CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 9 - MCQExams.com

Let $$\displaystyle I_1 = \int_1^2 \frac{1}{\sqrt{1 + x^2}} dx$$ and $$I_2 \displaystyle = \int_1^2 \frac{1}{x} dx$$. Then
  • $$I_1 > I_2$$
  • $$I_2 > I_1$$
  • $$I_1 = I_2$$
  • $$I_1 > 2 I_2$$
The value (s) of  $$\displaystyle \int_{0}^{1}\frac{x^{4}\left ( 1-x \right )^{4}}{1+x^{2}} dx $$ is (are)
  • $$\displaystyle \frac{22}{7}-\pi $$
  • $$\displaystyle \frac{2}{105}$$
  • $$0$$
  • $$\displaystyle \frac{71}{15}-\frac{3\pi}{2}$$
If $$\displaystyle I= \int_{0}^{1}\frac{x dx}{8+x^{3}}$$ then the smallest interval in which $$I$$ lies is
  • $$\displaystyle \left ( 0,\frac{1}{8} \right )$$
  • $$\displaystyle \left ( 0,\frac{1}{9} \right )$$
  • $$\displaystyle \left ( 0,\frac{1}{10} \right )$$
  • $$\displaystyle \left ( 0,\frac{1}{7} \right )$$
If $$\displaystyle I_{n} =\int_{0}^{\frac{\pi }{4}}\tan ^{n}xdx$$ 

then $$\displaystyle \frac{1}{I_{2}+I_{4}},\frac{1}{I_{3}+I_{5}},\frac{1}{I_{4}+I_{6}}$$ are in?
  • $$A.P$$
  • $$H.P$$
  • $$G.P$$
  • None of these
If $$\displaystyle I_{t}=\int_{0}^{\dfrac{\pi }{2}}\frac{\sin^{2}tx}{\sin^{2}x}dx$$ then ,$$I_{1},I_{2},I_{3}$$ are in
  • A.P.
  • H.P.
  • G.P.
  • None of these
The value of $$ \displaystyle \int_{0}^{\pi /2}\sin \theta \log \left ( \sin \theta \right )\:d\theta  $$ equals
  • $$ \displaystyle \log_{e}\left ( \frac{1}{e} \right ) $$
  • $$ \displaystyle \log _{2}e $$
  • $$ \displaystyle \log_{e}{2}-1 $$
  • $$ \displaystyle \log_{e}\left ( \frac{e}{2} \right ) $$
If $$f(x) =\displaystyle  \underset{1}{\overset{x}{\int}} \dfrac{\tan^{-1} t}{t} dt \, \, \forall \in R$$, then the value of $$f(e^2) - f \left (\dfrac{1}{e^2}\right) $$is 
  • $$0$$
  • $$\dfrac{\pi}{2}$$
  • $$\pi$$
  • $$2 \pi$$
The value(s) of $$\int _{ 0 }^{ 1 }{ \cfrac { { x }^{ 4 }{ \left( 1-x \right)  }^{ 4 } }{ 1+{ x }^{ 2 } }  } dx$$ is (are)
  • $$\cfrac{22}{7}-\pi$$
  • $$\cfrac{2}{105}$$
  • $$0$$
  • $$\cfrac{71}{15}-\cfrac{3\pi}{2}$$
If $$\displaystyle 2f(x)+f(-x)=\frac{1}{x}\sin{\left(x-\frac{1}{x}\right)}$$, then the value of $$\displaystyle\int_{\frac{1}{e}}^{e}{f(x)dx}$$, is
  • $$1$$
  • $$0$$
  • $$e$$
  • $$-1$$
If $$\displaystyle I= \int_{1/\pi }^{\pi }\frac{1}{x}\cdot \sin \left ( x-\frac{1}{x} \right )dx$$ then I is equal to
  • $$0$$
  • $$\displaystyle \pi $$
  • $$\displaystyle \pi -\frac{1}{\pi }$$
  • $$\displaystyle \pi +\frac{1}{\pi }$$
If $$x$$ satisfies the equation $$\displaystyle\left(\int_0^1{\frac{dt}{t^2+2t\cos{\alpha}+1}}\right)x^2-\left(\int_{-3}^3{\frac{t^2\sin{2t}}{t^2+1}dt}\right)x-2=0$$ 
for $$(0<\alpha<\pi)$$
then the value of $$x$$ is?
