CBSE Questions for Class 12 Commerce Maths Integrals Quiz 10 - MCQExams.com

The value of $$\int_{a}^{b}{(x-a)^{3}(b-x)^{4}dx}$$ is
  • $$\dfrac{(b-a)^{4}}{6^{4}}$$
  • $$\dfrac{(b-a)^{8}}{280}$$
  • $$\dfrac{(b-a)^{7}}{7^{3}}$$
  • $$none\ of\ these$$
The value of $$\int_{-1}^{3}[tan^{-1}(\frac{x}{x^{2}+1})+tan^{-1}(\frac{x^{2}+1}{x})]dx$$ 
  • 2$$\pi $$
  • $$\pi $$
  • $$\frac{\pi }{2}$$
  • $$\frac{\pi }{4}$$
Evaluate : $$\int { \cfrac { dx }{ x\cos { x }  }  } $$
  • $$\ln { x } +\cfrac { { x }^{ 2 } }{ 2 } +\cfrac { { 3x }^{ 3 } }{ 3 } +...$$
  • $$\ln { x } -\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { 4x }^{ 4 } }{ 16 } +...$$
  • $$\ln { x } +\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { 5x }^{ 4 } }{ 96 } +...$$
  • $$\ln { x } -\cfrac { { x }^{ 2 } }{ 3 } +\cfrac { { x }^{ 4 } }{ 9 } +...$$
The value of the integral $$\underset{0}{\overset{\pi / 4}{\int}} \dfrac{\sin \, x + \cos \, x}{3 + \sin \, 2x} dx$$, is 
  • $$\log 2$$
  • $$\log 3$$
  • $$\dfrac{1}{4} \log \, 3$$
  • $$\dfrac{1}{8} \log \, 3$$
$$\int_{}^{} {\frac{{{x^2}}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}dx} $$ would be equal to 
  • $$\frac{{\sin x + x\cos x}}{{x\sin x + \cos x}} + c$$
  • $$\frac{{\sin x - x\cos x}}{{x\sin x + \cos x}} + c$$
  • $$\frac{{\sin x - x\cos x}}{{x\sin x - \cos x}} + c$$
  • none of these
$$\displaystyle \int_{}^{} {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx} $$ is equal to :
  • $$ - {e^x}\tan \frac{x}{2} + c$$
  • $$ - {e^x}\cot \frac{x}{2} + c$$
  • $$ - \frac{1}{2}{e^x}\tan \frac{x}{2} + c$$
  • $$ - \frac{1}{2}{e^x}\cot \frac{x}{2} + c$$
The value of $$\displaystyle \int _{0}^{\pi/2} \log{\sin{x}}dx$$ is 
  • $$-\pi \log{2}$$
  • $$\dfrac{-\pi}{2} \log{2}$$
  • $$\pi \log{2}$$
  • 0
The value of $$\cfrac { \int _{ 0 }^{ \pi /2 }{ { \left( \sin { x }  \right)  }^{ \sqrt { 3 } +1 } } dx }{ \int _{ 0 }^{ \pi /2 }{ { \left( \sin { x }  \right)  }^{ \sqrt { 3 } -1 } }  } $$ is
  • $$\cfrac { \sqrt { 3 } +1 }{ \sqrt { 3 } -1 } $$
  • $$\cfrac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } $$
  • $$\cfrac { \sqrt { 3 } +1 }{ \sqrt { 3 } } $$
  • none of these
The value of $$\displaystyle \int _{-1}^{1} \log{\left(\dfrac{2-x}{2+x}\right)}\sin^{2}{x}dx$$
  • $$1$$
  • $$-1$$
  • $$2$$
  • $$0$$
$$\displaystyle\int _{ -1 }^{ 1 }{ x\ell n\left( 1+{ e }^{ x } \right) dx } =$$
  • $$0$$
  • $$\ell n(1+e)$$
  • $$\ell n(1+e)-1$$
  • $$1/3$$
$$\displaystyle\int^{2+\sqrt{3}}_{2-\sqrt{3}}\dfrac{xdx}{(1+x)(1+x^2)}=?$$
