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CBSE Questions for Class 12 Commerce Maths Integrals Quiz 4 - MCQExams.com
CBSE
Class 12 Commerce Maths
Integrals
Quiz 4
The value of $$\displaystyle \int_{0}^{\pi }\displaystyle \frac{dx}{1-2\alpha \cos x+\alpha ^{2}}$$ is
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$$\displaystyle \frac{\pi }{1+\alpha ^{2}}$$ if $$\alpha > 1$$
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$$\displaystyle \frac{\pi }{\alpha ^{2}-1}$$ if $$\alpha > 1$$
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$$\displaystyle \frac{\pi }{1+\alpha ^{2}}$$ if $$\alpha < 1$$
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$$\displaystyle \frac{\pi }{\alpha ^{2}-1}$$ if $$\alpha < 1$$
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ \pi } \frac { dx }{ 1-2\alpha \cos x+\alpha ^{ 2 } } =\int _{ 0 }^{ \pi } \frac { dx }{ 1-2\alpha \cos x+\alpha ^{ 2 } } $$
Substitute $$\displaystyle \tan { \frac { x }{ 2 } } =t\Rightarrow \frac { 1 }{ 2 } \sec ^{ 2 }{ \frac { x }{ 2 } } dx=dt$$
$$\displaystyle I=\int _{ 0 }^{ \infty } \dfrac { 1 }{ 1-2\alpha \left( \dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } \right) +\alpha ^{ 2 } } .\dfrac { 2t }{ 1+{ t }^{ 2 } } dt$$
$$\displaystyle =\int _{ 0 }^{ \infty } \dfrac { 2t }{ 1+{ t }^{ 2 }-2\alpha \left( 1-{ t }^{ 2 } \right) +\alpha ^{ 2 }\left( 1+{ t }^{ 2 } \right) } dt$$
$$\displaystyle =\frac { \pi }{ { \alpha }^{ 2 }-1 } $$ if $$\alpha >1$$
Given $$\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { x }^{ 2 } } } dx=a,$$ the value of $$\displaystyle \int _{ e }^{ { e }^{ 4 } }{ \sqrt { \ln { \left( x \right) } } dx } $$ is?
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$${ e }^{ 4 }-e$$
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$${ e }^{ 4 }-a$$
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$$2{ e }^{ 4 }-a$$
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$$2{ e }^{ 4 }-e-a$$
Explanation
Given $$\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { x }^{ 2 } } } dx=a,$$
Let $$I=\displaystyle \int _{ e }^{ { e }^{ 4 } }{ \sqrt { \ln { \left( x \right) } } dx } $$
Put $$\displaystyle \ln { \left( x \right) } ={ t }^{ 2 }\Rightarrow \frac { 1 }{ x } dx=2tdt$$
$$\therefore I=\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { t }^{ 2 } } } .{ 2t }^{ 2 }dt={ \left( t{ e }^{ { t }^{ 2 } } \right) }_{ 1 }^{ 2 }-\displaystyle \int _{ 1 }^{ 2 }{ { e }^{ { t }^{ 2 } } } dt={ 2e }^{ 4 }-e-a.$$
The value of $$\displaystyle \int_{0}^{1}\displaystyle \frac{dx}{\left ( x+1 \right )\sqrt{x^{2}+2x}}$$ is
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$$\pi /6$$
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$$\pi /3$$
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$$\pi /2$$
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$$\pi $$
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( x+1 \right) \sqrt { { x }^{ 2 }+2x } } dx } =\int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( x+1 \right) \sqrt { { \left( x+1 \right) }^{ 2 }-1 } } dx } $$
Substitute $$u=x+1\Rightarrow du=dx$$
$$\displaystyle\therefore I=\int _{ 1 }^{ 2 }{ \frac { 1 }{ u\sqrt { { u }^{ 2 }-1 } } du } $$
Substitute
$$\displaystyle s=\sqrt { { u }^{ 2 }-1 } \Rightarrow ds=\frac { u }{ \sqrt { { u }^{ 2 }-1 } } du$$
$$\displaystyle\therefore I=\int _{ 0 }^{ \sqrt { 3 } }{ \frac { 1 }{ { s }^{ 2 }+1 } ds } ={ \left[ \tan ^{ -1 }{ s } \right] }_{ 0 }^{ \sqrt { 3 } }=\frac { \pi }{ 3 } $$
Value of $$\displaystyle \int_{0}^{\pi /4}\left ( \sqrt{\tan x}-\sqrt{\cot x} \right )\: dx$$ is
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$$\sqrt{2}\log \left ( \sqrt{2}-1 \right )$$
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$$\sqrt{2}\log \left ( \sqrt{2}+1 \right )$$
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$$\log \left ( \sqrt{2}+1 \right )$$
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$$\log \left ( \sqrt{2}-1 \right )$$
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \left( \sqrt { \tan { x } } -\sqrt { cot{ x } } \right) } dx=-\int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \frac { cos{ x }-sin{ x } }{ \sqrt { cos{ x }sin{ x } } } dx } $$
Substitute $$\left( sin{ x+cos{ x } } \right) =t\Rightarrow 2sin{ x }cos{ x }={ t }^{ 2 }-1$$
$$\displaystyle \therefore I=\sqrt { 2 } \int _{ 1 }^{ \sqrt { 2 } }{ \frac { dt }{ \sqrt { { t }^{ 2 }-1 } } } \\ =\left[ \sqrt { 2 } \log { \left( t+\sqrt { { t }^{ 2 } } -1 \right) } \right] _{ 0 }^{ \sqrt { 2 } }\\ =\sqrt { 2 } \log { \left( \sqrt { 2 } -1 \right) } $$
Value of $$\displaystyle \int_{0}^{2a}\dfrac{x^{3/2}}{\sqrt{2a-x}} dx$$ is
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$$\displaystyle \frac{3\pi a^{2}}{2}$$
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$$\pi a^{3}$$
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$$\sqrt{2}\pi a^{3}$$
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$$2\pi a^{3}$$
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ 2a }{ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \sqrt { 2a-x } } dx } $$
Substitute
$$\displaystyle u=\sqrt { x } \Rightarrow du=\frac { 1 }{ 2\sqrt { x } } dx$$
$$\displaystyle I=2\int _{ 0 }^{ \sqrt { 2a } }{ \frac { { u }^{ 4 } }{ \sqrt { 2a-{ u }^{ 2 } } } du } $$
Substitute
$$u=\sqrt { 2a } \sin { t } \Rightarrow du=\sqrt { 2a } \cos { t } dt$$
$$\displaystyle I=2\sqrt { 2a } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2\sqrt { 2 } { a }^{ \frac { 3 }{ 2 } }\sin ^{ 4 }{ t } \right) dt } =8{ a }^{ 2 }\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \sin ^{ 4 }{ t } dt } $$
Using reduction formulae
$$\displaystyle \int { \sin ^{ m }{ x } dx } =\frac { -\cos { x } \sin ^{ m-1 }{ x } }{ m } +\frac { m-1 }{ m } \int { \sin ^{ m-2 }{ x } dx } $$