  • $$\displaystyle\pm\sqrt{\frac{\alpha}{2\sin{\alpha}}}$$
  • $$\displaystyle\pm\sqrt{\frac{2\sin{\alpha}}{\alpha}}$$
  • $$\displaystyle\pm\sqrt{\frac{\alpha}{\sin{\alpha}}}$$
  • $$\displaystyle\pm2\sqrt{\frac{\sin{\alpha}}{\alpha}}$$
The tangent to the graph of the function $$\displaystyle y = f(x)$$ at the point with abscissa x = a forms with the x-axis an angle of $$\displaystyle \pi/3$$ and at the point with abscissa x = b at an angle of $$\displaystyle \pi/4$$, then the value of the integral,
$$\displaystyle \int_{a}^{b} f'(x).f''(x)dx$$ is equal to
  • 1
  • 0
  • $$\displaystyle -\sqrt 3$$
  • -1
Let $$\displaystyle F\left ( x \right )=f\left ( x \right )+f\left ( \frac{1}{x} \right )$$ where $$\displaystyle f\left ( x \right )=\int_{1}^{x}\frac{\log t}{1+t}dt$$ 
Then $$F(e)$$ is equal to?
  • $$1$$
  • $$2$$
  • $$1/2$$
  • $$0$$
The value of $$\displaystyle \int _{ 0 }^{ \pi /2 }{ \frac { d\theta  }{ 5+3\cos { \theta  }  }  }$$ is?
  • $$\displaystyle \tan ^{ -1 }{ \frac { 1 }{ 2 }  } $$
  • $$\displaystyle \tan ^{ -1 }{ \frac { 1 }{ 3 }  } $$
  • $$\displaystyle \frac{1}{2}\tan ^{ -1 }{ \frac { 1 }{ 2 }  } $$
  • $$\displaystyle \frac{1}{3}\tan ^{ -1 }{ \frac { 1 }{ 3 }  } $$
Evaluate : $$\displaystyle \underset{0}{\overset{\infty}{\int}} \dfrac{dx}{(1 + x^2)^4}$$
  • $$\dfrac{\pi}{32}$$
  • $$\dfrac{3 \pi}{32}$$
  • $$\dfrac{5 \pi}{32}$$
  • $$\dfrac{7 \pi}{32}$$
Let $$ f:R\rightarrow R^{+}$$ and $$I_{I}=\int^{k}_{1-k}\,xf(x(1-x))\,dx,I_2=\int^{k}_{1-k}f(x(1-x))\,dx$$ where $$2k-1>0$$. Then $$\dfrac{I_I}{I_2}$$ is 
  • 2
  • k
  • 1/2
  • 1
The value of $$\int _{ 0 }^{ \infty  }{ \cfrac { \log { x }  }{ { a }^{ 2 }+{ x }^{ 2 } }  } $$
  • $$\cfrac{2\pi\log{a}}{a}$$
  • $$\cfrac{\pi\log{a}}{2a}$$
  • $$\pi \log{a}$$
  • $$0$$
$$\displaystyle \int_{-1}^{1} \cot^{-1} \left(\dfrac{x+x^{3}}{1+x^{4}}\right)dx$$ is equal to 
  • $$2\pi$$
  • $$\dfrac{\pi}{2}$$
  • $$0$$
  • $$\pi$$
$$\displaystyle \int_{0}^{\pi /2}\sin x\log \left ( \sin x \right )dx= $$
  • $$\displaystyle \log _{e}e$$
  • $$\displaystyle \log _{e}2$$
  • $$\displaystyle \log _{e}\left ( e/2 \right )$$
  • $$\displaystyle \log _{e}\left ( 2/e \right )$$
$$\displaystyle \int_{0}^{\infty}\frac{1}{1+x^{n}}dx,\:\forall\:n\:> 1$$ is equal to?