  • $$\dfrac{\pi}{4}$$
  • $$\dfrac{\pi}{6}$$
  • $$\dfrac{\pi}{12}$$
  • $$\dfrac{\pi}{24}$$
The value of the integral $$\displaystyle \int _ { 0 } ^ { 1 } \dfrac { d x } { x ^ { 2 } + 2 x \cos \alpha + 1 }$$ , where $$0 < \alpha < \dfrac { \pi } { 2 } ,$$ is equal to
  • $$\sin{\alpha}$$
  • $$\alpha \sin {\alpha}$$
  • $$\dfrac { \alpha } { 2 \sin{\alpha} }$$
  • $$\dfrac { \alpha } { 2 } \sin {\alpha}$$
If $$\displaystyle \int _{0}^{1}\cot^{-1}(1+x^{2}-x)dx=k\left(\dfrac {\pi}{4}-\log_{e}\sqrt {2}\right)$$, then the value of $$k$$ is equal to
  • $$0$$
  • $$1$$
  • $$-1$$
  • $$2$$
$$\int _{ \pi /4 }^{ 3\pi /4 }{ \dfrac { dx }{ 1+\cos { x }  }  }$$ is equal to 
  • $$-2$$
  • $$2$$
  • $$4$$
  • $$-1$$
$$\int_{0}^{\frac{1}{2}}\frac{xsin^{-1}x}{\sqrt{1-x^{2}}}dx$$ is equal to
  • $$\frac{1}{2}+\frac{\pi }{2\sqrt{3}}$$
  • $$\frac{1}{2}-\frac{\pi }{2\sqrt{3}}$$
  • $$\frac{1}{2}+\frac{\pi }{4\sqrt{3}}$$
  • $$\frac{1}{2}-\frac{\pi }{4\sqrt{3}}$$
The integral $$\int { \dfrac { { sin }^{ 2 }\times { cos }^{ 2 }\times  }{ { (sin }^{ 5 }\times +{ cos }^{ 3 }\times { sin }^{ 2 }\times +{ sin }^{ 3 }\times { cos }^{ 2 }\times +{ cos }^{ 5 }{ \times ) }^{ 2 } }  } $$dx is equal to:
  • $$\dfrac { 1 }{ { 1+cos }^{ 3 }x } +c$$
  • $$\dfrac { -1 }{ { 1+cos }^{ 3 }x } +c$$
  • $$\dfrac { 1 }{ 3\left( 1+{ tan }^{ 3 }x \right) } +c$$
  • $$\dfrac { -1 }{ 3\left( 1+{ tan }^{ 3 }x \right) } +c$$
The value of $$\int_{0}^{[x]} (x-[x])dx$$, where $$[x]$$ is the greatest integer $$|le x$$ is equal to
  • $$4[x]$$
  • $$2[x]$$
  • $$\dfrac{1}{2} [x]$$
  • $$\dfrac{1}{5} [x]$$
The value of the definite integral $$\int _{ 2 }^{ 3 }{ \left[ \sqrt { 2x-\sqrt { 5\left( 4x-5 \right)  }  } +\sqrt { 2x+\sqrt { 5\left( 4x-5 \right)  }  }  \right]  } dx=$$ 
  • $$\dfrac {7\sqrt {3}+3\sqrt {5}}{3\sqrt {2}}$$
  • $$4\sqrt {2}$$
  • $$4\sqrt {3}+\dfrac {4}{3}$$
  • $$\dfrac {7\sqrt {7}-2\sqrt {5}}{3\sqrt {2}}$$
The value of the integral $$\displaystyle \int _{ { e }^{ -1 } }^{ { e }^{ 2 } }{ \left| \frac { \log _{ e }{ x }  }{ x }  \right| dx } $$ is
  • $$\dfrac {3}{2}$$
  • $$\dfrac {5}{2}$$
  • $$3$$
  • $$5$$
The integral $$\displaystyle \int_{\dfrac{\pi}{12}}^{\dfrac{\pi}{4}}{\dfrac{8\cos 2x}{(\tan x+\cot x)^{3}}dx}$$ equals :
  • $$\dfrac{15}{128}$$
  • $$\dfrac{15}{64}$$
  • $$\dfrac{13}{32}$$
  • $$\dfrac{13}{256}$$
Let $$f:R\longrightarrow R,g : R\longrightarrow R$$ be continuous functions. then the value of integeral. 
