi }{ 4 } -\log { \sqrt { 2 } } -0=\dfrac { \pi }{ 4 } -\log { { 2 }^{ { 1 }/{ 2 } } } =\dfrac { \pi }{ 4 } -\dfrac { 1 }{ 2 } \log { 2 } $$
Evaluate $$\displaystyle \int_{0}^{\pi /4}\frac{\cos x-\sin x}{10+\sin 2x}dx$$
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$$\displaystyle \frac{1}{3}\left (\tan^{-1} \frac{\sqrt{2}}{3}+\tan^{-1} \frac{1}{3} \right )$$
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$$\displaystyle \frac{1}{3}\left (\tan^{-1} \frac{\sqrt{1}}{3}-\cot^{-1} \frac{2}{3} \right )$$
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$$\displaystyle \frac{1}{3}\left ( \tan^{-1} \frac{\sqrt{2}}{3}-\tan^{-1} \frac{1}{3} \right )$$
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$$\displaystyle \frac{1}{3}\left ( \tan^{-1} \frac{\sqrt{1}}{3}-\cot^{-1} \frac{1}{3} \right )$$
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { \cos { x } -\sin { x } }{ 10+\sin { 2x } } } dx$$
Put $$\\ \cos { x } +\sin { x } =t\Rightarrow \left( -\sin { x } +\cos { x } \right) dx=dt$$
$$\displaystyle I=\int _{ 1 }^{ \sqrt { 2 } }{ \frac { 1 }{ 10+\left( { t }^{ 2 }-1 \right) } dt } =\int _{ 1 }^{ \sqrt { 2 } }{ \frac { 1 }{ { t }^{ 2 }+9 } } dt$$
$$\displaystyle =\frac { 1 }{ 3 } \left[ \tan ^{ -1 }{ \frac { t }{ 3 } } \right] _{ 1 }^{ \sqrt { 2 } }=\frac { 1 }{ 3 } \left( \tan ^{ -1 }{ \frac { \sqrt { 2 } }{ 3 } -\tan ^{ -1 }{ \frac { 1 }{ 3 } } } \right) $$
The value of $$\displaystyle \int_{3}^{4}\sqrt {(4 - x)(x - 3)}dx$$ is
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$$\dfrac {\pi}{16}$$
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$$\dfrac {\pi}{8}$$
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$$\dfrac {\pi}{4}$$
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$$\dfrac {\pi}{2}$$
Explanation
Let
$$\displaystyle I = \int_{3}^{4} \sqrt {(4 - x)(x - 3)}\ dx$$
$$\displaystyle = \int_{3}^{4} \sqrt {-x^{2} + 7x - 12}\ dx$$
$$\displaystyle = \int_{3}^{4} \sqrt {-\left (x^{2} - 7x + \dfrac {49}{4} - \dfrac {49}{4}\right ) - 12}\ dx$$
$$\displaystyle = \int_{3}^{4} \sqrt {- \left (x - \dfrac {7}{2}\right )^{2} + \dfrac {49}{4} - 12}\ dx$$
$$\displaystyle = \int_{3}^{4} \sqrt {\dfrac {1}{4} - \left (x - \dfrac {7}{2}\right )^{2}}\ dx$$
Let
$$t = x - \dfrac {7}{2}\rightarrow dt = dx$$
$$\therefore$$ Upper limit $$= \dfrac {1}{2}$$ and lower limit $$= -\dfrac {1}{2}$$
$$\displaystyle = \int_{-1/2}^{1/2} \sqrt {\left (\dfrac {1}{2}\right )^{2} - t^{2}}dt$$
$$\displaystyle = 2\int_{0}^{1/2} \sqrt {\left (\dfrac {1}{2}\right )^{2} - t^{2}}dx$$
$$\displaystyle = 2\left [\dfrac {t}{2}\sqrt {\dfrac {1}{4} - t^{2}} + \dfrac {1}{8} \sin^{-1} 2t\right ]_{0}^{\frac12}$$
$$= 2\left [0 + \dfrac {1}{8}\times \dfrac {\pi}{2}\right ]$$
$$= \dfrac {\pi}{8}$$
The value of the integral $$\displaystyle \int_{\frac {1}{3}}^1\frac {(x-x^3)^{\frac {1}{3}}}{x^4}dx$$