$$\displaystyle I={ \left[ -2{ a }^{ 2 }\sin ^{ 3 }{ t } \cos { t } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }+6{ a }^{ 2 }\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin ^{ 2 }{ t } dt } $$
$$\displaystyle =0+6{ a }^{ 2 }\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \frac { 1 }{ 2 } -\frac { 1 }{ 2 } \cos { 2t } \right) dt } $$
$$\displaystyle ={ \left[ 3{ a }^{ 2 }t-3{ a }^{ 2 }\sin { t } \cos { t } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=\frac { 3\pi { a }^{ 2 } }{ 2 } $$
The value of $$\displaystyle \int_{a}^{b}\displaystyle \frac{\log x}{x}\: dx$$ is
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$$\log \left ( ab \right )\log \displaystyle \left ( \frac{b}{a} \right )$$
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$$\displaystyle \frac{1}{2}\log \left ( ab \right )\log \displaystyle \left ( \frac{b}{a} \right )$$
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$$\log \left ( a^{2}-b^{2} \right )$$
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$$\left ( a+b \right )\log \left ( a+b \right )$$
Explanation
$$\displaystyle I=\int _{ a }^{ b }{ \frac { \log { x } }{ x } } dx={ \left[ \log { x } .\log { x } \right] }_{ a }^{ b }-\int _{ a }^{ b }{ \frac { \log { x } }{ x } } dx\\ \Rightarrow 2I={ \left[ { \left( \log { x } \right) }^{ 2 } \right] }_{ a }^{ b }={ \left( \log { b } \right) }^{ 2 }-{ \left( \log { a } \right) }^{ 2 }$$
$$\displaystyle \Rightarrow I=\frac { 1 }{ 2 } \left( \log { b } +\log { a } \right) \left( \log { b } -\log { a } \right) =\frac { 1 }{ 2 } \log { ab } \log { \frac { b }{ a } } $$
Value of $$\displaystyle \int_{0}^{25}\displaystyle \frac{1}{\sqrt{4+\sqrt{x}}}\: dx$$ is
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$$2\left ( \sqrt{29}-1 \right )$$
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$$2\left ( \sqrt{29}-5 \right )$$
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$$3 \sqrt{29}-1 $$
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none of these
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ 25 }{ \frac { 1 }{ \sqrt { \sqrt { x } +4 } } dx } $$
Substitute
$$\displaystyle u=\sqrt { x } \Rightarrow du=\frac { 1 }{ 2\sqrt { x } } dx$$
$$\displaystyle I=2\int _{ 0 }^{ 5 }{ \frac { u }{ \sqrt { u+4 } } du } $$
Substitute
$$ s=u+4\Rightarrow ds=du$$
$$\displaystyle I=2\int _{ 4 }^{ 9 }{ \frac { s-4 }{ \sqrt { s } } ds } =2\int _{ 4 }^{ 9 }{ \left( \sqrt { s } -\frac { 4 }{ \sqrt { s } } \right) ds } $$
$$\displaystyle={ \left[ \frac { 4{ s }^{ 3/2 } }{ 3 } \right] }_{ 4 }^{ 9 }-{ \left[ 16\sqrt { s } \right] }_{ 4 }^{ 9 }=36-\frac { 32 }{ 3 } -48-32=-\frac { 164 }{ 3 } $$
$$\displaystyle \int_{0}^{\infty }f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x}dx$$
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is equal to zero
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is equal to one
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is equal to $$\displaystyle \frac{1}{2}$$
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can not be evaluated
Explanation
$$I = \displaystyle \int_{0}^{\infty }f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x}dx $$
Substitute
$$\displaystyle x = \frac{1}{t}\Rightarrow dx =-\frac{1}{t^2}dt$$
$$I = \displaystyle \int_{\infty }^0 f\left ( t+\frac{1}{t} \right )\frac{\ln (1/t)}{1/t} .\frac{-dt}{t^2} $$
$$\quad \displaystyle =\int_{\infty }^0 f\left ( t+\frac{1}{t} \right )\frac{\ln t}{t} dt$$
$$\quad \displaystyle =\int_{\infty }^0 f\left ( x+\frac{1}{x} \right )\frac{\ln x}{x} dx=-I$$
$$\Rightarrow 2I = 0\Rightarrow I = 0$$
If$$\displaystyle \int_{0}^{\pi /3}\frac{\cos }{3+4\sin x}dx=K\log \frac{\left ( 3+2\sqrt{3} \right )}{3}$$ then K is
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$$\displaystyle \frac{1}{2}$$
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$$\displaystyle \frac{1}{3}$$
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$$\displaystyle \frac{1}{4}$$
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$$\displaystyle \frac{1}{8}$$
Explanation
$$\displaystyle I=\int _{ 0 }^{ \pi /3 }{ \cfrac { \cos { x } }{ 3+4\sin { x } } dx } $$
Substituting $$t=3+4\sin { x } \Rightarrow dt=4\cos { x } $$
$$\displaystyle I=\frac { 1 }{ 4 } \int _{ 3 }^{ 3+2\sqrt { 3 } }{ \frac { 1 }{ t } dt } ={ \left[ \frac { \log { t } }{ 4 } \right] }_{ 3 }^{ 3+2\sqrt { 3 } }$$
$$\displaystyle=\frac { \log { \left( 3+2\sqrt { 3 } \right) } -\log { 3 } }{ 4 } =\frac { 1 }{ 4 } \log { \left( \frac { 3+2\sqrt { 3 } }{ 3 } \right) } \Rightarrow K=\frac { 1 }{ 4 } $$
Suppose that F(x) is an antiderivative of f(x)$$\displaystyle =\frac{\sin x}{x},x> 0$$ then $$\displaystyle \int_{1}^{3}\frac{\sin 2x}{x}$$ can be expressed as
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$$F(6) - F(2)$$
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$$\displaystyle \frac{1}{2}\left ( F\left ( 6 \right )-F\left ( 2 \right ) \right )$$
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$$\displaystyle \frac{1}{2}\left ( F\left ( 3 \right )-F\left ( 1 \right ) \right )$$
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$$\displaystyle 2\left ( F\left ( 6 \right )-F\left ( 2 \right ) \right )$$
Explanation
$$\displaystyle F\left ( x \right )=\int \frac{\sin x}{x}dx$$ Now $$\displaystyle I = \int_{1}^{3}\frac{\sin 2x}{x}dx\left [ put2x=t \right ]=\int_{2}^{6}\frac{2}{2}\frac{\sin }{t}dt=\left [ F\left ( x \right ) \right ]_{2}^{6}=F(6)-F(2)$$
The value of $$\displaystyle \int_{0}^{1}\displaystyle \frac{dx}{e^{x}+e^{-x}}$$ is
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$$\tan ^{-1}e$$
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$$\tan ^{-1}\left ( e \right )-\pi /4$$
