CBSE Questions for Class 12 Commerce Maths Integrals Quiz 6 - MCQExams.com

The value of the integral $$\displaystyle \int_0^{1} {{1 + 2 x}}dx$$ is 
  • $$2$$
  • $$3$$
  • $$4$$
  • $$\frac{ - 1}{4}$$
The value of the integral $$\displaystyle \int\limits_0^1 {\dfrac{{{x^3}}}{{1 + {x^8}}}\,\,dx} $$ is 
  • $$\frac{\pi }{{16}}$$
  • $$\frac{\pi }{{4}}$$
  • $$\frac{\pi }{{8}}$$
  • none of these
Solve:
$$\int\limits_0^{\pi /6} {\dfrac{{\cos 2x}}{{{{\left( {\cos x - \sin x} \right)}^2}}}dx} $$
  • $$ - \log \left( {\dfrac{{\sqrt 3 - 1}}{2}} \right)$$
  • $$ - \log \left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)$$
  • $$\log \left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)$$
  • None of these
Solve $$\displaystyle\int\limits_0^{\pi /2} {{{\sin }^4}x{{\cos }^3}xdx} $$ 
  • $$\dfrac {6}{35}$$
  • $$\dfrac {2}{21}$$
  • $$\dfrac {2}{15}$$
  • $$\dfrac {2}{35}$$
Evaluate :
$$\displaystyle\int {\dfrac{9cosx-sinx}{4sinx+5cosx}dx}$$
  • $$x+\ln(4\sin x+\cos x)+c$$
  • $$x+\ln(\sin x-5\cos x)+c$$
  • $$x+\ln(4\sin x+5\cos x)+c$$
  • None of these
If $$\int { \cfrac { \sin { x }  }{ \sin { \left( x-\alpha  \right)  }  }  } dx=Ax+B\log { \sin { \left( x-\alpha  \right)  }  } +c$$, then the value of $$(A,B)$$ is-
  • $$\left( \sin { \alpha } ,\cos { \alpha } \right) $$
  • $$\left( \cos { \alpha } ,\sin { \alpha } \right) $$
  • $$\left( -\sin { \alpha } ,\cos { \alpha } \right) $$
  • $$\left(- \cos { \alpha } ,\sin { \alpha } \right) $$
Evaluate: $$\displaystyle\int _ { 0 } ^ { \pi / 2 } \dfrac { \sin x \cos x } { \cos ^ { 2 } x + 3 \cos x + 2 }dx$$
  • $$\ln\left(\dfrac {5}{3}\right)$$
  • $$\ln\left(\dfrac {4}{3}\right)$$
  • $$\ln\left(\dfrac {1}{3}\right)$$
  • None of these
$$\displaystyle \int_{0}^{2}\left( \sqrt {\dfrac{4-x}{x} }-  \sqrt {\dfrac{x}{4-x}} \right) dx $$  is equal to:
  • $$0$$
  • $$8$$
  • $$4$$
  • $$16$$
If $$f(x)=\int { \left( \dfrac { x^2+\sin^2x }{ 1+x^2 }  \right)  } \sec^2 x dx $$ and f(0)=0,then  f (1) equals:
  • $$1-\dfrac { \pi }{ 4 } $$
  • $$-\dfrac { \pi }{ 4 } $$
  • $$\tan 1-\dfrac { \pi }{ 4 } $$
  • $$\tan 1+1$$
$$\int _ { - 4 } ^ { - 5 } e ^ { ( x + 5 ) ^ { 2 } } d x + 3 \int _ { 1 / 3 } ^ { 2 / 3 } e ^ { 9 ( x - 2 / 3 ) ^ { 2 } } d x$$ is equal to-
  • $$e ^ { 5 }$$
  • $$e ^ { 4 }$$
  • 3$$e ^ { 2 }$$
  • 0
The integrating factor of the differential equation $$\dfrac{dy}{dx}\left(x\log _e\:x\right)+y=2\log _e\:x$$ is given by
  • $$x$$
  • $$e^x$$
  • $$\log _e\:x$$