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$$6$$
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$$0$$
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$$3$$
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$$4$$
Explanation
Let $$\displaystyle I=\int_{ {1}/{3}}^1\frac {(x-x^3)^{\frac {1}{3}}}{x^4}dx$$
Put $$x=\sin\theta $$
$$\Rightarrow dx=\cos\theta \, d\theta $$
When $$x=\cfrac {1}{3},\theta =\sin^{-1}\left (\cfrac {1}{3}\right )$$ and when $$x=1, \theta =\cfrac {\pi}{2}$$
$$\Rightarrow \displaystyle I=\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\sin\theta-\sin^3\theta)^{\frac {1}{3}}}{\sin^4\theta}cos \theta \,d\theta$$
$$\displaystyle =\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\sin\theta)^{\frac {1}{3}}(1-\sin^2\theta)^{\frac {1}{3}}}{\sin^4\theta}\cos \theta \, d\theta$$
$$\displaystyle =\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\sin\theta)^{\frac {1}{3}}(\cos\theta)^{\frac {2}{3}}}{\sin^4\theta}\cos \theta \, d\theta$$
$$\displaystyle =\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\sin\theta)^{\frac {1}{3}}(\cos\theta)^{\frac {2}{3}}}{\sin^2\theta \sin^2\theta}\cos \theta d\theta$$
$$\displaystyle =\int_{\sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}\frac {(\cos\theta)^{\frac {5}{3}}}{(\sin\theta)^{\frac {5}{3}}}cosec^2\theta d\theta$$
$$\displaystyle=\int_{sin^{-1}\left (\frac {1}{3}\right )}^{\frac {\pi}{2}}(\cot\theta)^{\frac {5}{3}}cosec^2\theta d\theta$$
Put $$\cot \theta=t $$
$$\Rightarrow -cosec^2\theta \,d\theta=dt$$
When $$\theta=\sin^{-1}\left (\dfrac {1}{3}\right ),t=2\sqrt 2$$ and when $$\theta=\dfrac {\pi}{2}, t=0$$
$$\therefore I=-\int_{2\sqrt 2}^0(t)^{\frac {5}{3}}dt$$
$$\Rightarrow \displaystyle I=\int_0^{2\sqrt 2} (t)^{\frac {5}{3}}dt$$
$$\displaystyle =\left [\frac {3}{8}(t)^{\frac {8}{3}}\right ]_0^{2\sqrt 2}$$
$$\displaystyle =\frac {3}{8}[(2\sqrt 2)^{\frac {8}{3}}]$$
$$=\dfrac {3}{8}[(\sqrt 8)^{\frac {8}{3}}]$$
$$=\cfrac {3}{8}[(8)^{\frac {4}{3}}]$$
$$=\cfrac {3}{8}[16]$$
$$=3\times 2$$
$$=6$$
Hence, the correct Answer is A.
If $${ I }_{ m }=\overset { e }{ \underset { 1 }{ \int } } (lnx)^{ m }dx,$$ where $$m\epsilon N,$$then $${ I }_{ 10 }+10{ I }_{ 9 }$$ is equal to-
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$${ e }^{ 10 }$$
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$$\frac { { e }^{ 10 } }{ 10 } $$
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e
0%
e-1
If I = $$\overset { 2 }{ \underset { -3 }{ \int } }$$ (|x + 1 | + |x + 2| + |x - 1|)dx, then I equal
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$$\dfrac{31}{2}$$
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$$\dfrac{35}{2}$$
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$$\dfrac{47}{2}$$
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None of these
$$\displaystyle\int^{\pi /3}_0\frac{\cos \theta}{5-4\sin \theta}d\theta$$ equal to.