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$$\tan ^{-1}\left ( e \right )-\tan ^{-1}\left ( 1/e \right )$$
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$$\tan ^{-1}\left ( 1/e \right )+\pi /4$$
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ 1 } \frac { dx }{ e^{ x }+e^{ -x } } =\int _{ 0 }^{ 1 } \frac { { e }^{ x }dx }{ e^{ 2x }+1 } $$
Substitute $${ e }^{ x }=t\Rightarrow { e }^{ x }dx=dt$$
$$\displaystyle I=\int _{ 1 }^{ e }{ \frac { t }{ { t }^{ 2 }+1 } dt } ={ \left[ \tan ^{ -1 }{ t } \right] }_{ 1 }^{ e }=\tan ^{ -1 }{ e } -\frac { \pi }{ 4 } $$
$$\displaystyle \int_{1/2}^{2}\frac{1}{x}\sin \left ( x-\frac{1}{x} \right )dx$$ has the value equal to
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$$0$$
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$$\displaystyle \frac{3}{4}$$
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$$\displaystyle \frac{5}{4}$$
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$$2$$
Explanation
$$\displaystyle I=\int _{ \frac { 1 }{ 2 } }^{ 2 }{ \frac { 1 }{ x } } \sin { \left( x-\frac { 1 }{ x } \right) dx } $$
Put $$\displaystyle x=\frac { 1 }{ t } \Rightarrow dx=\frac { 1 }{ { t }^{ 2 } } dt$$
$$\displaystyle I=\int _{ 2 }^{ \frac { 1 }{ 2 } }{ t\sin { \left( \frac { 1 }{ t } -t \right) \left( \frac { -1 }{ { t }^{ 2 } } \right) dt } } =\int _{ 2 }^{ \frac { 1 }{ 2 } }{ \frac { 1 }{ t } \sin { \left( t-\frac { 1 }{ t } \right) dt } } $$
$$\displaystyle =-\int _{ \frac { 1 }{ 2 } }^{ 2 }{ \frac { 1 }{ t } } \sin { \left( t-\frac { 1 }{ t } \right) dt } =-I.$$
$$\Rightarrow 2I=0\quad \Rightarrow I=0$$
If $$\displaystyle f\left ( x \right )=\int_{1}^{x}\frac{\ln t}{1+t}dt$$ where $$x > 0$$, then the value(s) of $$x$$ satisfying the equation, $$f(x) +f(1/x)=2$$ is
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$$2$$
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$$e$$
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$$\displaystyle e^{-2}$$
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$$\displaystyle e^{2}$$
Explanation
$$\displaystyle f\left( x \right)=\int _{ 1 }^{ x }{ \frac { \ln { t } }{ 1+t } dt } $$
$$\displaystyle \Rightarrow f\left( \frac { 1 }{ x } \right) =\int _{ 1 }^{ 1/x }{ \cfrac { \ln { t } }{ 1+t } dt } $$
Substituting $$t=\dfrac { 1 }{ u } \Rightarrow dt=\left( -\dfrac { 1 }{ { u }^{ 2 } } \right) du$$
Therefore $$\displaystyle f\left( \dfrac { 1 }{ x } \right) =\int _{ 1 }^{ x }{ \dfrac { \ln { \left( \dfrac { 1 }{ u } \right) \left( -1 \right) } }{ \left( 1+\dfrac { 1 }{ u } \right) { u }^{ 2 } } du }$$
$$\displaystyle=\int _{ 1 }^{ x }{ \frac { \ln { u } }{ u\left( u+1 \right) } du } =\int _{ 1 }^{ x }{ \frac { \ln { t } }{ t\left( 1+t \right) } dt } $$
Now, $$\displaystyle f\left( x \right) +f\left( \frac { 1 }{ x } \right) =\int _{ 1 }^{ x }{ \cfrac { \ln { t } }{ \left( 1+t \right) } dt } +\int _{ 1 }^{ x }{ \frac { \ln { t } }{ t\left( 1+t \right) } dt }$$
$$\displaystyle=\int _{ 1 }^{ x }{ \frac { \left( 1+t \right) \ln { t } }{ t\left( 1+t \right) } dt } =\int _{ 1 }^{ x }{ \frac { \ln { t } }{ t } dt } =\frac { 1 }{ 2 } { \left( \ln { x } \right) }^{ 2 }$$
Hence $$\displaystyle f\left( x \right) +f\left( \frac { 1 }{ x } \right) =2\Rightarrow x={ e }^{ \pm 2 }$$
Solve $$\displaystyle \int_{2}^{-13}\frac{dx}{\sqrt[5]{\left ( 3-x \right )^{4}}}$$
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$$\displaystyle -5\left ( \sqrt[5]{16}-1 \right )$$
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$$\displaystyle 5\left ( \sqrt[5]{16}-1 \right )$$
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$$\displaystyle -5\left ( \sqrt[5]{16}+1 \right )$$
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None of these
Explanation
Let $$\displaystyle I=\int _{ 2 }^{ -13 }{ \frac { dx }{ 5\sqrt { { \left( 3-x \right) }^{ 4 } } } } $$
Put $$3-x=t\Rightarrow -dx=dt$$
$$\displaystyle I=\int _{ 1 }^{ 16 }{ \frac { dt }{ { t }^{ \frac { 4 }{ 5 } } } } ={ \left[ \frac { { t }^{ \frac { 1 }{ 5 } } }{ \frac { 1 }{ 5 } } \right] }_{ 1 }^{ 16 }$$
$$=-5\left( \sqrt [ 5 ]{ 16 } -1 \right) $$
Choose a function $$f(x)$$ such that it is integrable over every interval on the real line
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$$f(x) = [x]$$
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$$f(x)=x|x|$$
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$$f(x)=[sinx]$$
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$$f(x)=\dfrac{|x-1|}{x-1}$$
$$\displaystyle \int_{0}^{\infty }\frac{x}{\left ( 1+x \right )\left ( 1+x^{2} \right )}dx$$
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$$\displaystyle \frac{\pi }{4}$$
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$$\displaystyle \frac{\pi }{2}$$
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is sme as $$\displaystyle \int_{0}^{\infty }\frac{dx}{\left ( 1+x \right )\left ( 1+x^{2} \right )}$$
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cannot be evaluated
Explanation
Let $$I=\int _{ 0 }^{ \infty }{ \cfrac { xdx }{ \left( 1+x \right) \left( 1+{ x }^{ 2 } \right) } } $$
Using partial fraction
$$=\int _{ 0 }^{ \infty }{ \left( \cfrac { x+1 }{ 2\left( 1+{ x }^{ 2 } \right) } -\cfrac { 1 }{ 2\left( 1+x \right) } \right) dx } \\ ={ \left( \lim _{ b\rightarrow \infty }{ \cfrac { 1 }{ 2 } \log { \left( 1+{ x }^{ 2 } \right) } -\cfrac { 1 }{ 2 } \log { \left( 1+x \right) } +\cfrac { 1 }{ 2 } { tan }^{ -1 }x } \right) }_{ 0 }^{ b }\\ =\lim _{ b\rightarrow \infty }{ \left( \cfrac { 1 }{ 2 } \log { \left( 1+b \right) } -\cfrac { 1 }{ 2 } \log { \left( 1+b \right) } +\cfrac { 1 }{ 2 } { tan }^{ -1 }b \right) } \\ =\cfrac { \pi }{ 4 } $$
State true or false:
The average value of the function $$f(x) = sin^2xcos^3x$$ on the interval $$[ -\pi ,\pi ]$$ is 0.