  • $$\log _e\left(\log _e\:x\right)$$
If $$\displaystyle\int {{{{2^x}} \over {\sqrt {1 - {4^x}} }}dx = K{{\sin }^{ - 1}}} ({2^x}) + C,$$ then K is equal to 
  • $$\ell n2$$
  • $$\displaystyle{1 \over 2}\ell n2$$
  • $$\displaystyle{1 \over 2}$$
  • $$\displaystyle{1 \over {\ell n2}}$$
$$\int \{1 + 2 \tan x (\tan x + \sec x)\}^{\dfrac{1}{2}} dx =$$
  • $$log (\sec x + \tan x) + c$$
  • $$log (\sec x + \tan x)^{\dfrac{1}{2}} + c$$
  • $$log \,\sec x (\sec x + \tan x) + c$$
  • None of these
The value of the defined integral $$\displaystyle \int^{\pi/2}_{0}(\sin x+\cos x)\sqrt {\dfrac {e^{x}}{\sin x}}dx$$ equals
  • $$2\sqrt {e^{\pi/2}}$$
  • $$\sqrt {e^{\pi/2}}$$
  • $$2\sqrt {e^{\pi/2}}.\cos 1$$
  • $$\dfrac {1}{2}e^{\pi/4}$$
$$\int \log _ { 10 } x d x =$$
  • $$x \log _ { 10 } x + c$$
  • $$x \left( \log _ { 10 } x + \log _ { 10 } e \right) + c$$
  • $$\log _ { 10 } x + c$$
  • $$x \left( \log _ { 10 } x - \log _ { 10 } e \right) + c$$
If $$\int { f\left( x \right)dx=f\left( x \right) } ,$$ then $$\int { \left\{ f\left( x \right) \right\}^2  }$$ dx is equal to :
  • $$\frac { 1 }{ 2 } \left\{ f\left( x \right) \right\}^2 $$
  • $${ \left\{ f\left( x \right) \right\}^3 } $$
  • $$\dfrac { { \left\{ f\left( x \right) \right\} ^{ 3 } } }{ 3 } $$
  • $${ \left\{ f\left( x \right) \right\}^2 } $$
Evaluate: $$\displaystyle \int _ { 0 } ^ { \pi / 4 } \sec ^ { 7 } {\theta} \sin ^ { 3 } {\theta} {d \theta} =$$
  • $$\dfrac { 1 } { 12 }$$
  • $$\dfrac { 3 } { 12 }$$
  • $$\dfrac { 5 } { 12 }$$
  • $$\dfrac { 7 } { 12 }$$
$$\displaystyle \int \frac { 1 - x ^ { 2 } } { \left( 1 + x ^ { 2 } \right) \sqrt { 1 + x ^ { 4 } } } d x$$ is equal to 
  • $$\sqrt { 2 } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$$
  • $$\frac { 1 } { \sqrt { 2 } } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$$
  • $$\frac { 1 } { 2 } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$$
  • $$\frac { 1 } { \sqrt { 2 } } \sin ^ { - 1 } \left\{ \frac { x ^ { 2 } + 1 } { \sqrt { 2 } x } \right\} + c$$
Integral of $$f ( x ) = \sqrt { 1 + x ^ { 2 } }$$ with respect to $$x ^ { 2 }$$ is
  • $$\frac { 2 } { 3 } \frac { \left( 1 + x ^ { 2 } \right) ^ { 3 / 2 } } { x } + k$$
  • $$\frac { 2 } { 3 } \left( 1 + x ^ { 2 } \right) ^ { 3 / 2 } + k$$
  • $$\frac { 2 } { 3 } x \left( 1 + x ^ { 2 } \right) ^ { 3 / 2 } + k$$
  • None of these
$$\int { { e }^{ x^{ 3 } }+{ x }^{ 2-1 }(3{ x }^{ 4 }+{ 2x }^{ 3 }+{ 2x }^{ 2 }\quad x=h(x)+c } $$ then the value of $$h(1)h(-1)$$.
