CBSE Questions for Class 12 Commerce Maths Integrals Quiz 9 - MCQExams.com

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
  • $$-\displaystyle \frac{4}{3}\tan \alpha \sec \alpha $$
  • $$-\displaystyle \frac{4}{3}\cot \alpha cosec\alpha $$
  • $$-\displaystyle \frac{4}{3}\tan \alpha cosec\alpha $$
  • $$-\displaystyle \frac{4}{3}\cot \alpha \sec \alpha $$
$$\displaystyle \int_{e^{e^{e}}}^{e^{e^{e^{e}}}}\frac{dx}{xlnx\cdot ln\left ( lnx \right )\cdot ln\left ( ln\left ( lnx \right ) \right )}$$ equals
  • 1
  • 1/e
  • e-1
  • 1+e
Value of $$\displaystyle \int_{0}^{\pi /2}\displaystyle \frac{\sin 4\Theta }{\sin \Theta }\: d\Theta $$ is
  • $$1/3$$
  • $$2/3$$
  • $$1$$
  • $$4/3$$
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 $$P=$$
  • 2
  • 1
  • 3
  • None of these
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:
  • v > u
  • 6v = $$\displaystyle \pi $$
  • $$\displaystyle 3u+2v=5\pi /6$$
  • $$\displaystyle u+v=\pi /3$$
$$\displaystyle\int { \cfrac { { x }^{ 2 }-1 }{ { x }^{ 4 }+{ x }^{ 2 }+1 }  } dx$$ is equal to
  • $$\log { \left( { x }^{ 4 }+{ x }^{ 2 }+1 \right) } +c$$
  • $$\log { \cfrac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+x+1 } +c } $$
  • $$\cfrac { 1 }{ 2 } \log { \cfrac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+x+1 } +c } $$
  • $$\cfrac { 1 }{ 2 } \log { \cfrac { { x }^{ 2 }+x+1 }{ { x }^{ 2 }-x+1 } +c } $$
Evaluate $$\displaystyle \int_{0}^{\pi /4}\frac{\cos x-\sin x}{10+\sin 2x}dx$$
  • $$\displaystyle \frac{1}{3}\left (\tan^{-1} \frac{\sqrt{2}}{3}+\tan^{-1} \frac{1}{3} \right )$$
  • $$\displaystyle \frac{1}{3}\left (\tan^{-1} \frac{\sqrt{1}}{3}-\cot^{-1} \frac{2}{3} \right )$$
  • $$\displaystyle \frac{1}{3}\left ( \tan^{-1} \frac{\sqrt{2}}{3}-\tan^{-1} \frac{1}{3} \right )$$
  • $$\displaystyle \frac{1}{3}\left ( \tan^{-1} \frac{\sqrt{1}}{3}-\cot^{-1} \frac{1}{3} \right )$$
Let $$f(x)$$ be a positive function. Let
$$I_{1} = \int_{1 - k}^{k} xf\left \{x(1 - x)\right \} dx$$,
$$I_{2} = \int_{1 - k}^{k} f\left \{x(1 - x) \right \} dx$$,
where $$2k - 1 > 0$$, then $$\dfrac {I_{1}}{I_{2}}$$ is
  • $$2$$
  • $$k$$
  • $$\dfrac {1}{2}$$
  • $$1$$
The value of $$\displaystyle \int_{3}^{4}\sqrt {(4 - x)(x - 3)}dx$$ is
  • $$\dfrac {\pi}{16}$$
  • $$\dfrac {\pi}{8}$$
  • $$\dfrac {\pi}{4}$$
  • $$\dfrac {\pi}{2}$$
The value of the integral $$\displaystyle \int_{\frac {1}{3}}^1\frac {(x-x^3)^{\frac {1}{3}}}{x^4}dx$$
  • $$6$$
  • $$0$$
  • $$3$$
  • $$4$$
$$\displaystyle\int^{\pi /3}_0\frac{\cos \theta}{5-4\sin \theta}d\theta$$ equal to.
  • $$\displaystyle\frac{1}{4}log\left(\displaystyle\frac{5}{5+2\sqrt{3}}\right)$$
  • $$\displaystyle\frac{1}{4}log\left(\displaystyle\frac{5}{5-2\sqrt{3}}\right)$$
  • $$\displaystyle\frac{1}{4}log\left(\displaystyle\frac{5+2\sqrt{3}}{5}\right)$$
  • $$\displaystyle\frac{1}{4}log\left(\displaystyle\frac{5-2\sqrt{3}}{5}\right)$$
Consider $$I=\displaystyle \int^{\pi}_{0}\displaystyle\frac{xdx}{1+\sin x}$$. What is I equal to?