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True
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False
Explanation
$$\int _{ -\pi }^{ \pi }{ f\left( x \right) dx } =\int _{ -\pi }^{ \pi }{ \sin ^{ 2 }{ x } \cos ^{ 3 }{ x } dx } \\ =\int _{ -\pi }^{ \pi }{ \sin ^{ 2 }{ x } \left( 1-\sin ^{ 2 }{ x } \right) \cos { x } dx } \\ =\int _{ -\pi }^{ \pi }{ \sin ^{ 2 }{ x } \cos { x } dx } -\int _{ -\pi }^{ \pi }{ \sin ^{ 4 }{ x } \cos { x } dx } $$
$$\displaystyle ={ \left[ \frac { \sin ^{ 3 }{ x } }{ 3 } \right] }_{ -\pi }^{ \pi }-{ \left[ \frac { \sin ^{ 5 }{ x } }{ 3 } \right] }_{ -\pi }^{ \pi }=0$$
$$I_{1}$$ is equal to
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$$\displaystyle \frac {2}{3} \int_{0}^{\pi/2}(sin^{2}\theta)(cos\theta)^{-1/3}d\theta$$
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$$\displaystyle \frac {3}{2} \int_{0}^{\pi/2}(sin^{2}\theta)(cos\theta)^{-1/3}d\theta$$
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$$\displaystyle \frac {2}{3} \int_{0}^{\pi/2}(sin\theta)^{2/3}(cos\theta)^{-1/3}d\theta$$
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$$\displaystyle \frac {3}{2} \int_{0}^{\pi/2}(sin\theta)^{2/3}(cos\theta)^{-1/3}d\theta$$
Explanation
$$\displaystyle I_{1}=\int_{0}^{1}(1-x^{2})^{1/3}dx=(1-x^{2})^{3}_{0}+\frac {2}{3} \int_{0}^{1}\frac {x^{2}}{(1-x^{2})^{2/3}}dx$$
$$x=sin\theta, dx=cos\theta d\theta$$
$$\displaystyle I_{1}=\frac {2}{3} \int_{0}^{\pi/2}\frac {sin^{2}\theta cos\theta}{(cos\theta)^{4/3}}d\theta$$
$$\displaystyle I_{1}=\frac {2}{3} \int_{0}^{\pi/2}sin^{2}\theta (cos\theta)^{-1/3}d\theta$$
Ans: $$A$$
Evaluate $$\displaystyle \int_{0}^{\pi /2} \frac{dx}{2+\sin 2x}$$
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$$\displaystyle \frac{2\pi }{{3}}$$
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$$\displaystyle \frac{\pi }{{3}}$$
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$$\displaystyle \frac{2\pi }{{5}}$$
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None of these
Explanation
$$\displaystyle \int _{ 0 }^{ 2\pi }{ \frac { dx }{ 2+\sin { 2x } } } =\int _{ 0 }^{ 2\pi }{ \frac { dx }{ 2+\frac { 2\tan { x } }{ 1+\tan ^{ 2 }{ x } } } } dx$$
$$\displaystyle =\frac { 1 }{ 2 } \int _{ 0 }^{ 2x }{ \frac { \sec ^{ 2 }{ x } }{ \tan ^{ 2 }{ x } +\tan { x+1 } } dx } $$
Put $$\tan { x } =t\Rightarrow \sec ^{ 2 }{ x } dx=dt$$
$$\displaystyle I=\frac { 1 }{ 2 } \int { \frac { 1 }{ { { t }^{ 2 }+t+1 } } } dt.=\frac { 1 }{ 2 } \int { \frac { 1 }{ { \left( t+\frac { 1 }{ 2 } \right) }^{ 2 }-\frac { 3 }{ 2 } } } =\frac { 2\pi }{ 3 } $$
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Both Assertion and Reason are incorrect
Explanation
Reason is true
Assertion
Put $$\displaystyle x=\frac{1}{t}\Rightarrow dx=-\frac{1}{t^{2}}dt$$
$$\displaystyle l=-\int_{3}^{1/3}tcosec ^{99}\left ( \frac{1}{t}-t \right )\frac{1}{t^{2}}dt$$
$$\displaystyle l=-\int_{1/3}^{3}\frac{1}{t}cosec^{99}\left ( t-\frac{1}{t} \right )dt$$
$$\displaystyle l=-1$$
$$\Rightarrow 2l=0$$
$$\Rightarrow l=0$$
$$\displaystyle \int_{\tfrac{1}{\sqrt{3}}}^{0}\dfrac{dx}{\left ( 2x^{2}+1 \right )\sqrt{x^{2}+1}}$$
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$$\displaystyle - \tan^{-1} \dfrac{1}{2}$$
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$$\displaystyle \tan^{-1} 1$$
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$$\displaystyle - \tan^{-1} \dfrac{1}{3}$$
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$$\displaystyle \tan^{-1} \dfrac{1}{\sqrt 2}$$
Explanation
$$I=\displaystyle \int_{\frac{1}{\sqrt{3}}}^{0}\frac{dx}{\left ( 2x^{2}+1 \right )\sqrt{x^{2}+1}}$$
Substitute
$$x=\displaystyle \frac{1}{z}$$
$$\Rightarrow\displaystyle dx=-\frac{1}{z^2}dz$$
$$I=\displaystyle \int _{ \sqrt { 3 } }^{ \infty } \dfrac { -\dfrac { 1 }{ { z }^{ 2 } } dz }{ \left( \dfrac { 2 }{ { z }^{ 2 } } +1 \right) \sqrt { \dfrac { 1 }{ { z }^{ 2 } } +1 } } $$
$$=\displaystyle \int _{ \sqrt { 3 } }^{ \infty } \dfrac { -zdz }{ \left( 2+{ z }^{ 2 } \right) \sqrt { 1+{ z }^{ 2 } } } $$
Substitute
$$\sqrt { 1+{ z }^{ 2 } } =t$$
$$\Rightarrow z^2+1=t^2$$
$$\Rightarrow zdz=tdt$$
$$I=\displaystyle \int _{ 2 }^{ \infty } \frac { -tdt }{ \left( 1+{ t }^{ 2 } \right) t } $$
$$I=[\cot^{-1}t]_{2}^{\infty}$$
$$\Rightarrow I=-\cot^{-1}2$$
$$\Rightarrow I=-\tan^{-1}{\dfrac{1}{2}}$$
The value of definite integral $$\displaystyle \int_{\infty }^{0}\frac{Ze^{-z}}{\sqrt{1-e^{-2z}}}dz$$
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$$\displaystyle -\frac{\pi }{2}ln2$$
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$$\displaystyle \frac{\pi }{2}ln2$$
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$$\displaystyle -\pi ln2$$
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$$\displaystyle \pi ln\frac{1}{\sqrt{2}}$$
Explanation
$$\displaystyle l=\int_{\infty }^{0}\frac{ze^{-z}}{\sqrt{1-e^{-2z}}}dz$$ put $$\displaystyle e^{-z}=\sin \theta $$
$$\displaystyle l=-\int_{0}^{\pi /2}\frac{ln\left ( \sin \theta \right )\left ( -\cos \theta \right )d\theta }{\sqrt{1-\sin ^{2}\theta }}=\int_{0}^{\pi /2}ln\sin \theta d\theta $$
$$\displaystyle \frac{-\pi }{2}ln2$$
If $$\displaystyle \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } }\left( x-\alpha \right) } dx=0$$, then
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$$1<\alpha <2$$
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$$\alpha <2$$
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$$0<\alpha<1$$
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$$\alpha=0$$
Explanation
$$\displaystyle\because \int _{ 0 }^{ 1 }{ { e }^{ { x }^{ 2 } }\left( x-\alpha \right) } dx=0,$$
$$\therefore { e }^{ { x }^{ 2 } }\left( x-\alpha \right) $$ must be $$+ve$$ and $$-ve$$ both for $$x\in (0,1)$$
i.e. $${ e }^{ x }\left( x-\alpha \right) =0$$ for one $$x\in (0,1)$$
$$\therefore \alpha \in(0,1)$$
$$\displaystyle \int _{ 0 }^{ a }{ \frac { dx }{ a+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } } $$ is equal to
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$$\displaystyle \frac { \pi }{ 2 } +1$$
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$$\displaystyle \frac { \pi }{ 2 } -1$$
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$$\displaystyle 1-\frac { \pi }{ 2 } $$
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none of these
Explanation
Let $$\displaystyle I=\int _{ 0 }^{ a }{ \frac { dx }{ a+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } } =\int _{ 0 }^{ \pi /2 }{ \frac { a\cos { \theta } d\theta }{ a+a\cos { \theta } } } $$
By putting $$x=a\sin { \theta } \Rightarrow dx=a\cos { \theta } d\theta $$
$$\displaystyle I=\int _{ 0 }^{ \pi /2 }{ \frac { \cos { \theta } }{ 1+\cos { \theta } } d\theta } =\int _{ 0 }^{ \pi /2 }{ \left( 1-\frac { 1 }{ 1+\cos { \theta } } \right) d\theta } $$
$$\displaystyle =\int _{ 0 }^{ \pi /2 }{ d\theta } -\frac { 1 }{ 2 } \int _{ 0 }^{ \pi /2 }{ \sec ^{ 2 }{ \frac { \theta }{ 2 } } d\theta } $$
$$\displaystyle ={ \left[ \theta -\tan { \frac { \theta }{ 2 } } \right] }_{ 0 }^{ \pi /2 }=\frac { \pi }{ 2 } -\tan { \frac { \pi }{ 4 } } =\left( \frac { \pi }{ 2 } -1 \right) $$
What is $$\displaystyle\int _{ 1 }^{ 2 }{ \ln { x } dx } $$ equal to?