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$$\displaystyle\frac{1}{4}log\left(\displaystyle\frac{5}{5+2\sqrt{3}}\right)$$
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$$\displaystyle\frac{1}{4}log\left(\displaystyle\frac{5}{5-2\sqrt{3}}\right)$$
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$$\displaystyle\frac{1}{4}log\left(\displaystyle\frac{5+2\sqrt{3}}{5}\right)$$
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$$\displaystyle\frac{1}{4}log\left(\displaystyle\frac{5-2\sqrt{3}}{5}\right)$$
Explanation
$$I=\int _{ 0 }^{ { \pi }/{ 3 } }{ \cfrac { \cos { \theta } }{ 5-4\sin { \theta } } d\theta } $$
Let $$5-4\sin { \theta =t } $$
$$\Rightarrow \cos { \theta } d\theta =-\cfrac { dt }{ 4 } $$
$$I=\int _{ 5 }^{ 5-2\sqrt { 3 } }{ \cfrac { -dt }{ 4t } } $$
$$I=\cfrac { 1 }{ 4 } \int _{ 5-2\sqrt { 3 } }^{ 5 }{ \cfrac { dt }{ t } } $$
$$I=\cfrac { 1 }{ 4 } { \left[ \log { t } \right] }^{ 5 }_{ 5-2\sqrt { 3 } }$$
$$I=\cfrac { 1 }{ 4 } { \left[ \log { 5 } - \log { (5-2\sqrt { 3 } ) } \right] }$$
$$I=\cfrac { 1 }{ 4 } { \left[ \log \left({ \cfrac { 5 }{ 5-2\sqrt { 3 } } }\right) \right] }$$
$$\therefore I=\cfrac { 1 }{ 4 } \log { \left( \cfrac { 5 }{ 5-2\sqrt { 3 } } \right) } $$
The value of $$\int _{ 1/3 }^{ 3 }{ \cfrac { \tan { \left( { x }^{ 2 }+\cfrac { 1 }{ { x }^{ 2 } } \right) } }{ \sin { \left( x+\cfrac { 1 }{ x } \right) } } } \cfrac { dx }{ x } $$ is
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$$0$$
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$$3/2$$
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$$1/2$$
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$$4/3$$
Let $$I_{1} =\displaystyle \int_{0}^{1}\dfrac {e^{x}dx}{1 + x}$$ and $$I_{2} = \displaystyle \int_{0}^{1} \dfrac {x^{2}dx}{e^{x^{3}}(2 - x^{3})}$$, then $$\dfrac {I_{1}}{I_{2}}$$ is
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$$3/e$$
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$$e/3$$
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$$3e$$
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$$1/3e$$
Explanation
Given : $${ I }_{ 1 }=\int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ (1+x) } } \quad { I }_{ 2 }=\int _{ 0 }^{ 1 }{ \cfrac { { x }^{ 2 }dx }{ { e }^{ x^{ 3 } }(2-{ x }^{ 3 }) } } $$
$${ I }_{ 1 }=\int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ (1+x) } } \\ { I }_{ 2 }=\int _{ 0 }^{ 1 }{ \cfrac { { x }^{ 2 }dx }{ { e }^{ x^{ 3 } }(2-{ x }^{ 3 }) } } $$
Let : $$1-{ x }^{ 3 }=t\quad x\rightarrow 0\quad t\rightarrow 1$$
$$-3{ x }^{ 2 }=dt\quad x\rightarrow 1\quad t\rightarrow 0\\ { I }_{ 2 }=-\int _{ 0 }^{ 1 }{ \cfrac { 1 }{ 3 } \cfrac { { e }^{ t-1 } }{ (1+t) } dt } =\cfrac { 1 }{ 3e } \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ t } }{ (1+t) } dt } \\ \cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =\cfrac { \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ (1+x) } } }{ \cfrac { 1 }{ 3e } \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ t } }{ (1+t) } dt } } =3e$$
Hence the correct answer is $$3e$$
The value of $$\int^3_{1/3}\dfrac{tan(x^2-\dfrac{1}{x^2})}{sin(x+\dfrac{1}{x}) }\dfrac{dx}{x}$$ is
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0
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$$\dfrac{3}{2}$$
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$$\dfrac{1}{2}$$
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$$\dfrac{4}{3}$$
$$\int _{ 0 }^{ \pi /2 }{ \sin ^{ 8 }{ x } \cos ^{ 2 }{ x } dx } $$ is equal to
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$$\cfrac { \pi }{ 512 } $$
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$$\cfrac {3 \pi }{ 512 } $$
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$$\cfrac { 5\pi }{ 512 } $$
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$$\cfrac {7 \pi }{ 512 } $$
The value of the integral $$\displaystyle \int_0^1 \dfrac{x^{\alpha}-1}{\log x}dx$$ is
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$$\log (\alpha+1)$$
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$$2\log (\alpha+1)$$
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$$3\log \alpha$$
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none of these
Explanation
$$I(\alpha)=\displaystyle\int_0^1\dfrac{x^{\alpha}-1}{\text{log }x}dx$$
Differentiating with respect to $$\alpha,$$ we get$$,$$
$$I'(\alpha)=\displaystyle\int_0^1x^\alpha dx=\dfrac{1}{\alpha +1}$$
We need to solve this simple differential equation $$:I'(\alpha)=\dfrac{1}{\alpha +1}$$
On solving, we get,
$$I(\alpha)=\text{log}(\alpha +1)+C$$
Now, from the initial equation, we can find that $$I(0)=0$$
$$I(0)=\text{log}(1)+C=0$$ $$\Rightarrow C=0$$
So, $$I(\alpha)=\text{log}(\alpha +1)$$
The value of the definite integral $$\displaystyle\int^{\pi/2}_{0}\left(\cos ^{10}x\cdot \sin 12x\right)dx$$, is equal to.