  • $$\displaystyle2 \int_{0}^{\infty}\frac{1}{1+x}dx$$
  • $$\displaystyle \int_{-\infty}^{\infty}\frac{1}{1+x^{n}}dx$$
  • $$\displaystyle \int_{1}^{\infty}\frac{dx}{(x^{n}-1)^{1/n}}$$
  • $$\displaystyle \int_{0}^{1}\frac{1}{(1-x^{n})^{1/n}}$$
If $$\displaystyle I_1=\int_{0}^{\pi /2}\frac{x}{\sin x}dx $$ and $$\displaystyle I_2=\int_{0}^{\pi /2}\frac{\tan ^{-1}x}{x}dx, $$ then $$\displaystyle \frac{I_{1}}{I_{2}}= $$ 

  • $$\displaystyle \frac{1}{2} $$
  • 1
  • 2
  • $$\displaystyle \frac{\pi }{2} $$
$$\displaystyle If \int_{-1}^{1}\frac{g\left ( x \right )}{1+t^{2}}dt= f\left ( x \right ) , where,  g\left ( x \right )= \sin x$$ , then $$ {f}'\left ( \frac{\pi }{3} \right )$$ equals

  • $$\displaystyle\frac{\pi }{4}$$
  • does not exist
  • $$ \displaystyle \frac{\pi \sqrt{3}}{4} $$
  • None of these
The value of $$\displaystyle \int_{1}^{1/e}f(x)dx+\int_{1}^{e}f(x)dx$$ where $$f(x)$$ is given as $$\displaystyle \frac{log\:x}{1+x}$$ equals 
  • $$0$$
  • $$\displaystyle -\frac{1}{2}$$
  • $$\displaystyle \frac{1}{2}$$
  • $$1$$
The value of $$\displaystyle \int_{1/e}^{\tan x}\displaystyle \frac{t}{1+t^{2}}\, dt\, +\, \displaystyle \int_{1/e}^{\cot x}\displaystyle \frac{dt}{t\left ( 1+t^{2} \right )}$$ is
  • $$1/2$$
  • $$1$$
  • $$\pi /4$$
  • none of these
Evaluate : $$\displaystyle \int_{-\frac{1}{\sqrt2}}^{\frac{1}{\sqrt2}}\frac{x^{8}}{1-x^{4}}\times \left [ \sin ^{-1}\left ( 1-2x^{2} \right ) +\cos ^{-1}\left ( 2x\sqrt{1-x^{2}} \right )\right ]dx$$
  • $$\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}\right ]$$
  • $$\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}+\frac{21}{10\sqrt{2}}\right ]$$
  • $$\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}\right ]$$
  • $$\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}+\frac{21}{10\sqrt{2}}\right ]$$
Solve :-
$$\int_0^1 {\frac{{dx}}{{{{({x^2} + 1)}^{3/2}}}} = } $$

  • $$1$$
  • $$\frac{1}{{\sqrt 2 }}$$
  • $$\frac{1}{2}$$
  • $$\sqrt 2 $$
Solve:-
$$\int\limits_0^1 {\frac{{dx}}{{{{({x^2} + 1)}^{3/2}}}}} $$
  • $$1$$
  • $$\sqrt 2 $$
  • $$1/2$$
  • $$\dfrac{1}{\sqrt 2}$$
If $$I= \displaystyle \int_{0}^{1/\sqrt{3}}\displaystyle \frac{dx}{\left ( 1+x^{2} \right )\sqrt{1-x^{2}}}$$ then $$I$$ is equal to
  • $$\pi /2$$
  • $$\pi /2\sqrt{2}$$
  • $$\pi /4\sqrt{2}$$
  • $$\pi /4$$
$$\displaystyle \int_{0}^{\pi /3}\frac{\cos \theta }{3+4\sin \theta }d\theta =\lambda \log \frac{3+2\sqrt{3}}{3}$$ then $$\displaystyle \lambda $$ equals
  • $$\dfrac{1}{2}$$
  • $$\dfrac{1}{3}$$
  • $$\dfrac{1}{4}$$
  • $$\dfrac{1}{8}$$
The value of $$\displaystyle \int ^{\tan x}_{1/e}\displaystyle \frac{t\, dt}{1+t^{2}}+\displaystyle \int ^{\cot x}_{1/e}\displaystyle \frac{dt}{t\left ( 1+t^{2} \right )}$$ is
  • $$\displaystyle \frac{1}{2+tan^{2}x}$$
  • $$1$$
  • $$\pi /4$$
  • $$\displaystyle \frac{2}{\pi }\displaystyle \int ^{1}_{-1}\displaystyle \frac{dt}{1+t^{2}}$$
If $$\displaystyle I_{n}= \int_{0}^{\pi /4}\tan ^{n}x\times \sec ^{2}x dx,$$ then $$\displaystyle I_{1}, I_{2}, I_{3},$$ .......are in
  • A.P.