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0%
$$\ln { 2 } $$
0%
$$1$$
0%
$$\ln { \left( \dfrac { 4 }{ e } \right) } $$
0%
$$\ln { \left( \dfrac { e }{ 4 } \right) } $$
Explanation
Let, $$y=\ln x$$
then, $$ x={ e^{ y } }$$
Now at, $$ x=1,y=0$$
$$ x=2,y=\ln 2$$
$$\Rightarrow dx={ e }^{ y }.dy$$
Putting the values, we get
$$\displaystyle \int_{1}^{2} \ln x dx=\int _{ 0 }^{ \ln2 }{ { e }^{ y }.y.dy }$$
$$ =\displaystyle { \left[ y.{ e }^{ y } \right] }_{ 0 }^{\ln2 }-\int _{ 0 }^{\ln2 }{ { e }^{ y }dy }$$
$$ =2\ln 2-1$$
$$=2\ln 2-\ln(e)$$
$$ =\ln\left(\dfrac { 4 }{ e }\right)$$
Hence, C is correct.
$$\displaystyle\int_0^\pi \frac{1}{1\, +\,\sin x}\ dx$$ is equal to
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$$1$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
Explanation
Let $$I = \displaystyle\int_0^\pi \frac{1}{1\, +\,\sin x}\, dx\, $$
$$=\, \displaystyle \int_0^\pi \frac{1}{1\, +\,\displaystyle \frac{2 \tan \frac {x}{2}}{1\, +\, \tan^2 \frac {x}{2}}}\, dx$$
$$=\, \displaystyle\int_0^\pi \frac{\sec^2 \dfrac {x}{2}}{\left(1\, +\,\tan \dfrac {x}{2}\right)^2}\, dx$$
Put $$1+\tan$$ $$\dfrac{x}{2}\,= t\,\Rightarrow\,\dfrac{1}{2}\,\sec^2\,\dfrac{x}{2}\,=\,dt$$
$$\therefore\,I\,=\,\displaystyle\int_1^\infty\dfrac{2dt}{1\,+\,t^2}\,=-\dfrac2t\bigg]_1^\infty = 2$$
The value of $$\displaystyle\int _{ 0 }^{ { x }/{ 4 } }{ \dfrac { \sec { x } }{ { \left( \sec { x } +\tan { x } \right) }^{ 2 } } dx } $$ is
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$$1+\sqrt{2}$$
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$$-11+\sqrt{2}$$
0%
$$-\sqrt{2}$$
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None of these
Explanation
$$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \cfrac { \sec { x } }{ { \left( \sec { x } +\tan { x } \right) }^{ 2 } } } dx$$
$$=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \cfrac { { 1 }/{ \cos { x } } }{ { \left[ \cfrac { 1 }{ \cos { x } } +\cfrac { \sin { x } }{ \cos { x } } \right] }^{ 2 } } } dx$$
$$=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \cfrac { { \cos ^{ 2 }{ x } }/{ \cos { x } } }{ { \left( 1+\sin { x } \right) }^{ 2 } } } dx$$
$$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { \left( 1+\sin { x } \right) }^{ -2 } } \cos { x } dx$$
Take $$1+\sin x=t \implies \cos x dx=dt$$
When $$x=0, t=1$$ and $$x=\dfrac{\pi}{4}, t=1+\dfrac{1}{\sqrt{2}}$$
$$\therefore \quad I=\int_{1}^{1+\frac{1}{\sqrt{2}}} t^{-2}\ dt$$
$$\therefore \quad I=\left|\dfrac{t^{-1}}{-1}\right|_{1}^{1+\frac{1}{\sqrt{2}}}$$
$$=-\left[ \cfrac { 1 }{ 1+\frac { 1 }{ \sqrt { 2 } } } -1 \right] $$
$$=-\cfrac { \sqrt { 2 } }{ \sqrt { 2 } +1 } +1$$$$=\cfrac { -\sqrt { 2 } +\sqrt { 2 } +1 }{ \sqrt { 2 } +1 } =\cfrac { 1 }{ \sqrt { 2 } +1 } $$
$$=\cfrac { 1 }{ \sqrt { 2 } +1 } \ast \cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } -1 } =\cfrac { \sqrt { 2 } -1 }{ 2-1 } $$
$$=\sqrt { 2 } -1$$
$$\therefore $$ None of these is correct option
If $$\quad f(x)=\begin{cases} 2{ x }^{ 2 }+1,x\le 1 \\ 4{ x }^{ 2 }-1,x>1 \end{cases}$$, then $$\int _{ 0 }^{ 2 }{ f(x)dx } $$ is
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0%
$$10$$
0%
$$50/3$$
0%
$$1/3$$
0%
$$47/2$$
Explanation
$$\quad f(x)=\begin{cases} 2{ x }^{ 2 }+1,x\le 1 \\ 4{ x }^{ 2 }-1,x>1 \end{cases}$$
from 0 to 1 we have $$2x^2+1$$ and from 1 to 2 we have $$4x^2-1$$
$$=\int _{ 0 }^{ 1 }{ \left( 2{ x }^{ 2 }+1 \right) dx+\int _{ 1 }^{ 2 }{ \left( 4{ x }^{ 2 }-1 \right) dx } } $$
$$={ \left[ \frac { 2{ x }^{ 3 } }{ 3 } +x \right] }_{ 0 }^{ 1 }+{ \left[ \frac { 4{ x }^{ 3 } }{ 3 } -x \right] }_{ 1 }^{ 2 }$$
$$=\frac { 2 }{ 3 } +1-0-0+\frac { 4{ (2) }^{ 3 } }{ 3 } -2-\left( \frac { 4 }{ 3 } -1 \right) $$
$$=\frac { 2 }{ 3 } +1+\frac { 32 }{ 3 } -2-\frac { 4 }{ 3 } +1$$
$$=\frac { 2+32-4 }{ 3 } $$
$$=\dfrac{30}{3}$$
=10.
What is $$\displaystyle \int_0^1 {\frac{\tan^{-1}x}{1 + x^2} dx}$$ equal to ?
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$$\dfrac{\pi}{4}$$
0%
$$\dfrac{\pi}{8}$$
0%
$$\dfrac{\pi^2}{8}$$
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$$\dfrac{\pi^2}{32}$$
Explanation
Let $$I=\displaystyle \int _{ 0 }^{ 1 } \dfrac { {\tan^{ -1 } }x }{ { 1+x^{ 2 } } } dx$$
Let, $$y={ {\tan^{ -1 } }x }$$
$$\Rightarrow dy=\dfrac { 1 }{ { 1+x^{ 2 } } } dx$$
Now for $$ x=1, y=\dfrac { \pi }{ 4 } $$ and for $$x=0, y=0$$
Putting the values in the integral we have,
$$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ y.dy }$$
$$= { \left[\dfrac { y^{ 2 } }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }$$
$$ =\dfrac { \pi ^{ 2 } }{ 32 } $$
Hence, D is correct.
If $$\dfrac {dy}{dt} = ky$$ and $$k\neq 0$$, which of the following could be the equation of $$y$$?
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$$y = kx - 7$$
0%
$$y = 95e^{kt}$$
0%
$$y = 5 + ln\ k$$
0%
$$y = (x - k)^{2}$$
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$$y = \sqrt [k]{x}$$
Explanation
Given, $$\dfrac{dy}{dt} = ky$$ and $$k \ne 0$$
We have $$\dfrac{dy}{y} = kdt$$
Apply integral on both sides, so we get
$$\displaystyle \int \dfrac{dy}{y} = \int kdt$$
$$\Rightarrow \ln(y) = kt+c$$ (where $$c$$ is constant)
$$\Rightarrow y = {e}^{c}{e}^{kt} = a{e}^{kt}$$ ($$a = {e}^{c}$$)
The possible solution is $$y = 95{e}^{kt}$$
Solve $$\int_{0}^{\dfrac {\pi}{2}}\sqrt {\sin \phi}\cos^{5}\phi d\phi$$.