  • $$-\pi$$
  • $$0$$
  • $$\pi$$
  • $$2\pi$$
$$\int { { x }^{ 4 }{ e }^{ 2x } } dx=$$
  • $$\cfrac { { e }^{ 2x } }{ 4 } \left( 2{ x }^{ 4 }-4{ x }^{ 3 }+6{ x }^{ 2 }-6x+3 \right) +C$$
  • $$\cfrac { { e }^{ 2x } }{ 2 } \left( 2{ x }^{ 4 }-4{ x }^{ 3 }+6{ x }^{ 2 }-6x+3 \right) +C$$
  • $$\cfrac { { e }^{ 2x } }{ 8 } \left( 2{ x }^{ 4 }+4{ x }^{ 3 }+6{ x }^{ 2 }+6x+3 \right) +C$$
  • $$\cfrac { { e }^{ 2x } }{ 4 } \left( 2{ x }^{ 4 }+4{ x }^{ 3 }+6{ x }^{ 2 }+6x+3 \right) +C$$
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 
  • $$0$$
  • $$3/2$$
  • $$1/2$$
  • $$4/3$$
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
  • $$3/e$$
  • $$e/3$$
  • $$3e$$
  • $$1/3e$$
The value of $$\int^3_{1/3}\dfrac{tan(x^2-\dfrac{1}{x^2})}{sin(x+\dfrac{1}{x}) }\dfrac{dx}{x}$$ is 
  • 0
  • $$\dfrac{3}{2}$$
  • $$\dfrac{1}{2}$$
  • $$\dfrac{4}{3}$$
The value of the integral $$\displaystyle \int_0^1 \dfrac{x^{\alpha}-1}{\log x}dx$$ is
  • $$\log (\alpha+1)$$
  • $$2\log (\alpha+1)$$
  • $$3\log \alpha$$
  • none of these
The value of the definite integral $$\displaystyle\int^{\pi/2}_{0}\left(\cos ^{10}x\cdot \sin 12x\right)dx$$, is equal to.
  • $$\displaystyle\frac{1}{10}$$
  • $$\displaystyle\frac{1}{11}$$
  • $$\displaystyle\frac{1}{12}$$
  • $$\displaystyle\frac{1}{22}$$
If $$I=\int _{ 0 }^{ \pi  }{ \frac { x\sin { x }  }{ 1+{ cos }^{ 2 }x } dx } $$, then the value of $$\sin { \sqrt { I }  } $$, is
  • $$\frac { 1 }{ 2 } $$
  • $$0$$
  • $$\frac { \sqrt { 2 } }{ 2 } $$
  • $$1$$
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
  • $$4\pi$$
  • $$2\pi +\dfrac { 1 }{ \sqrt { 3 } }$$
  • $$2\pi -\dfrac { 1 }{ \sqrt { 3 } }$$
  • $$4\pi +\sqrt { 3 } -\dfrac { 1 }{ \sqrt { 3 } }$$
If, $$\int^{\frac{\pi}{2}} _0 \frac{sin^2x}{(1 + cosx^2)} dx = 0$$
  • $$\frac{4 - \pi}{2}$$
  • $$\frac{\pi - 4}{2}$$
  • $$4 - frac{\pi}{2}$
  • $$\frac{4 + \pi}{2}$$
$${I}_{n}=\int_{1}^{e}{\left(logx\right)^{n}dx}$$ and $${I}_{n}=A+{BI}_{n-1}$$ then
A=........., B=............