  • 1
  • -1
  • 2
  • -2
$$\int _{ 0 }^{ 1 }{ \dfrac { dx }{ \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+2 \right)  }  } $$=
  • $$\dfrac { \pi }{ 4 } +\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } } $$
  • $$\dfrac { \pi }{ 2 } -\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } } $$
  • $$\dfrac { \pi }{ 4 } -\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } } $$
  • $$\dfrac { \pi }{ 3 } -\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } } $$
The value of the integral $$\displaystyle \int_{-\pi/2}^{\pi/2} \left(x^{2}+\log \dfrac{\pi-x}{\pi+x}\right) \cos x dx $$ is 
  • $$0$$
  • $$\dfrac{\pi^{2}}{2}-4$$
  • $$\dfrac{\pi^{2}}{2}+4$$
  • $$\dfrac{\pi^{2}}{2}$$
The value of $$\int _{ -1 }^{ 1 }{ \dfrac { { cot }^{ -1 }x }{ \pi  }  } dx$$
  • 1
  • 2
  • 3
  • 0
$$\displaystyle \int _{ 0 }^{ { \pi  }^{ 2 } }{ \dfrac { \sin { \sqrt { x }  }  }{ \sqrt { x }  }  }  dx$$ is equal to
  • $$2$$
  • $$1$$
  • $$1/2$$
  • $$4$$
The integral $$\displaystyle{\int}_{\pi/12}^{\pi/4}\dfrac{8\cos 2x}{\left(\tan x+\cot x\right)^{3}}dx$$ equals:
  • $$\dfrac{15}{128}$$
  • $$\dfrac{13}{156}$$
  • $$\dfrac{15}{64}$$
  • $$\dfrac{13}{32}$$
Solve:$$\int {\dfrac{{dx}}{{\left( {x - 3} \right)\sqrt {x + 1} }}} $$
  • $$\cos { { h }^{ -1 } } \left( \dfrac { 1-x }{ \sqrt { 3 } \left( 1+x \right) } \right) +c$$
  • $$\sin { { h }^{ -1 } } \left( \dfrac { 1-x }{ \sqrt { 3 } \left( 1+x \right) } \right) +c$$
  • $$-\sin { { h }^{ -1 } } \left( \dfrac { 1-x }{ \sqrt { 3 } \left( 1+x \right) } \right) +c$$
  • $$-\cos { { h }^{ -1 } } \left( \dfrac { 1-x }{ \sqrt { 3 } \left( 1+x \right) } \right) +c$$
The integral $$\int _{ 2a/4 }^{ a/2 }{ (2\quad cosecx{ ) }^{ 17 } } $$ dx is equal to:
  • $$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ 2({ e }^{ u }+{ e }^{ -u }{ ) }^{ 16 } } du$$
  • $$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ ({ e }^{ u }+{ e }^{ -u }{ ) }^{ 17 } } du$$
  • $$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ ({ e }^{ u }-{ e }^{ -u }{ ) }^{ 17 } } du$$
  • $$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ 2({ e }^{ u }-{ e }^{ -u }{ ) }^{ 16 } } du$$
$$\begin{matrix} lim \\ n\rightarrow \infty  \end{matrix}\int _{ 0 }^{ 1 }{ \frac { { nx }^{ { n- }1 } }{ { 1+x }^{ 2 } } dx= } $$
  • 0
  • 1
  • 2
  • $$\frac { 1 }{ 2 } $$
The value of the integral $$\int _{ -\pi /2 }^{ \pi /2 }{ \left[ { x }^{ 2 }+log\frac { \pi -x }{ \pi +x }  \right]  } $$ cos x dx is 
  • 0
  • $$\frac { { \pi }^{ 2 } }{ 2 } -4$$
  • $$\frac { { \pi }^{ 2 } }{ 2 } +4$$
  • $$\frac { { \pi }^{ 2 } }{ 2 } $$
$$\int _ { 0 } ^ { 1 } \frac { d x } { \sqrt { x + 1 } + \sqrt { x } } d x =$$