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$$\dfrac{64}{231}$$
0%
$$\dfrac{24}{231}$$
0%
$$\dfrac{54}{231}$$
0%
None of these
Explanation
Let
$$\begin{matrix} I=\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \sqrt { \sin \phi } { { \cos }^{ 5 } }\phi d\phi } \\ =\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \sqrt { \sin \phi } { { \cos }^{ 4 } }\phi \cos \phi d\phi } \\ =\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \sqrt { \sin \phi } { { \left( { 1-{ { \sin }^{ 2 } }\phi } \right) }^{ 2 } }\cos \phi d\phi } \\ put\, \, \, t=\sin \phi \\ \dfrac { { dt } }{ { d\phi } } =\cos \phi \\ dt=\cos \phi d\phi \\ When\, \, \phi =0,\, \, t=0\, \, when\, \, \phi =\dfrac { \pi }{ 2 } ,\, t=1 \\ \therefore I=\int _{ 0 }^{ 1 }{ \sqrt { t } { { \left( { 1-{ t^{ 2 } } } \right) }^{ 2 } }dt } \\ =\int _{ 0 }^{ 1 }{ \sqrt { t } \left( { 1-2{ t^{ 2 } }+{ t^{ 4 } } } \right) dt } \\ =\int _{ 0 }^{ 1 }{ { t^{ \dfrac { 1 }{ 2 } } }+{ t^{ \dfrac { 9 }{ 2 } } }-2{ t^{ \dfrac { 5 }{ 2 } } }dt } \\ =\left[ { \dfrac { 2 }{ 3 } { t^{ \dfrac { 3 }{ 2 } } }+\dfrac { 2 }{ { 11 } } { t^{ \dfrac { { 11 } }{ 2 } } }-\dfrac { 4 }{ 7 } { t^{ \dfrac { 7 }{ 2 } } } } \right] _{ 0 }^{ 1 } \\ =\dfrac { 2 }{ 3 } +\dfrac { 2 }{ { 11 } } -\dfrac { 4 }{ 7 } \\ =\dfrac { { 154+42-132 } }{ { 3\times 11\times 7 } } =\dfrac { { 64 } }{ { 231 } } \\ \end{matrix}$$
$$\displaystyle\int _{ 0 }^{ { \sqrt { \pi } }/{ 2 } }{ 2{ x }^{ 3 }\sin { \left( { x }^{ 2 } \right) } dx } $$ is equal to
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$$\dfrac { 1 }{ \sqrt { 2 } } \left( 1+\dfrac { \pi }{ 4 } \right) $$
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$$\dfrac { 1 }{ \sqrt { 2 } } \left( 1-\dfrac { \pi }{ 4 } \right) $$
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$$\dfrac { 1 }{ \sqrt { 2 } } \left( \dfrac { \pi }{ 2 } -1 \right) $$
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$$\dfrac { 1 }{ \sqrt { 2 } } \left( 1-\dfrac { \pi }{ 2 } \right) $$
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$$\dfrac { 1 }{ \sqrt { 2 } } \left( \dfrac { \pi }{ 4 } -1 \right) $$
Explanation
$$I=\displaystyle\int _{ 0 }^{ { \sqrt { \pi } }/{ 2 } }{ 2{ x }^{ 3 }\sin { \left( { x }^{ 2 } \right) } dx } $$
$$=\displaystyle\int _{ 0 }^{ { \sqrt { \pi } }/{ 2 } }{ 2{ x }^{ 2 }\cdot x\sin { \left( { x }^{ 2 } \right) } dx } $$
Put $${ x }^{ 2 }=t$$
$$\Rightarrow 2xdx=dt$$
Also, when $$x=0$$, then $$t=0$$
and when $$x=\dfrac { \sqrt { \pi } }{ 2 } $$, then $$t=\dfrac { \pi }{ 4 }$$
$$\Rightarrow I=\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ \underset { I }{ t } \underset { II }{ \sin { t } } dt } $$
$$={ \left[ t\left( -\cos { t } \right) \right] }_{ 0 }^{ { \pi }/{ 4 } }-\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ -\cos { t } \left( 1 \right) dt } $$
$$={ \left[ -t\cos { t } \right] }_{ 0 }^{ { \pi }/{ 4 } }+\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ \cos { t } dt } $$
$$={ \left[ -t\cos { t } +\sin { t } \right] }_{ 0 }^{ { \pi }/{ 4 } }$$
$$=\left[ -\dfrac { \pi }{ 4 } \cdot \dfrac { 1 }{ \sqrt { 2 } } +\dfrac { 1 }{ \sqrt { 2 } } \right] $$
$$=\dfrac { 1 }{ \sqrt { 2 } } \left( 1-\dfrac { \pi }{ 4 } \right) $$
$$\displaystyle\int { \cfrac { 1 }{ 7 } \sin { \left( \cfrac { x }{ 7 } +10 \right) } dx } $$ is equal to
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$$\cfrac { 1 }{ 7 } \cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$$
0%
$$-\cfrac { 1 }{ 7 } \cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$$
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$$-\cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$$
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$$-7\cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$$
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$$\cos { \left( x+70 \right) } +C$$
Explanation
Let $$I=\int { \cfrac { 1 }{ 7 } \sin { \left( \cfrac { x }{ 7 } +10 \right) } dx } $$
$$=\cfrac { 1 }{ 7 } \int { \sin { \left( \cfrac { x }{ 7 } +10 \right) } } =\cfrac { 1 }{ 7 } \cfrac { -\cos { \left( \cfrac { x }{ 7 } +10 \right) } }{ \cfrac { 1 }{ 7 } } $$
$$=-\cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$$
If $$ \int _0^1 xdx = \dfrac {\pi}{4} - \dfrac {1}{2} ln 2 $$ then the value of definite integral $$ \int _0^1 \tan^{-1} (1-x+x^2) dx $$ equals :
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$$ ln2 $$
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$$ \dfrac {\pi}{4} + ln 2 $$
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$$ \dfrac {\pi}{4} - ln2 $$
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$$ 2 ln 2 $$
The value of $$\int_{0}^{1} \dfrac {8\log (1 + x)}{1 + x^{2}} dx$$ is
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$$\dfrac {\pi}{2}\log 2$$
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$$\pi\log 2$$
0%
$$2\pi\log 2$$
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None of these
Explanation
Let
$$\begin{matrix} I=\int _{ 0 }^{ 1 }{ \frac { { 8\log \left( { 1+x } \right) } }{ { 1+{ x^{ 2 } } } } } dx \\ \, \, \, \, =8\int _{ 0 }^{ 1 }{ \frac { { \log \left( { 1+x } \right) } }{ { 1+{ x^{ 2 } } } } } dx \\ put\, \, \, x=\tan \theta \\ \, \, \, \, \, \Rightarrow dx={ \sin ^{ 2 } }\theta d\theta \\ x,\, \frac { \pi }{ 4 } ,\, \, \, \, x=0 \\ \quad \quad \quad \quad I=8\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { { \log \left( { 1+x } \right) } }{ { { { \sin }^{ 2 } }\theta } } } { \sin ^{ 2 } }\theta d\theta \\ \, \, \, \, \, \quad \quad \quad \quad =8\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \log \left( { 1+\tan x } \right) } dx \\ \quad \quad =8\times \frac { \pi }{ 8 } \log 2 \\ \, \, \, =\pi \log 2 \\ \end{matrix}$$
Hence option (A) is true.