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$$\displaystyle\frac{1}{10}$$
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$$\displaystyle\frac{1}{11}$$
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$$\displaystyle\frac{1}{12}$$
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$$\displaystyle\frac{1}{22}$$
Explanation
Let $$ I =\displaystyle \int_{0}^{\pi/2}(\cos^{10}x.\sin\, 12x) dx$$
$$\cos^{10}x.\sin\,12x = \cos^{10}x.\sin(11x+x) = \cos^{10}x.(\sin\,11x.\cos\,x + \cos\, 11x.\sin\,x)$$
$$= - \dfrac{1}{11}(-11\,\sin\,11x.\cos^{11}x - 11\,\cos\,11x.\cos^{10}x.\sin\,x)$$
$$= -\dfrac{1}{11} \times$$ $$\dfrac{d}{dx}(\cos\,11x.\cos^{11}x)$$ ... (1)
Therefore,
$$I = \left (-\dfrac {1}{11}\right)\displaystyle \int_{0}^{\pi/2} \dfrac{d}{dx}(\cos\,11x.\cos^{11}x) dx$$
$$I =- \dfrac {1}{11}\times \cos\,11x.\cos^{11}x]_{0}^{\pi/2}$$
$$I = -\dfrac {1}{11}\times (0-1)$$
$$I = \dfrac{1}{11}$$
Correct answer is option
B.
If $$\displaystyle I=\int _{ { -\pi }/{ 6 } }^{ { \pi }/{ 6 } }{ \dfrac { \pi +4{ x }^{ 5 } }{ 1-\sin { \left( \left| x \right| +\dfrac { \pi }{ 6 } \right) } } } dx$$, then I equals to
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$$4\pi$$
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$$2\pi +\dfrac { 1 }{ \sqrt { 3 } }$$
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$$2\pi -\dfrac { 1 }{ \sqrt { 3 } }$$
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$$4\pi +\sqrt { 3 } -\dfrac { 1 }{ \sqrt { 3 } }$$
If, $$\int^{\frac{\pi}{2}} _0 \frac{sin^2x}{(1 + cosx^2)} dx = 0$$
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$$\frac{4 - \pi}{2}$$
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$$\frac{\pi - 4}{2}$$
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$$4 - frac{\pi}{2}$
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$$\frac{4 + \pi}{2}$$
$${I}_{n}=\int_{1}^{e}{\left(logx\right)^{n}dx}$$ and $${I}_{n}=A+{BI}_{n-1}$$ then
A=........., B=............
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$$e,-n$$
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$$1/e,n$$
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$$-e,n$$
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$$-e-n$$
$$\displaystyle \int _{ 0 }^{ x }{ \cfrac { \sin { x } }{ 1+\cos ^{ 2 }{ x } } } dx=\pi \cfrac { \cos { \alpha } }{ 1-\sin ^{ 2 }{ \alpha } } $$
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0%
for no value of $$\alpha$$
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for exactly two values of $$\alpha$$ in $$\left( 0,\pi \right) $$
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for atleast one $$\alpha$$ in $$\left( \pi /2,\pi \right) $$
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for exactly one $$\alpha$$ in $$\left( 0,\pi /2 \right) $$
What is $$\int _{ 0 }^{ 1 }{ x{ \left( 1-x \right) }^{ 9 }dx } $$ equal to?