  • $$\frac { 4 } { 3 } ( \sqrt { 2 } + 1 )$$
  • $$\frac { 4 } { 3 } ( \sqrt { 2 } - 1 )$$
  • $$\frac { 3 } { 4 } ( \sqrt { 2 } - 1 )$$
  • $$\frac { 3 } { 4 } ( \sqrt { 2 } - 2 )$$
The value of the definite integral
$$\overset { { a }_{ 1 } }{ \underset { { a }_{ 2 } }{ \int { \frac { d\theta  }{ 1+tan\theta  }  }  }  } =\frac { 501\pi  }{ K } $$ where $$\ a _{ 2 }=\quad \frac { 1003\pi  }{ 2008 } $$ and $${ \ a  }_{ 1 }=\frac { \pi  }{ 2008 } $$ The value of K equalls
  • 2007
  • 2006
  • 2009
  • 2008
For $$x\in R,\ f(x)=|\log 2-\sin x|$$ and $$g(x)=f(f(x))$$, then 
  • $$g'(0)=\cos (\log 2)$$
  • $$g'(0)=-\cos (\log 2)$$
  • $$g$$ is differentible at $$x=0$$ and $$g'(0)=-\sin (\log 2)$$
  • $$g$$ is not differentiable at $$x=0$$
The integral $$\displaystyle \int { \left( 1+2{ x }^{ 2 }+\frac { 1 }{ x }  \right)  } { e }^{ { x }^{ 2-\frac { 1 }{ x }  } }dx$$ is equal to
  • $$(2x-1).e^{x^{2-\dfrac {1}{x}}}+c$$
  • $$(2x+1).e^{x^{2-\dfrac {1}{x}}}+c$$
  • $$xe^{x^{2-\dfrac {1}{x}}}+c$$
  • $$-xe^{x^{2-\dfrac {1}{x}}}+c$$
$$\displaystyle \int \dfrac {1}{\sqrt {\sin^{3}x\sin(x+a)}}dx$$ is equal to
  • $$2\cos ec\alpha \sqrt {\cos \alpha +\sin \alpha \tan x}+c$$
  • $$-2\cos ec\alpha \sqrt {\cos \alpha +\sin \alpha \cot x}+c$$
  • $$\cos ec\alpha \sqrt {\cos \alpha +\sin \alpha \tan x}+c$$
  • $$-\cos ec\alpha \sqrt {\cos \alpha +\sin \alpha \tan x}+c$$
$$\int {\dfrac{1}{{9{x^2} - 25}}dx = \_\_\_\_\_\_ + c.} $$
  • $$\dfrac{1}{{30}}\log \left| {\dfrac{{3x + 5}}{{3x - 5}}} \right|$$
  • $$\log \left| {x + \sqrt {3x - 5} } \right|$$
  • $$\dfrac{1}{{30}}\log \left| {\dfrac{{3x - 5}}{{3x + 5}}} \right|$$
  • $$\log \left| {x - \sqrt {3x - 5} } \right|$$
$$\int {{e^{3{{\log }_e}x}}.{{\left( {{x^4} + 1} \right)}^{ - 1}}dx = \_\_\_\_\_\_\_\_\_ + C.} $$
  • $$\log \left( {{x^4} + 1} \right)$$
  • $$\frac{1}{4}\log \left( {{x^4} + 1} \right)$$
  • $$-\log \left( {{x^4} + 1} \right)$$
  • $$\frac{{ - 3}}{{{{\left( {{x^4} + 1} \right)}^3}}}$$
$$\int {\sqrt {1 - \cos x} \,dx = \_\_\_\_\_\_\_ + C;\,2\pi < x < 3\pi } $$
  • $$ - 2\sqrt 2 \cos \dfrac{x}{2}$$
  • $$ - \sqrt 2 \cos \dfrac{x}{2}$$
  • $$ 2\sqrt 2 \cos \dfrac{x}{2}$$
  • $$\dfrac{{ - 1}}{2}\sqrt 2 \cos \dfrac{x}{2}$$
$$\int _{ 0 }^{ 400\pi  }{ \sqrt { 1-\cos { 2x }  }  }$$
  • $$200\sqrt 2$$
  • $$400\sqrt 2$$
  • $$800\sqrt 2$$
  • $$none$$
If $$g\left( x \right) =\int { { x }^{ x }\log _{ e }{ (ex)dx }  } $$ then  $$g\left( \pi \right) $$ equals
  • $$\pi \log _{ e }{ \pi } $$
  • $${ \pi }^{ \pi }\log _{ e }{ (e\pi } )$$
  • $${ \pi }^{ \pi }\log _{ e }{ (\pi } )$$
  • $${\pi}^\pi$$
If $$f(a-x)=-f(x)$$, then $$\displaystyle \int_{0}^{a}f(x)dx=0$$.