$$\int _{ \ell n\lambda  }^{ \ell n/\lambda  }{ \frac { f\left( \dfrac { { x }^{ 2 } }{ 4 }  \right) \left[ f\left( x \right) -f\left( -x \right)  \right]  }{ g\left( \dfrac { { x }^{ 2 } }{ 4 }  \right) \left[ g\left( x \right) +g\left( -x \right)  \right]  }  } dx$$ is:
  • depend on $$\lambda$$
  • a non-zero constant
  • zero
  • 1
If $$P=\displaystyle \lim _{ n\rightarrow \infty  }{ \frac { { \left( \prod _{ r=1 }^{ n }{ \left( { n }^{ 3 }+{ r }^{ 3 } \right)  }  \right)  }^{ 1/n } }{ { n }^{ 3 } }  }$$  and $$\lambda =\displaystyle \int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 3 } }  } $$ then $$In P$$ is equal to
  • $$In 2-1+\lambda$$
  • $$In 2-3+3\lambda$$
  • $$2In 2-\lambda$$
  • $$In 4-3+3\lambda$$
The value of $$\displaystyle \int _{0}^{\infty}\dfrac{\sqrt{x^2+1}}{(x+\sqrt{x^{2}+1})^{n+1}} .dx  \ \forall \ n\ \in \ N- \{ \pm 1 \}$$ is
  • $$0$$
  • $$n(n^{2}-1)$$
  • $$\dfrac{n}{(n^{2}-1)}$$
  • $$n^{2}$$
$$\displaystyle \int_{-3\pi}^{3\pi}{\sin^{2}\theta\sin^{2} 2\theta d \theta}$$ is equal to-
  • $$\pi$$
  • $$\dfrac{3\pi}{2}$$
  • $$\dfrac{5\pi}{2}$$
  • $$6\pi$$
$$\displaystyle \int _ { 0 } ^ { \pi / 4 } \frac { x \cdot \sin x } { \cos ^ { 3 } x } d x$$ equals to :
  • $$\displaystyle \frac { \pi } { 4 } + \frac { 1 } { 2 }$$
  • $$\displaystyle \frac { \pi } { 4 } - \frac { 1 } { 2 }$$
  • $$\dfrac { \pi } { 4 }$$
  • $$\dfrac { \pi } { 4 } + 1$$
Value of the definite integral $$\displaystyle \int_{-1}^{1}\dfrac {dx}{(1+x^{3}+\sqrt {1+x^{6}})}$$
  • $$is\dfrac {1}{2}$$
  • $$is\ 1$$
  • $$is\ 2$$
  • $$is\ zero$$
The value of 
$$\displaystyle\int _{ 5 }^{ 10 }{ \left( \sqrt { x+\sqrt { 20x-100 }  } +\sqrt { x-\sqrt { 20x-100 }  }  \right)  } dx$$ is 
  • $$10\sqrt{5}$$
  • $$5\sqrt{5}$$
  • $$10\sqrt{2}$$
  • $$8\sqrt{2}$$
Let $$f(x)= \left| \begin{matrix} \sec { x }  & \cos { x }  & \sec { ^{ 2 }x+\cot { x\csc { x }  }  }  \\ \cos { ^{ 2 }x }  & \cos { ^{ 2 }x }  & \csc { ^{ 2 }x }  \\ 1 & \cos { ^{ 2 }x }  & \cos { ^{ 2 }x }  \end{matrix} \right|$$. Then $${\int}_{0}^{\pi/2}f(x)dx$$ is equal to 
  • $$\dfrac{15\pi}{60}$$
  • $$\dfrac{15\pi+32}{60}$$