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$$\dfrac{1}{240}$$
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$$\dfrac{1}{110}$$
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$$\dfrac{1}{132}$$
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$$\dfrac{1}{148}$$
Explanation
We need to find $$\int_{0}^{1}x(1-x)^9dx$$
$$\int_{0}^{1}x(1-x)^9dx=\int_{0}^{1}(1-x)x^9dx$$
$$=\int_{0}^{1}(x^9-x^{10})dx$$
$$=\left[\dfrac{x^{10}}{10}-\dfrac{x^{11}}{11}\right]_{0}^{1}$$
$$=\dfrac{1}{10}-\dfrac{1}{11}$$
$$=\dfrac{11-10}{110}$$
$$=\dfrac{1}{110}$$
$$\therefore \int_{0}^{1}x(1-x)^9dx=\dfrac{1}{110}$$
The value of $$\int _{ 0 }^{ 14 }{ \dfrac { \left[ { x }^{ 2 } \right] }{ \left[ { x }^{ 2 }-28x+196 \right] +\left[ { x }^{ 2 } \right] } } $$
where [x] is the integral part of real x, is
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14
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0
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7
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49
If $$I_1 =\int^1_0 2x^2 dx, I_2 =\int^1_0 2^{x3} dx, I_3 =\int^2_1 2x^{x2} dx$$ then
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$$I_1 > I_2$$
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$$I_2 > I_1$$
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$$I_3 > I_4$$
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$$I_3 = I_4$$
$$\displaystyle E = \int_{R}^{\infty}\dfrac{GMm}{x^2}$$ dx , (where $$G$$ , $$M$$ , $$m$$ are constants ) equal to
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$$-\dfrac{GMm}{R^2}$$
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$$+\dfrac{GMm}{R^2}$$
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$$-\dfrac{GMm}{R}$$
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$$+\dfrac{GMm}{R}$$
$$\int _{ 0 }^{ 4036 }{ \dfrac { { 2 }^{ x } }{ { 2 }^{ x }+{ 1 }^{ 4036-x } } } dx=............$$
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2018
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4035
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2017
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-2015
If $$I_{1} = \int_{0}^{1} 2^{x^{3}} dx, I_{2} = \int_{0}^{1}2^{x^{2}}dx, I_{3} = \int_{1}^{2}2^{x^{2}}dx$$ and $$I_{4} = \int_{1}^{2}2^{x^{3}}dx$$, then
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$$I_{1} > I_{2}$$
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$$I_{2} > I_{1}$$
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$$I_{3} > I_{4}$$
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$$I_{1} > I_{3}$$
Explanation
Given :-
$$I_{1} = \int_{0}^{1} 2^{x^{3}} dx, I_{2} = \int_{0}^{1}2^{x^{3}}dx, I_{3} = \int_{1}^{2}2^{x^{2}}dx$$ and $$I_{4} = \int_{1}^{2}2^{x^{3}}dx$$,
As we know $$ x^2 < x^3 $$ for $$ x\in(1,2) $$ ,
So $$ 2^{x^{3}}$$ will also be grater than $$2^{x^{2}}$$ for $$ x\in (0,1)$$ and
$$ x\in (1,2)$$
$$\therefore I_{1}>I_2$$
hence
$$ I_{3} = \int_{1}^{2}2^{x^{2}}dx$$ < $$I_{4} = \int_{1}^{2}2^{x^{3}}dx$$
And also
$$ 2^{x^{2}} $$ when $$x \in (1,2)$$ is grater than $$ 2^{x^{3}}$$ for $$ x\in (0,1)$$ hence
$$ I_{1} = \int_{0}^{1}2^{x^{3}}dx$$ < $$I_{3} = \int_{1}^{2}2^{x^{2}}dx$$
Hence,
$$I_{1}>I_2$$ is correct.