  • $$e,-n$$
  • $$1/e,n$$
  • $$-e,n$$
  • $$-e-n$$
$$\displaystyle \int _{ 0 }^{ x }{ \cfrac { \sin { x }  }{ 1+\cos ^{ 2 }{ x }  }  } dx=\pi \cfrac { \cos { \alpha  }  }{ 1-\sin ^{ 2 }{ \alpha  }  } $$
  • for no value of $$\alpha$$
  • for exactly two values of $$\alpha$$ in $$\left( 0,\pi \right) $$
  • for atleast one $$\alpha$$ in $$\left( \pi /2,\pi \right) $$
  • for exactly one $$\alpha$$ in $$\left( 0,\pi /2 \right) $$
$$\int _{ 0 }^{ \pi /2 }{ \sin ^{ 8 }{ x } \cos ^{ 2 }{ x } dx } $$ is equal to
  • $$\cfrac { \pi }{ 512 } $$
  • $$\cfrac {3 \pi }{ 512 } $$
  • $$\cfrac { 5\pi }{ 512 } $$
  • $$\cfrac {7 \pi }{ 512 } $$
Let $$y = y(x), y(1)=1\ and\ y(e) ={e^2}$$ . Consider
$$J = \int {{{x + y} \over {xy}}} dy$$, $$I = \int {{{x + y} \over {{x^2}}}} dx$$, $$J - I = g\left( x \right)$$ and g(1) = 1, then the value of g(e) is
  • $$2e+1$$
  • $$e+1$$
  • $${e^2}$$-e+1
  • $${e^2}$$+e-1
If $$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$$, then
  • $$I_{1} > I_{2}$$
  • $$I_{2} > I_{1}$$
  • $$I_{3} > I_{4}$$
  • $$I_{1} > I_{3}$$
Value of the definite integral $$\displaystyle \int _{ 0 }^{ 1 }{ \cot ^{ -1 }{ \left( 1-x+{ x }^{ 2 } \right)  } dx }$$ is:
  • $$\pi -\log { 2 }$$
  • $$\dfrac{\pi}{2} -\log { 2 }$$
  • $$\pi +\log { 2 }$$
  • $$\dfrac{\pi}{2} +\log { 2 }$$
$$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}}$$ then k is equal to
  • n
  • -n
  • m-1
  • None of these
Evaluate the integral, $$\int _{ 0 }^{ 1 }{ \cos { \left( 2\cot ^{ -1 }{ \sqrt { \dfrac { 1-x }{ 1+x }  }  }  \right)  }  } dx=$$
  • $$-1/2$$
  • $$1/2$$
  • $$0$$
  • $$1$$
$$\displaystyle\int^{1}_0\sqrt{x(1-x)}dx=$$.
  • $$\dfrac{\pi}{8}$$
  • $$\dfrac{3\pi}{8}$$
  • $$\dfrac{5\pi}{4}$$
  • $$\dfrac{\pi}{2}$$
Solve $$\displaystyle \int _{ 0 }^{ 1/2 }{ \dfrac { x\sin ^{ -1 }{ x }  }{ \sqrt { 1-{ x }^{ 2 } }  } dx= }$$
  • $$\dfrac { 1 }{ 2 } +\dfrac { \sqrt { 3 } \pi }{ 12 }$$
  • $$\dfrac { 1 }{ 2 } -\dfrac { \sqrt { 3 } \pi }{ 12 }$$
  • $$\dfrac { 1 }{ 2 } -\dfrac { \sqrt { 2 } \pi }{ 12 }$$
  • $$None\ of\ these$$
One of the roots of the equation $$2000x^6+100x^5+10x^3+x-2=0$$ is of the form $$\dfrac{m+\sqrt{n}}{r}$$. When 'm' is non zero integer and n and r relatively prime natural numbers. Then $$\dfrac{m+n+r}{100}=?$$
  • $$100$$
  • $$2$$
  • $$3$$
  • $$0$$
$$\int _{ -\pi /4 }^{ \pi /4 }{ ln\sqrt { 1+\sin { 2x }  } dx }$$ has the value equal to:
  • $$-\dfrac {\pi}{4}l\ n2$$
  • $$-\dfrac {\pi}{2}l\ n2$$
  • $$-\dfrac {\pi}{8}l\ n2$$
  • $$-\dfrac {\pi}{16}l\ n2$$
The value of the integer $$I=\int_{1}^{\infty} \dfrac{(x^{2}-x)}{x^{3}\sqrt{(x^{2}-1)}}dx$$ is
  • $$0$$
  • $$2/3$$
  • $$4/3$$
  • $$none\ of\ these$$
$$\int _{ 0 }^{ 2\pi  }{ \sqrt { 1+\sin { \dfrac { x }{ 2 }  }  } dx } =$$
  • $$0$$
  • $$2$$
  • $$8$$
  • $$4$$