  • $$\dfrac{15\pi-32}{60}$$
  • $$none\ of\ these$$
$$\displaystyle \int_{0}^{\pi/2}{(\sin x-\cos x).\log(\sin x+\cos x)dx}$$=
  • $$\dfrac{\pi}{4}$$
  • $$\dfrac{\pi}{2}$$
  • $$0$$
  • $$2$$
Let $$f\left(x\right)+f\left(\dfrac{1}{x}\right)=F\left(x\right)$$ where $$f\left(x\right)=\displaystyle\int_{1}^{x}{\dfrac{\ln{t}}{1+t}dx}$$.Then $$F\left(e\right)=$$
  • $$\dfrac{-1}{2}$$
  • $$\dfrac{1}{2}$$
  • $$1$$
  • $$-1$$
Evaluate:
$$\displaystyle\int_{-2}^{3}{\left|1-{x}^{2}\right|dx}$$
  • $$\dfrac{28}{3}$$
  • $$\dfrac{2}{3}$$
  • $$\dfrac{8}{3}$$
  • $$\dfrac{5}{3}$$
If $${ I }_{ 1 }=\int _{ 0 }^{ \pi /2 }{ \cos(\sin x)dx, { I }_{ 2 }= } \int _{ 0 }^{ \pi /2 }{ \sin(\cos x)dx } $$ and $${ I }_{ 3 }=\int _{ 0 }^{ \pi /2 }{ \cos xdx } $$, then
  • $${ I }_{ 1 }>{ I }_{ 2 }>{ I }_{ 3 }$$
  • $${ I }_{ 3 }>{ I }_{ 2 }>{ I }_{ 1 }$$
  • $${ I }_{ 3 }>{ I }_{ 1 }>{ I }_{ 2 }$$
  • $${ I }_{ 1 }>{ I }_{ 3 }>{ I }_{ 2 }$$
$$\displaystyle \int_{2}^{8}\dfrac {\sqrt {10-x}}{\sqrt {x}+\sqrt {10-x}}dx$$ is
  • $$1$$
  • $$2$$
  • $$3$$
  • $$4$$
$${\int}_{0}^{\pi}\dfrac{\sqrt{1-x}}{1+x}dx=$$
  • $$\dfrac{\pi}{2}$$
  • $$\dfrac{\pi}{2}-1$$
  • $$\dfrac{\pi}{2}+1$$
  • $$\pi+1$$
$$\int _ { 0 } ^ { \pi / 4 } \tan ^ { 2 } x d x$$ equals -
  • $$\pi / 4$$
  • $$1 + ( \pi / 4 )$$
  • $$1 - ( \pi / 4 )$$
  • $$1 - ( \pi / 2 )$$
Evaluate
$$\int \dfrac{dx}{\sqrt{1-x}}$$
  • $$\sin^{-1} \sqrt{x}$$
  • $$-\sin^{-1} \sqrt{x}+c$$
  • $$2\sqrt{1-x}+c$$
  • $$-2\sqrt{1-x}+c$$
$$\displaystyle \overset{2\pi}{\underset{0}{\int}} x \,log \left(\dfrac{3 + \cos x}{3 - \cos x}\right)dx$$
  • $$\dfrac{\pi}{2} \,log \,3$$
  • $$\dfrac{\pi}{12} \,log \,3$$
  • $$\dfrac{\pi}{3} \,log \,3$$
  • $$0$$
The smallest interval in which value of $$\int_{0}^{1} \dfrac{xdx}{x^{3}+3}$$ lies 
  • $$[0,1]$$
  • $$[0,\dfrac{1}{2}]$$
  • $$[0,\dfrac{1}{4}]$$
  • $$[0,\dfrac{1}{8}]$$
$$\displaystyle \overset{3}{\underset{0}{\int}} \dfrac{dx}{\sqrt{5 - x^2}}$$
  • $$\dfrac{\pi}{6}$$
  • $$\dfrac{\pi}{2}$$
  • $$-\dfrac{\pi}{2}$$
  • $$-\dfrac{\pi}{6}$$
$$\displaystyle\int _{ 0 }^{ \infty  }{ \dfrac { dx }{ \left( x+\sqrt { { x }^{ 2 }+1 }  \right) ^{ 3 } }  } =$$