  • G.P.
  • H.P.
  • none
The value of the integral $$\displaystyle \int_{\alpha }^{\beta }\displaystyle \dfrac{dx}{\sqrt{\left ( x-\alpha  \right )\left ( \beta -x \right )}}$$ for $$\beta > \alpha $$, is
  • $$\sin ^{-1}\: \alpha /\beta $$
  • $$\pi /2$$
  • $$\sin ^{-1}\beta /2\alpha $$
  • $$\pi $$
$$\displaystyle \int_{\pi /6}^{\pi /4}\frac{dx}{\sin 2x}$$is equal to
  • $$\displaystyle \frac{1}{2}\log \left ( -1 \right )$$
  • $$\displaystyle \log \left ( -1 \right )$$
  • $$\displaystyle \log 3$$
  • $$\displaystyle \frac {1}{2} \log \sqrt{3}$$
Value of $$\displaystyle \int_{0}^{1}\displaystyle \frac{dx}{\left ( 1+x^{2} \right )\sqrt{1-x^{2}}}$$ is?
  • $$\dfrac{\pi}{2\sqrt{2}}$$
  • $$\dfrac{\pi}{\sqrt{2}}$$
  • $$\sqrt{2}\pi $$
  • $$2\sqrt{2\pi }$$
If $$3+2\displaystyle\int _{ 0 }^{ 1 }{ { x }^{ 2 }{ e }^{ -x^{ 2 } } } dx=\displaystyle\int _{ 0 }^{ 1 }{ { e }^{ -x^{ 2 } } } dx$$ then the value of $$\beta$$ is
  • $$\dfrac{1}{e}$$
  • $$e$$
  • $$\dfrac{1}{2e}$$
  • $$2e$$
If $$I_n=\displaystyle\int^1_0\dfrac{dx}{(1+x^2)^n}; n\in N$$, then which of the following statements hold good?
  • $$2n I_{n+1}=2^{-n}+(2n-1)I_n$$
  • $$I_2=\dfrac{\pi}{8}+\dfrac{1}{4}$$
  • $$I_2=\dfrac{\pi}{8}-\dfrac{1}{4}$$
  • $$I_3=\dfrac{\pi}{16}-\dfrac{5}{48}$$
If $$\int { \frac { { x }^{ 1/2 } }{ { x }^{ 1/2 }-{ x }^{ 1/3 } }  } dx=Ax\frac { 6 }{ 5 } { x }^{ 5/6 }+\frac { 3 }{ 2 } { x }^{ 2/3 }+B\sqrt { x } +{ Cx }^{ 1/3 }+{ Dx }^{ 1/6 }+E$$ In$$ \left( { x }^{ 1/6 }-1 \right) =k,$$ then A+B+C+D+E=
  • 6
  • 12
  • 18
  • 17
The value of $$\displaystyle \int_{1/2}^{1}\displaystyle \frac{dx}{x\sqrt{3x^{2}+2x-1}}$$ is?