If $$\displaystyle\int { \sqrt { 1+\sin { x } } \cdot f\left( x \right) dx } =\dfrac { 2 }{ 3 } { \left( 1+\sin { x } \right) }^{ { 3 }/{ 2 } }+C$$, then $$f\left( x \right) $$ is equal to
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$$\cos { x } $$
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$$\sin { x } $$
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$$\tan { x } $$
0%
$$1$$
Explanation
$$\int { \sqrt { 1+\sin { x } } } .f\left( x \right) dx=\dfrac { 2 }{ 3 } { (1+\sin { x } ) }^{ { 3 }/{ 2 } }+c$$
On differentiating both sides, we get
$$\sqrt { 1+\sin { x } } .f\left( x \right) =\dfrac { 2 }{ 3 } .\dfrac { 3 }{ 2 } { (1+\sin { x } ) }^{ { 1 }/{ 2 } }.\cos { x } +0\\ \Rightarrow f\left( x \right) =\cos { x } $$
$$\displaystyle \int _0^{\pi /2} f(\sin 2x)\sin x\, dx = K\int_0^{\pi/2} f(\cos 2x) \cos x\,dx$$ where $$k$$ equals to
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$$2$$
0%
$$4$$
0%
$$\sqrt{2}$$
0%
$$2\sqrt{2}$$
Explanation
$$\displaystyle \int_0^{\dfrac{\pi}{2}} \sin 2x \sin x dx = K \int_0^{\dfrac{\pi}{2}} \cos 2x \cos x dx$$
$$\displaystyle \int_0^{\dfrac{\pi}{2}} 2 \sin x \cos x \sin x dx = K \int_0^{\dfrac{\pi}{2}} (1 - 2 \sin^2 x) \cos x dx$$
$$\displaystyle \int_0^{\dfrac{\pi}{2}} 2 \sin^2 x \cos x dx = K \left[\int_0^{\dfrac{\pi}{2}} \cos x dx - \int_0^{\dfrac{\pi}{2}} 2 \sin^2 x \cos x dx \right]$$
$$\left\{\dfrac{2 \sin^3 x}{3} \right\}_0^{\dfrac{\pi}{2}} = K \left[\{\sin x\}_0^{\dfrac{\pi}{2}} - \left\{\dfrac{2 \sin^3 x}{3} \right\}_0^{\dfrac{\pi}{2}}\right]$$
$$\dfrac{2}{3} = K \left[1 - \dfrac{2}{3} \right]$$
$$K = 2$$
$$\int_{0}^{1}{\frac{dx}{x\sqrt{x}}}$$
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$$2$$
0%
$$-2$$
0%
$$1$$
0%
$$3$$
Explanation
$$\int _{ 0 }^{ 1 }{ \cfrac { 1 }{ x\sqrt { x } } dx } \\ \int _{ 0 }^{ 1 }{ { x }^{ \cfrac { -3 }{ 2 } }dx } ={ [\cfrac { { x }^{ \cfrac { -3 }{ 2 } +1 } }{ \cfrac { -3 }{ 2 } +1 } ] }_{ 0 }^{ 1 }\\ ={ [\cfrac { { x }^{ \cfrac { -1 }{ 2 } } }{ \cfrac { -1 }{ 2 } } ] }_{ 0 }^{ 1 }=\cfrac { 1-0 }{ \cfrac { -1 }{ 2 } } =-2$$
What is $$\displaystyle \int_{0}^{2\pi}\sqrt {1 + \sin \dfrac {x}{2}}dx$$ equal to?
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$$8$$
0%
$$4$$
0%
$$2$$
0%
$$0$$
Explanation
$$1+\sin\frac { x }{ 2 } ={ \sin }^{ 2 }\frac { x }{ 4 } +{ \cos }^{ 2 }\frac { x }{ 4 } +2\sin\frac { x }{ 4 } \cos\frac { x }{ 4 } \\ { (\sin\frac { x }{ 4 } +\cos\frac { x }{ 4 } ) }^{ 2 }\\ \sqrt { 1+\sin\frac { x }{ 2 } } =\sqrt { { (\sin\frac { x }{ 4 } +\cos\frac { x }{ 4 } ) }^{ 2 } } =\sin\frac { x }{ 4 } +\cos\frac { x }{ 4 } $$
$$\displaystyle \int _{ 0 }^{ 2\pi }{ \left(\sin\frac { x }{ 4 } +\cos\frac { x }{ 4 }\right )dx } =4\left(-\cos\frac { x }{ 4 } +\sin\frac { x }{ 4 }\right )+c\\ After\quad putting\quad the\quad limit\quad we\quad get\\\Rightarrow 4(1-(-1))=8\\ $$
So correct answer will be option A
Let $$I=\displaystyle \int _{ \pi /4 }^{ \pi /3 }{ \cfrac { \sin { x } }{ x } } dx$$. Then?
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$$\cfrac { 1 }{ 2 } \le I\le 1\quad $$
0%
$$4\le I\le 2\sqrt { 30 } $$
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$$\cfrac { \sqrt { 3 } }{ 8 } \le I\le \cfrac { \sqrt { 2 } }{ 6 } $$
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$$1\le I\le \cfrac { 2\sqrt { 3 } }{ \sqrt { 2 } } $$
Explanation
$$I=\displaystyle \int^{\tfrac{\pi}{3}}_{\tfrac{\pi}{4}}\dfrac{sinx}{x}dx$$
$$\dfrac{sinx}{x}$$ is a decreasing function in given interval
difference of limits $$=\dfrac{\pi}{3}-\dfrac{\pi}{4}=\dfrac{\pi}{12}$$
so, $$\dfrac{\pi}{12}\cdot\dfrac{sin\dfrac{\pi}{3}}{\dfrac{\pi}{3}} \le I \le \dfrac{\pi}{12}
\cdot\dfrac{sin\dfrac{\pi}{4}}{\dfrac{\pi}{4}}$$
$$\dfrac{\sqrt3}{8} \leq I \leq \dfrac{\sqrt2}{6}$$
$$\int { { \left( ex \right) }^{ x }\left( 2+\log { x } \right) } dx=....+c,x\in { R }^{ + }-\left\{ 1 \right\} $$
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0%
$${ x }^{ x }$$
0%
$${ \left( ex \right) }^{ x }\quad $$
0%
$${ e }^{ x }$$
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$$\left( 1+\log { x } \right) { \left( ex \right) }^{ x }$$
Explanation
$$\int (ex)^x(2+\log x)dx=\int (ex)^xdx+\int e^x.x^x(1+\log x)dx$$
Integrating by parts ,
$$\Rightarrow \int(ex)^x +e^x(\int x^x(1+\log x)) -\int(\dfrac{d}{dx}(e^x).(\int x^x(1+\log x)dx))dx$$
$$\Rightarrow \int (ex)^x+ e^xx^x-\int (ex)^x$$
$$\Rightarrow (ex)^x$$
If $$I=\displaystyle\overset{1}{\underset{0}{\displaystyle\int}}x(1-x)^{1/2}dx$$ and $$60I+k=25$$ then $$k=$$ _________. $$(k\in R)$$.
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0%
$$9$$
0%
$$25$$
0%
$$60$$
0%
$$41$$
Explanation
We know,
$$\displaystyle \int_0^af(x)dx=\int_0^af(a-x)dx$$
$$\therefore I=\displaystyle \int_0^1x(1-x)^{1/2}dx=\int_0^1(1-x)x^{1/2}dx$$
$$\Rightarrow I=\displaystyle \int_0^1(x^{1/2}-x^{3/2})dx$$
$$\Rightarrow I=\left[\dfrac{2}{3}x^{3/2}-\dfrac{2}{5}x^{5/2} \right]_0^1$$
$$\Rightarrow I=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{4}{15}$$
$$60I+k=25$$
$$k=25-60\times \dfrac{4}{15}=9$$
$$\displaystyle \int_1^{32}\dfrac{dx}{x^{1/5}\sqrt{1+x^{4/5}}}$$
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0%
$$\dfrac{2}{5}(\sqrt{17}+\sqrt{2})$$
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$$\dfrac{2}{5}(\sqrt{17}-\sqrt{2})$$
0%
$$\dfrac{5}{2}(\sqrt{17}-\sqrt{2})$$
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$$\dfrac{5}{2}(\sqrt{17}+\sqrt{2})$$
$$\int _{ 0 }^{ \pi /2 }{ \sin { 2x } .\sin { x } } dx=.....$$
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0%
$$\cfrac{1}{3}$$
0%
$$\cfrac{2}{3}$$
0%
$$-\cfrac{2}{3}$$
0%
$$\cfrac{4}{3}$$
Explanation
$$I=\int _{ 0 }^{ \pi /2 }{ \sin { 2x } .\sin { x } } dx$$
$$=\int _{ 0 }^{ \pi /2 }{ 2\sin { x } \cos { x } .\sin { x } } dx=2\int _{ 0 }^{ \pi /2 }{ \sin ^{ 2 }{ x } \cfrac { d }{ dx } \sin { x } } dx$$
$$=\cfrac { 2 }{ 3 } { \left[ \sin ^{ 2 }{ x } \right] }_{ 0 }^{ \pi /2 }=\cfrac { 2 }{ 3 } (1-0)=\cfrac { 2 }{ 3 } $$
Evaluate $$\displaystyle \int_{-2\pi}^{5\pi} \cot^{-1} (\tan x) dx$$.
