Interpretation of define integral as an area - class-XII
| Description: interpretation of define integral as an area | |
| Number of Questions: 26 | |
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| Tags: maths physics integrals business maths mathematical tools elementary mathematics integral calculus definite integral definite integrals definite integration applications of integration definite integral and it applications to areas |
The value of $\displaystyle \int _0^1\tan^{-1}\left (\frac {2x-1}{1+x-x^2}\right )dx$ is
-
$1$
-
$0$
-
$-1$
-
$\dfrac {\pi}{4}$
Let $I=\int _0^1\tan^{-1}\left (\dfrac {2x-1}{1+x-x^2}\right )dx$
$\Rightarrow I=\int _0^1 \tan^{-1}\left (\dfrac {x-(1-x)}{1+x(1-x)}\right )dx$
$\Rightarrow I=\int _0^1[\tan^{-1}x-\tan^{-1}(1-x)]dx$ ................ (1)
$\Rightarrow I=\int _0^1[\tan^{-1}(1-x)-\tan^{-1}(1-1+x)]dx$
$\Rightarrow I=\int _0^1[\tan^{-1}(1-x)-\tan^{-1}(x)]dx$
$\Rightarrow I=\int _0^1[\tan^{-1}(1-x)-\tan^{-1}(x)]dx$ ........... (2)
Adding (1) and (2), we obtain
$2I=\int _0^1(\tan^{-1}x+\tan^{-1}(1-x)-\tan^{-1}(1-x)-\tan^{-1}x)dx=0$
$\Rightarrow I=0$
Hence, the correct Answer is B.
$\int _{0}^{\pi /2}sin2xtan^{-1}\left ( sinx \right )dx=$
-
$\dfrac{\pi }{2}$-1
-
$\dfrac{\pi }{2}$+1
-
$\dfrac{3\pi }{2}$+1
-
$\dfrac{3\pi }{2}$-1
We have,
$I=\int _{0}^{\dfrac{\pi }{2}}{\sin 2x{{\tan }^{-1}}\left( \sin x \right)dx}$
$=\int _{0}^{\dfrac{\pi }{2}}{2\sin x\cos x{{\tan }^{-1}}\left( \sin x \right)dx}$
Let
$ \sin x=t $
$ \cos xdx=dt $
Change limit
$ \sin 0=t $
$ t=0 $
And,
$ \sin \dfrac{\pi }{2}=t $
$ t=1 $
Then,
$ \int _{0}^{1}{2t{{\tan }^{-1}}t\cos xdx} $
$ =\int _{0}^{1}{2t{{\tan }^{-1}}tdt} $
$ =2\int _{0}^{1}{t{{\tan }^{-1}}tdt} $
On integrating and we get,
$ 2\left[ {{\tan }^{-1}}t\int _{0}^{1}{t}dt-\int _{0}^{1}{\left( \dfrac{d\left( {{\tan }^{-1}}t \right)}{dt}\int _{0}^{1}{tdt} \right)}dt \right] $
$ =2\left[ {{\tan }^{-1}}t\left[ {{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1} \right]-\int _{0}^{1}{\dfrac{1}{1+{{t}^{2}}}}\dfrac{{{t}^{2}}}{2}dt \right] $
$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{\dfrac{{{t}^{2}}}{1+{{t}^{2}}}}dt $
$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{\dfrac{{{t}^{2}}+1-1}{1+{{t}^{2}}}}dt $
$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{\dfrac{{{t}^{2}}+1}{1+{{t}^{2}}}}dt+\int _{0}^{1}{\dfrac{1}{1+{{t}^{2}}}}dt $
$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{1}dt+\int _{0}^{1}{\dfrac{1}{1+{{t}^{2}}}}dt $
$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-{{\left[ t \right]} _{0}}^{1}+{{\left[ {{\tan }^{-1}}t \right]} _{0}}^{1}+C $
$ =2\left[ {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right]\left[ \dfrac{{{1}^{2}}}{2}-\dfrac{{{0}^{2}}}{2} \right]-\left[ 1-0 \right]+\left[ {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right]+C $
$ =2\left[ \dfrac{\pi }{4}-0 \right]\left[ \dfrac{1}{2} \right]-1+\left[ \dfrac{\pi }{4}-0 \right] $
$ =\dfrac{\pi }{4}+\dfrac{\pi }{4}-1 $
$ =\dfrac{2\pi }{4}-1 $
$ =\dfrac{\pi }{2}-1 $
Hence, this is the answer.
Evaluate: $\displaystyle \int _{0}^{\sqrt{3}}[x^{3} -1] dx$
-
$\dfrac{1}{4}-\sqrt3$
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$\dfrac{1}{4}-\sqrt2$
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$\dfrac{9}{4}-\sqrt3$
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$\dfrac{9}{4}-\sqrt2$
Consider, $\displaystyle I= \int _{0}^{\sqrt{3}}[x^{3} -1] dx$
$\displaystyle\int^{100} _0[\tan^{-1}x]dx$.