  • $$\pi /2$$
  • $$\pi /3$$
  • $$\pi /6$$
  • $$\pi /\sqrt{2}$$
If $$I= \displaystyle \int_{1}^{\infty }\displaystyle \frac{x^{2}-2}{x^{3}\sqrt{x^{2}-1}}\: dx$$, then $$I$$ equals
  • $$-1$$
  • $$0$$
  • $$\pi /2$$
  • $$\pi -\sqrt{3}$$
If $$0< \alpha < \pi /2$$ then the value of $$\displaystyle \int_{0}^{\alpha }\displaystyle \frac{dx}{1-\cos x\cos \alpha }$$ is
  • $$\pi /\alpha $$
  • $$\pi /2\sin \alpha $$
  • $$\pi /2\cos \alpha $$
  • $$\pi /2\alpha $$
Value of $$\displaystyle \int_{a}^{\infty }\displaystyle \frac{dx}{x^{4}\sqrt{a^{2}+x^{2}}}$$ is
  • $$\displaystyle \frac{2+\sqrt{2}}{3a^{4}}$$
  • $$\displaystyle \frac{2-\sqrt{2}}{3a^{2}}$$
  • $$\displaystyle \frac{2-\sqrt{2}}{3a^{4}}$$
  • $$\displaystyle \frac{\sqrt{2}+1}{3a^{2}}$$
The value of $$\displaystyle \int_{-4}^{-5}e^{\left ( x+5 \right )^{2}}dx+3\displaystyle \int_{1/3}^{2/3}e^{9\left ( x-2/3 \right )^{2}}dx$$ is
  • $$2/5$$
  • $$1/5$$
  • $$1/2$$
  • none of these
Value of $$\displaystyle \int_{0}^{16}\displaystyle \frac{x^{1/4}}{1+x^{1/2}}\: dx$$ is
  • $$\displaystyle \frac{8}{3}$$
  • $$\displaystyle \frac{4}{3}\tan ^{-1}\: 2$$
  • $$4\displaystyle \left ( \displaystyle \frac{2}{3}+\tan ^{-1}\: 2 \right )$$
  • $$4\displaystyle \left ( \displaystyle \frac{2}{3}-\tan ^{-1}\: 2 \right )$$
The solution of $$x$$ of the equation $$\displaystyle \int_{\sqrt{2}}^{x}{\dfrac{dt}{t\sqrt{t^{2}-1}}}=\dfrac{\pi}{2}$$ is 
  • $$2\sqrt{2}$$
  • $$2$$
  • $$\pi$$
  • $$-\sqrt{2}$$
$$\int_{0}^{1}\frac{1}{ax+b}dx is$$
  • zero
  • $$log_{e}\frac{a+b}{b}$$
  • $$log_{e}(ax+b)$$
  • $$\frac{1}{a}log_{e}\left ( \frac{a+b}{b} \right )$$
$$\int _{ 0 }^{ 8 }{ \left[ \sqrt { t }  \right]  } dt$$ at equals to (where [.] greatest integer function.)
  • 28
  • 11
  • 2
  • 8
If $$\displaystyle \frac{\pi }{4}< \alpha < \displaystyle \frac{\pi }{2}$$, value of $$\displaystyle \int_{-\pi /2}^{\pi /2}\displaystyle \frac{\sin 2x}{\sqrt{1+\sin 2\alpha \sin x}}$$ is
  • $$-\displaystyle \frac{4}{3}\tan \alpha \sec \alpha $$
  • $$-\displaystyle \frac{4}{3}\cot \alpha cosec\alpha $$
  • $$-\displaystyle \frac{4}{3}\tan \alpha cosec\alpha $$
  • $$-\displaystyle \frac{4}{3}\cot \alpha \sec \alpha $$
If $$I_{1}= \displaystyle \int_{x}^{1}\displaystyle \frac{dt}{1+t^{2}}$$ and $$I_{2}= \displaystyle \int_{1}^{1/x}\displaystyle \frac{dt}{1+t^{2}}$$ for $$x> 0$$, then
  • $$I_{1}= I_{2}$$
  • $$I_{1}> I_{2}$$
  • $$I_{2}> I_{1}$$
  • $$I_{2}= \left ( \pi /2 \right )-\tan ^{-1}x$$
If $$I_{1}= \displaystyle \int_{0}^{\infty }\displaystyle \frac{dx}{1+x^{4}}$$ and $$I_{2}= \displaystyle \int_{0}^{\infty }\displaystyle \frac{x^{2}}{1+x^{4}}\: dx$$, then
  • $$I_{1}= I_{2}$$
  • $$I_{1}=2 I_{2}$$
  • $$2I_{1}= I_{2}$$
  • none of these
If $$\displaystyle \int_{0}^{\infty }\displaystyle \frac{\log \left ( 1+x^{2} \right )}{1+x^{2}}\: dx= \lambda \displaystyle \int_{0}^{1}\displaystyle \frac{\log \left ( 1+x \right )}{1+x^{2}}\: dx$$ then $$\lambda $$ equals
  • $$4$$
  • $$\pi $$
  • $$8$$
  • $$2\pi $$
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