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0%
$$0$$
0%
$$-1$$
0%
$$1$$
0%
$$2$$
Explanation
$$I=\displaystyle \int_{2\pi} ^{5\pi}\cot ^{-1}(\tan x)dx$$ ...... $$(1)$$
Applying property of limits of integral formula ;
$$\Rightarrow I=\displaystyle \int_{2\pi} ^{5\pi}\cot ^{-1}(\tan (7\pi - x)) dx$$
$$\Rightarrow I=\displaystyle \int_{2\pi} ^{5\pi}\cot ^{-1}(-\tan x)dx$$
$$\Rightarrow I=\displaystyle -\int_{2\pi} ^{5\pi}\cot ^{-1}(\tan x)dx$$ ...... $$(2)$$
Adding both equations;
$$\Rightarrow 2I=0$$
$$\Rightarrow I=0$$
$$\Rightarrow \displaystyle \int_{2\pi} ^{5\pi}\cot ^{-1}(\tan x)dx=0$$
$$\int {\left( 3.{ x }^{ 2 }.\tan ^{ -1 }{ x } +\cfrac { { x }^{ 3 } }{ 1+{ x }^{ 2 } } \right) } dx=....+c$$.
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$${ x }^{ 3 }\tan ^{ -1 }{ x } $$
0%
$$\cfrac { { x }^{ 3 } }{ 3 } \tan ^{ -1 }{ x } $$
0%
$${ x }^{ 2 }\tan ^{ -1 }{ x } $$
0%
$$\cfrac { { x }^{ 2 } }{ 2 } \tan ^{ -1 }{ x } $$
Explanation
Given $$\int (3x^2.\tan^{-1}x+\dfrac{x^3}{1+x^2})dx$$=
$$\int (3x^2.\tan^{-1}x.dx)+\int(\dfrac{x^3}{1+x^2})dx$$
Integrating by parts
$$\int 3x^2tan^{-1}x.dx+x^3\int\dfrac{1}{1+x^2}dx-\int( \dfrac{d}{dx}(x^3)(\int \dfrac{1}{1+x^2}dx)dx)$$
$$\dfrac{d}{dx}x^3=3x^2;\int \dfrac{1}{1+x^2}=tan^{-1}x$$
$$\Rightarrow\int 3x^2tan^{-1}x.dx + x^3 \tan^{-1}x-\int 3x^2tan^{-1}x.dx$$
$$\Rightarrow x^3\tan^{-1}x+c$$
If $$\displaystyle \int _0^{\pi/2} \sin x \cos x dx $$ is equal to:
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$$\dfrac 1 2$$
0%
$$\dfrac 14$$
0%
$$2$$
0%
$$1$$
Explanation
$$\displaystyle \int _0^{\pi/2} \sin x \cos x dx $$
$$\sin x=t\implies \cos x dx=dt$$
$$x\to 0\to \dfrac \pi 2$$
$$t\to 0\to 1$$
$$\Rightarrow \displaystyle \int _0^{\pi/2} t dt$$
$$\Rightarrow\left.\dfrac {t^2}2\right|^1_0$$
$$\Rightarrow\dfrac 12-0=\dfrac 12$$
$$I= \int \frac{x+2}{(x+1)^2}dx;$$ then I is equal to
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$$\log (x+1)+\dfrac{1}{x+1}+c$$
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$$\log (x+2)-\dfrac{1}{x+1}+c$$
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$$\log (1+x)-\dfrac{1}{x+1}+c$$
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$$\log (x+2)+\dfrac{1}{x+1}+c$$
Explanation
$$\int { \dfrac { x+2 }{ { \left( x+1 \right) }^{ 2 } } } dx$$
$$=\int { \dfrac { x+1+1 }{ { \left( x+1 \right) }^{ 2 } } } dx$$
$$=\int { \left( \dfrac { x+1 }{ { \left( x+1 \right) }^{ 2 } } +\dfrac { 1 }{ { \left( x+1 \right) }^{ 2 } } \right) } dx$$
$$=\int { \dfrac { 1 }{ x+1 } } dx+\int { \dfrac { 1 }{ { \left( x+1 \right) }^{ 2 } } } dx$$ $$\left[ \because \int { { x }^{ n }dx=\dfrac { { x }^{ n+1 } }{ n+1 } +c } \right] $$
$$=\log \left(x+1\right) +\dfrac { { \left( x+1 \right) }^{ -2+1 } }{ -2+1 } +c$$ $$\left[ \because \int { \dfrac { 1 }{ x } dx=\log { x+c } } \right] $$
$$=\log\left(x+1\right) +\left(-1\right) \dfrac{1}{\left(x+1\right)}+c$$
$$=\log\left(x+1\right)-\dfrac{1}{x+1}+c$$
Hence, the answer is $$\log\left(x+1\right)-\dfrac{1}{x+1}+c.$$
If $$\int _{ 0 }^{ \pi /3 }{ \dfrac { \cos { x } }{ 3+4\sin { x } } dx } =k\log { \left( \dfrac { 3+2\sqrt { 3 } }{ 3 } \right) }$$, then, $$k$$ is equal to ?
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$$\dfrac{1}{2}$$
0%
$$\dfrac{1}{3}$$
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$$\dfrac{1}{4}$$
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$$\dfrac{1}{8}$$
Explanation
$$\sin x=t \Rightarrow dx.\cos x=dt$$
$$\displaystyle I=\int_{0}^{\frac{\pi }{3}}\frac{dt}{3+4t}$$
$$\displaystyle =\frac{1}{4}[ln(3+4t)]^{\frac{\pi }{3}}$$
When $$x=0\Rightarrow t=0$$
When $$\displaystyle x=\frac{\pi }{3}\Rightarrow t=\frac{\sqrt{3}}{2}$$
$$\displaystyle =\frac{1}{4}[ln(3+4t)]^{\frac{\sqrt{3}}{2}}-\frac{1}{4}[ln(3+4t)]^{0}$$
$$\displaystyle =\frac{1}{4}ln(3+2\sqrt{3})-\frac{1}{4}ln(3)$$
$$\displaystyle I=\frac{1}{4}ln\left[\frac{3+2\sqrt{3}}{3}\right]$$
$$K=\dfrac{1}{4}$$
Find proper substitution
$$\int _{ 0 }^{ 1 }{ \dfrac { { e }^{ -x } }{ 1+{ e }^{ -x } } dx }$$
Report Question
0%
$$1+{ e }^{ -x }\rightarrow t$$
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$$-{ e }^{ -x }dx\rightarrow dt$$
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$$-\int _{ 0 }^{ 1 }{ \dfrac { dt }{ t } }$$
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$$-\int _{ 0 }^{ 1 }{ ln\left| t \right| }$$
Explanation
$$\int _{ 0 }^{ 1 }{ \cfrac { { e }^{ -x } }{ 1+{ e }^{ -x } } dx } $$
$$1+{ e }^{ -x }=t$$
$$\Rightarrow { -e }^{ -x }dx=dt$$
$$\int _{ 0 }^{ 1 }{ \cfrac { -dt }{ t } } \quad \therefore 1+{ e }^{ -x }=t$$
0:0:1
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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