A function $$f(x)$$ which satisfies the relation $$f(x) = e^{x} + \int_{0}^{1} e^{x}f(t)dt$$, then $$f(x)$$ is
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$$\dfrac {e^{x}}{2 - e}$$
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$$(e - 2)e^{x}$$
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$$2e^{x}$$
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$$\dfrac {e^{x}}{2}$$
Value of the definite integral $$\displaystyle \int _{ 0 }^{ 1 }{ \cot ^{ -1 }{ \left( 1-x+{ x }^{ 2 } \right) } dx }$$ is:
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$$\pi -\log { 2 }$$
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$$\dfrac{\pi}{2} -\log { 2 }$$
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$$\pi +\log { 2 }$$
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$$\dfrac{\pi}{2} +\log { 2 }$$
If $$I=\int _{ 0 }^{ \pi }{ \frac { x\sin { x } }{ 1+{ cos }^{ 2 }x } dx } $$, then the value of $$\sin { \sqrt { I } } $$, is
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0%
$$\frac { 1 }{ 2 } $$
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$$0$$
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$$\frac { \sqrt { 2 } }{ 2 } $$
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$$1$$
$$if\,{l_n} = \int\limits_0^1 {{x^m}} {\left( {\ln x} \right)^n}dx,\,and\,{l_n} = \frac{1}{{m + 1}}{l_{n - 1}}$$ then k is equal to
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n
0%
-n
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m-1
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None of these
$$For x>0$$,let $$f\left( x \right) =\int _{ 1 }^{ 2 }{ \dfrac { \log { t } }{ 1+t } } dt$$,then$$f\left( x \right) +f\left( \dfrac { 1 }{ x } \right)$$is equal to:
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$$\dfrac { 1 }{ 4 } \left( \log { x } \right) ^{ 2 }$$
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$$\dfrac { 1 }{ 2 } \left( \log { x } \right) ^{ 2 }$$
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$$\log { x }$$
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$$\dfrac { 1 }{ 4 } \log { { x }^{ 2 } }$$
Evaluate the integral, $$\int _{ 0 }^{ 1 }{ \cos { \left( 2\cot ^{ -1 }{ \sqrt { \dfrac { 1-x }{ 1+x } } } \right) } } dx=$$
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$$-1/2$$
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$$1/2$$
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$$0$$
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$$1$$
$$\int _{ \pi /4 }^{ \pi /2 }{ \cos { \theta } \csc ^{ 2 }{ \theta } d\theta = }$$
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$$\sqrt {2}-1$$
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$$1-\sqrt {2}$$
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$$\sqrt {2}+1$$
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$$None\ of\ these$$
Explanation
$$\int _{ \pi /4 }^{ \pi /2 }{ \cfrac { \cos { \theta } }{ \sin ^{ 2 }{ \theta } } } $$
$$\sin { \theta } =t$$
$$\cos { \theta d\theta } =dt$$
$$\int _{ \pi /4 }^{ \pi /2 }{ { t }^{ -2 }dt } ={ \left( \cfrac { t-1 }{ -1 } \right) }_{ \pi /4 }^{ \pi /2 }={ -\left( \cfrac { 1 }{ t } \right) }_{ \pi /4 }^{ \pi /2 }={ \left( +\cfrac { 1 }{ \sin { \theta } } \right) }_{ \pi /4 }^{ \pi /2 }$$
$$=-\left( 1-\sqrt { 2 } \right) =\sqrt { 2 } -1$$
$$\displaystyle \int_2^ 3\frac{(x+2)^2}{2x^2- 10x +53}dx$$ is equal to
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$$2$$
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$$1$$