  • True
  • False
$$\displaystyle \int_{0}^{1}\sin^{-1}x dx=\dfrac {\pi}{2}-1$$
  • True
  • False
$$\displaystyle\int{{e}^{{\tan}^{-1}{x}}\left(\dfrac{1+x+{x}^{2}}{1+{x}^{2}}\right)dx}$$ is equal to
  • $${ -e }^{ { tan }^{ -1 }x }+c$$
  • $${ e }^{ { tan }^{ -1 }x }+c$$
  • $${ -xe }^{ { tan }^{ -1 }x }+c$$
  • $${ xe }^{ { tan }^{ -1 }x }+c$$
Integrate: $$\displaystyle \int \dfrac{x}{\sqrt{x+4}}dx$$
  • $$\dfrac 23(x+4)^{\tfrac 32}-8\sqrt{x+4}$$
  • $$\dfrac 23(x+4)^{\tfrac 32}+8\sqrt{x+4}$$
  • $$\dfrac 23(x+4)^{\tfrac 32}+4\sqrt{x+4}$$
  • None of these
$$\int _{ 0 }^{ 1 }{ \frac { x }{ { \left( { x }^{ 2 }+1 \right)  }^{ \frac { 3 }{ 2 }  } } dx } =........$$
  • $$\dfrac{1}{3}$$
  • $$\dfrac{2}{3}$$
  • $$\dfrac{3}{2}$$
  • $$1-\dfrac{1}{\sqrt{2}}$$
$$\int _{ -1 }^{ 1/2 }{ \dfrac { { e }^{ x }\left( 2-{ x }^{ 2 } \right) dx }{ \left( 1-x \right) \sqrt { 1-{ x }^{ 2 } }  }  } $$ is equal to
  • $$\dfrac { \sqrt { e } }{ 2 } \left( \sqrt { 3 } +1 \right) $$
  • $$\dfrac { \sqrt { 3e } }{ 2 } $$
  • $$\sqrt { 3e } $$
  • $$\sqrt { \dfrac { e }{ 3 } } $$
If $$\int _{ log2 }^{ x }{ \dfrac { dx }{ \sqrt { { e }^{ x }-1 }  }  } =\dfrac { \pi  }{ 6 } ,$$then x is equal to _________.
  • 4
  • in 8
  • in 4
  • None of these
If $$\displaystyle\int^{\dfrac{\pi}{2}}_0\dfrac{\cot x}{\cot x+cosec x}dx=m(\pi +n)$$, then $$mn$$ is equal to?
  • $$-1$$
  • $$1$$
  • $$\dfrac{1}{2}$$
  • $$-\dfrac{1}{2}$$
What is $$\displaystyle \int \dfrac{dx}{2x^2 - 2x + 1}$$ equal to ?
  • $$\dfrac{\tan^{-1} (2x - 1)}{2} + c$$
  • $$2 \tan^{-1} (2x - 1) += c$$
  • $$\dfrac{\tan^{-1} (2x + 1)}{2} + c$$
  • $$\tan^{-1} (2x - 1) + c$$
The value of $$\displaystyle\int^{2\pi}_{0}\dfrac{x\sin^8x}{\sin^8x+\cos^8x}dx$$ is equal to?
  • $$2\pi$$
  • $$\pi^2$$
  • $$2\pi^2$$
  • $$4\pi$$
Select and write the most appropriate answer from the given alternatives for question :
If $$\displaystyle \int^k_0 4x^3dx=16$$, then the value of $$k$$ is _____.
  • $$1$$
  • $$2$$
  • $$3$$
  • $$4$$
0:0:1


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