The integral $$\int _{ \tfrac { \pi  }{ 12 }  }^{ \tfrac { \pi  }{ 4 }  }{ \dfrac { 8\cos { 2x }  }{ { \left( \tan { x } +\cot { x }  \right)  }^{ 3 } } dx }$$ equals:
  • $$\dfrac {15}{128}$$
  • $$\dfrac {5}{64}$$
  • $$\dfrac {13}{32}$$
  • $$\dfrac {13}{256}$$
Solve $$\displaystyle\int_{0}^{\infty}\dfrac {x \tan^{-1}x}{(1+x^{2})^{2}}dx$$
  • $$\pi/2$$
  • $$\pi/6$$
  • $$\pi/4$$
  • $$\pi/8$$
$$\displaystyle \int_{0}^{100 \pi}\sqrt {1-\cos 2x}dx$$ is
  • $$200\sqrt {2}$$
  • $$100\sqrt {2}$$
  • $$0$$
  • $$none\ of\ these$$
$$\displaystyle \overset{\pi/2}{\underset{0}{\int}} \dfrac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}$$ equals-
  • $$\pi/ab$$
  • $$2\pi/ab$$
  • $$ab/ \pi$$
  • $$\pi/2ab$$
The value of $$\int_{-1}^{1}\dfrac {dx}{(2-x)\sqrt {1-x^{2}}}$$ is
  • $$0$$
  • $$\dfrac {\pi}{\sqrt {3}}$$
  • $$\dfrac {2\pi}{\sqrt {3}}$$
  • $$cannot\ be\ evaluated$$
$$\displaystyle \int^{\pi/2}_{-\pi/2}\sqrt {\cos x-\cos^{3}x}dx=$$
  • $$1$$
  • $$4/3$$
  • $$-1/3$$
  • $$0$$
If $$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$$
Let $$ 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}$$ is 
  • 2
  • k
  • 1/2
  • 1
$$\int {x\left( {\frac{{\ln {a^{{a^{\frac{x}{2}}}}}}}{{3{a^{\frac{{5x}}{2}}}{b^{3x}}}} + \frac{{\ln {b^{{b^{\frac{x}{2}}}}}}}{{2{a^{2x}}{b^{4x}}}}} \right)} dx\left( {where\,a,b \in R} \right)$$ is equal to
  • $$\frac{1}{{6\ln {a^2}{b^3}}}{a^{2x}}{b^{3x}}\ln \frac{{{a^{2x}}{b^{3x}}}}{e} + k$$
  • $$\frac{1}{{6\ln {a^2}{b^3}}}{a^{2x}}{b^{3x}}\ln \frac{1}{{e{a^{2x}}{b^{3x}}}} + k$$
  • $$\frac{1}{{6\ln {a^2}{b^3}}}\frac{1}{{{a^{2x}}{b^{3x}}}}\ln \left( {e{a^{2x}}{b^{3x}}} \right) + k$$
  • $$ - \frac{1}{{6\ln {a^2}{b^3}}}\frac{1}{{{a^{2x}}{b^{3x}}}}\ln \left( {e{a^{2x}}{b^{3x}}} \right) + k$$
$$\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$$
$$I=\int { \left\{ \log _{ e }( \log _{ e }{ x }) +\cfrac { 1 }{ { \left( \log _{ e }{ x }  \right)  }^{ 2 } }  \right\}  } dx$$ is equal to 
  • $$x\log _{ e }( \log _{ e }{ x } )+c$$
  • $$x\log _{ e }( \log _{ e }{ x })-\cfrac { x }{ { \left( \log _{ e }{ x } \right) }^{ } } +c$$
  • $$x\log _{ e }{ x } \log _{ e }{ x } +\cfrac { x }{ { \left( \log _{ e }{ x } \right) }^{ } } +c$$
  • none of these
$$\int (1+x-x^{-1})e^{x+x^{1}}dx=$$
  • $$(x+1)e^{x+x^{-1}}+c$$
  • $$(x-1)e^{x+x^{-1}}+c$$
  • $$-xe^{x+x^{-1}}+c$$
  • $$xe^{x+x^{-1}}+c$$
Solve: $$\int\dfrac{sin^32x}{cos^52x}dx$$
  • $$\dfrac{{{{\tan }^4}2x}}{8} + C$$
  • $$\dfrac{{{{\cos}^4}2x}}{8} + C$$
  • $$\dfrac{{{{\sin}^4}2x}}{8} + C$$
  • $$\dfrac{{{{\sec}^4}2x}}{8} + C$$
$$\int^{\pi/4}_{0}\dfrac {\sin^{2}x\cos^{2}x}{(\sin^{3}x+\cos^{3}x)^{2}}dx$$ is
  • $$1/3$$
  • $$1/2$$
  • $$1/6$$
  • $$1/4$$
$$\int _{ 1/2 }^{ 2 }{ \dfrac { 1 }{ x } \csc { ^{ 101 }\left( x-\dfrac { 1 }{ x }  \right) dx } \\  } $$ is equal to 
  • $$1/4$$
  • $$1$$
  • $$0$$
  • $$\dfrac{101}{2}$$
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