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$$\dfrac{1}{2}$$
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$$\dfrac{5}{2}$$
Solve $$\displaystyle \int _{ 0 }^{ 1/2 }{ \dfrac { x\sin ^{ -1 }{ x } }{ \sqrt { 1-{ x }^{ 2 } } } dx= }$$
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$$\dfrac { 1 }{ 2 } +\dfrac { \sqrt { 3 } \pi }{ 12 }$$
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$$\dfrac { 1 }{ 2 } -\dfrac { \sqrt { 3 } \pi }{ 12 }$$
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$$\dfrac { 1 }{ 2 } -\dfrac { \sqrt { 2 } \pi }{ 12 }$$
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$$None\ of\ these$$
For n >2,
$$\displaystyle \int _{ 0 }^{ \pi /2 }{ \dfrac { \sec ^{ 2 }{ x } dx }{ { \left( \sec { x } +\tan { x } \right) }^{ n } } } = ?$$
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$$\dfrac { 1 }{ { n }^{ 2 }-1 } $$
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$$\dfrac { n }{ { n }^{ 2 }-1 } $$
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$$\dfrac { n }{ { n }^{ 2 }+1 } $$
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$$\dfrac { 2 }{ { n }^{ 2 }-1 } $$
Explanation
$$\int_{0}^{\frac{\pi}{2}} \dfrac{\sec ^2xdx}{(\sec x+\tan x)^n}$$
substitute $$\sec x+\tan x = t\rightarrow \sec x dx = \dfrac{dt}{t}\Rightarrow x(0,\dfrac{\pi}{2})\rightarrow t(1,\infty )$$
$$\sec x = \dfrac{1}{2}\left ( t+\dfrac{1}{t} \right )$$
$$\int_{0}^{\frac{\pi}{2}} \dfrac{\sec ^2xdx}{(\sec x+\tan x)^n}=\int_{1}^{\infty } \dfrac{\frac{1}{2}\left ( t+\frac{1}{t} \right )\frac{dt}{t}}{t^n}$$
$$=\dfrac{1}{2}\int_{1}^{\infty }\left ( \dfrac{1}{t}+\dfrac{1}{t^{n+2}} \right )dt$$
$$=\left [ \dfrac{t^{-n+1}}{-n+1} +t^{-n+1}+\dfrac{t^{-(n+1)}(n+2)}{-(n+1)}\right ]_1^\infty $$
$$=-\dfrac{1}{2} \left [\dfrac{1}{-n+1}+1-\dfrac{n+2}{n+1} \right ]$$
$$=-\dfrac{1}{2} \left [ \dfrac{n^2+2n-1+1-n^2}{1-n^2} \right ]$$
$$=\dfrac{n}{n^2-1}$$
Let $$I=\displaystyle\int _{ 0 }^{ 1 }{ \sqrt { \dfrac { 1+\sqrt { x } }{ 1-\sqrt { x } } } dx } $$ and $$J=\displaystyle\int _{ 0 }^{ 1 }{ \sqrt { \dfrac { 1-\sqrt { x } }{ 1+\sqrt { x } } } dx }$$, then the correct statement is
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$$I+J=4$$
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$$I-J=\pi$$
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$$I=\dfrac{2+\pi}{2}$$
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$$J=\dfrac{4-\pi}{2}$$
$$\int _{ -\pi /4 }^{ \pi /4 }{ ln\sqrt { 1+\sin { 2x } } dx }$$ has the value equal to:
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$$-\dfrac {\pi}{4}l\ n2$$
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$$-\dfrac {\pi}{2}l\ n2$$
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$$-\dfrac {\pi}{8}l\ n2$$
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$$-\dfrac {\pi}{16}l\ n2$$
$$\int _{ -1 }^{ 1 }{ x } \tan { ^{ -1 } } xdx$$
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$$\left( \frac { \pi }{ 2 } -1 \right)$$
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$$\left( \frac { \pi }{ 2 } +1 \right)$$
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$$(\pi-1)$$
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$$None$$
$$\int _{ 0 }^{ 2\pi }{ \sqrt { 1+\sin { \dfrac { x }{ 2 } } } dx } =$$
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$$0$$
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$$2$$
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$$8$$
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$$4$$
$$\displaystyle\int^{1}_0\sqrt{x(1-x)}dx=$$.