  • $$\dfrac{3}{8}$$
  • $$\dfrac{1}{8}$$
  • $$-\dfrac{3}{8}$$
  • $$None\ of\ these$$
If I =$$\overset { 2 }{ \underset { -3 }{ \int }  } (|x + 1| + |x + 2| +|x -1|) dx$$, then i equals 
  • $$\dfrac{31}{2}$$
  • $$\dfrac{35}{2}$$
  • $$\dfrac{47}{2}$$
  • $$\dfrac{39}{2}$$
If $$I=\displaystyle\int _{ 8 }^{ 15 }{ \dfrac { dx }{ \left( x-3 \right) \sqrt { x+1 }  }  } $$, then $$I$$ equals 
  • $$\dfrac{1}{2}\log\dfrac{5}{3}$$
  • $$2\log\dfrac{1}{3}$$
  • $$\dfrac{1}{2}\log\dfrac{1}{5}$$
  • $$2\log\dfrac{5}{3}$$
The floor value of integral $$\displaystyle \int_\dfrac{\pi }{4}^{3\pi }\dfrac{x}{1+4x}dx$$ is 
  • $$1$$
  • $$2$$
  • $$3$$
  • $$4$$
Solve :-
$$\int_0^1 {\frac{{dx}}{{{{({x^2} + 1)}^{3/2}}}} = } $$
  • $$1$$
  • $$\frac{1}{{\sqrt 2 }}$$
  • $$\frac{1}{2}$$
  • $$\sqrt 2 $$
Value of $$ \displaystyle \int_{1}^{5} \left(\sqrt {x+2\sqrt {x-1}}+\sqrt {x-2(x-1)}\right)dx$$ is
  • $$\dfrac {8}{3}$$
  • $$\dfrac {16}{3}$$
  • $$\dfrac {32}{3}$$
  • $$\dfrac {34}{3}$$
Let $$[\cdot]$$ denote the greatest integer function then the value of $$\displaystyle\int^{1.5}_0x[x^2]dx$$ is?
  • $$0$$
  • $$\dfrac{3}{2}$$
  • $$\dfrac{7}{4}$$
  • $$\dfrac{5}{4}$$
  $$\int _ { 0 } ^ { 2 } \cfrac { \sqrt { x } } { \sqrt { x } + \sqrt { 2 - x } }$$ is equal to
  • $$-1$$
  • $$0$$
  • $$2$$
  • $$1$$
$$\int _{ 0 }^{ \infty  }{ \frac { { x }^{ 2 }+1 }{ { x }^{ 4 }+{ 7x }^{ 2 }+1 }  } $$ dx=
  • $$\pi $$
  • $$\frac { \pi }{ 2 } $$
  • $$\frac { \pi }{ 3 } $$
  • $$\frac { \pi }{ 6 } 4$$
The solution of $$x$$ of the equation $$\displaystyle \int_{\sqrt{2}}^{x}{\dfrac{dt}{t\sqrt{t^{2}-1}}}=\dfrac{\pi}{2}$$ is 
  • $$2\sqrt{2}$$
  • $$2$$
  • $$\pi$$
  • $$-\sqrt{2}$$
If the value of the definite integral $$\int _{ \dfrac { \pi  }{ 6 }  }^{ \dfrac { \pi  }{ 4 }  }{ \dfrac { 1+\cot { x }  }{ { e }^{ x }sin{ x } } dx } $$, is equal to $${ ae }^{ -\pi /6 }+{ be }^{ -\pi /4 }$$ then $$(a+b)$$ equals
  • $$2-\sqrt { 2 } $$
  • $$2+\sqrt { 2 } $$
  • $$2\sqrt { 2 } -2$$
  • $$2\sqrt { 3 } -\sqrt { 2 } $$
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