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$$\dfrac{\pi}{8}$$
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$$\dfrac{3\pi}{8}$$
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$$\dfrac{5\pi}{4}$$
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$$\dfrac{\pi}{2}$$
Explanation
We are given, $$I=\displaystyle \int \sqrt {x(1-x)}dx$$
let $$x=\sin^2 \theta \ \Rightarrow \ dx=2\sin \theta . \cos \theta \ d\theta =\sin 2\theta \ d\theta$$
(differentiating both side) & (chain rule)
now when $$x=0, \sin^2 \theta =0\ \Rightarrow \theta =0$$
& when $$x=1, \sin^2 \theta =1\ \Rightarrow \theta =\dfrac {\pi}{2}$$
$$I=\displaystyle \int_{0}^{\pi /2}\sqrt {\sin^2 \theta (1-\sin^2 \theta)} (\sin 2\theta \ d\theta)$$
$$=\displaystyle \int_{0}^{\pi /2}\sqrt {\sin^2 \theta .\cos^2 \theta} (\sin 2 \theta) d\theta$$
$$=\displaystyle \int_{0}^{\pi /2} |\sin \theta . \cos \theta| \sin 2\theta \ d\theta$$
(here $$\sin \theta . \cos \theta$$ are positive in $$[0, \pi /2$$])
$$I=\displaystyle \int_{0}^{\pi /2} (\sin \theta . \cos \theta). \sin 2\theta \ d\theta$$
$$=\dfrac {1}{2} \displaystyle \int_{0}^{\pi /2} (2\sin \theta \cos \theta ) \sin 2\theta \ d\theta $$ (multiply & divide the integration by $$2$$ to get $$2\cos \theta \sin \theta =\sin 2\theta$$)
$$I=\dfrac {1}{2} \displaystyle \int_{0}^{\pi /2} \sin^2 2\theta \ d\theta$$
$$I=\dfrac {1}{2} \displaystyle \int_{0}^{\pi /2} \left (\dfrac {1-\cos 4\theta}{2}\right) d\theta$$
$$=\dfrac {1}{2} \left (\displaystyle \int_{0}^{\pi /2} \left (\dfrac {1}{2} d\theta\right)-\displaystyle \int_{0}^{\pi /2}\cos 4\theta \ d\theta \right)$$
$$I=\dfrac {1}{2}\left [\left (\dfrac {\theta}{2}\right)_0^{\pi /2} -\left (\dfrac {\sin 4\theta}{4}\right)_0^{\pi /2} \right]$$
$$=\dfrac {1}{2} \left [\left (\dfrac {\pi}{4}-0\right)-\left (\dfrac {\sin 2\theta}{4}-\dfrac {\sin 0}{4}\right) \right]$$
$$=\dfrac {1}{4}\left (\dfrac {\pi}{4}-0 \right)$$
$$\boxed {I=\dfrac {\pi}{8}}$$
If $${ I }_{ 1 }=\int _{ x }^{ 1 }{ \cfrac { 1 }{ 1+{ t }^{ 2 } } } dt$$ and $${ I }_{ 2 }=\int _{ 1 }^{ 1/x }{ \cfrac { 1 }{ 1+{ t }^{ 2 } } } dt$$ for x > 0, then
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$${ I }_{ 1 }={ I }_{ 2 }$$
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$${ I }_{ 1 }>{ I }_{ 2 }$$
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$${ I }_{ 2 }>{ I }_{ 1 }$$
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None of these.
$$\displaystyle \overset{\pi/2}{\underset{0}{\int}} \dfrac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}$$ equals-
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$$\pi/ab$$
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$$2\pi/ab$$
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$$ab/ \pi$$
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$$\pi/2ab$$
Explanation
$$I=\displaystyle\int^{\pi/2}_0\dfrac{dx}{a^2\cos^2x+b^2\sin^2x}$$
$$=\displaystyle\int^{\pi/2}_0\dfrac{\sec^2xdx}{a^2+b^2\tan^2x}$$
Let $$\tan x=t$$ $$x=0, \tan x=0$$
$$\sec^2xdx=dt$$ $$x=\pi/2, \tan x=\infty$$
$$\Rightarrow \displaystyle\int^{\infty}_0\dfrac{dt}{a^2+b^2t^2}=\dfrac{1}{b^2}\displaystyle\int^{\infty}_0\dfrac{dt}{\left(\dfrac{a^2}{b^2}+t^2\right)}$$
$$\Rightarrow \dfrac{1}{b^2}\times \dfrac{1}{(a/b)}\left[\tan^{-1}\left(\dfrac{t}{(a/b)}\right)\right]^{\infty}_0$$
$$=\dfrac{1}{ab}\left[\tan^{-1}(\infty)-\tan^{-1}(0)\right]$$
$$=\dfrac{1}{ab}[\pi/2]$$
$$\Rightarrow \dfrac{\pi}{2ab}$$.
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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