-
$100+\tan 1$
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$100-\tan 1$
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$\tan 1$
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$99+\tan 1$
Solve $\displaystyle\int^{100} _0e^{x-[x]}dx=?$ where $[x]$ is greatest integer function.
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$100e$
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$100(e-1)$
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$100(e+1)$
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$100(1-e)$
$\int _0^\pi {{x^2}\,g\left( x \right)\,dx\, = } $
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0
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$\frac{\pi }{8}$
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$\frac{8}{{{\pi ^2}}}$
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$\frac{16}{{{\pi ^2}}}$
$\int _0^\pi {f\left( x \right)\,dx\, = } $
-
0
-
$\frac{8}{\pi }$
-
$\frac{8}{{{\pi ^2}}}$
-
$\frac{16}{{{\pi ^2}}}$
If $I _1 = \displaystyle \int^{2\pi /3} _{\pi / 2}\left|cos\dfrac{x}{2}cosx\right|dx,I _2=\left|\displaystyle \int _{\pi/2}^{2\pi/3} cos\dfrac{x}{2}cosxdx\right|$ then $I _1 - I _2$ equals
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$\dfrac{1}{3}(\sqrt{32}-\sqrt{27})$
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$\dfrac{1}{3}(\sqrt{32}-\sqrt{25})$
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$\dfrac{1}{3}(\sqrt{27}-\sqrt{25})$
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None
The value of the definite integral, $\displaystyle \int _0^{\pi/2} \dfrac{sin5x}{sinx}dx$ is
-
0
-
$\dfrac{\pi}{2}$
-
$\pi$
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$2\pi$
The value of the definite integral $\int _{ 0 }^{ \pi /2 }{ \sin { x } \sin { 2x } \sin { 3x } dx } $ is equal to:
-
$\cfrac{1}{3}$
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$-\cfrac{2}{3}$
-
$-\cfrac{1}{3}$
-
$\cfrac{1}{6}$
$\int _0^{\pi/2}\sin x\sin 2x\sin 3x dx$
The value of the integral $\displaystyle\int{\sin{x}{\cos}^{4}{x}dx}$ where $x\in\left[-1,\,1\right]$ is
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1
-
1\2
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0
-
4
If $\Delta (x)=\left| \begin{matrix} 1+x+2{ x }^{ 2 } & x+3 & 1 \ x+2{ x }^{ 2 } & x & 3 \ 3x+6{ x }^{ 2 } & 3x+11 & 9 \end{matrix} \right| $ then $\displaystyle \int^{1} _{0}\Delta (x)dx$ is
-
$\dfrac {176}{5}$
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$-\dfrac {176}{3}$
-
$\dfrac {186}{3}$
-
$-\dfrac {192}{3}$
-
${ \phi }^{ 2 }$
-
$2{ \phi }^{ 2 }$
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$3{ \phi }^{ 2 }$
-
$2{ \phi }^{ 3 }$
$\displaystyle \int _{1}^{4}\frac{\mathrm{x}\mathrm{d}\mathrm{x}}{\sqrt{2+4\mathrm{x}}}=$
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$\displaystyle \frac{1}{2}$
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$\displaystyle \frac{1}{\sqrt{2}}$
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$\displaystyle \frac{3}{2}$
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$\displaystyle \frac{3}{\sqrt{2}}$
$\int _{1}^{4}\dfrac{x
dx}{\sqrt{2 + 4x}}=\dfrac{1}{2}\int _{1}^{4}\dfrac{x dx}{\sqrt{x+\dfrac{1}{2}}}$
$=\dfrac{1}{2}\left [ \int _{1}^{4}\dfrac{(x+\dfrac{1}{2})dx}{\sqrt{x+\dfrac{1}{2}}}-\int _{1}^{4}\dfrac{\dfrac{1}{2}dx}{\sqrt{x+\dfrac{1}{2}}}
\right ]$
$=\dfrac{1}{2}\left [ \int _{1}^{4} \sqrt{x+\dfrac{1}{2}}
dx-\int _{1}^{4}(x+\dfrac{1}{2})^{\dfrac{1}{2}} \int _{1}^{4} \right ]$
$=\dfrac{1}{2} \left [ \dfrac{2}{3} (x+\dfrac{1}{2})^{\dfrac{3}{2}}
\int _{1}^{4}-(x+\dfrac{1}{2})^{\dfrac{4}{2}} \int _{1}^{4} \right ]$
$=\dfrac{1}{3} \left [ \left ( \dfrac{9}{2}
\right )^{\dfrac{3}{2}}-\left ( \dfrac{3}{2} \right )^{\dfrac{3}{2}} \right ]
-\dfrac{1}{2} \left [ \left ( \dfrac{9}{2} \right )^{\dfrac{1}{2}}-\left ( \dfrac{3}{2}
\right )^{\dfrac{1}{2}} \right ]$
$=\dfrac{1}{3} \left [ \left ( \dfrac{9}{2}
\right )\left ( \dfrac{9}{2} \right )^{\dfrac{1}{2}}-\left ( \dfrac{3}{2}
\right )\left ( \dfrac{3}{2} \right )^{\dfrac{1}{2}} \right ] - \dfrac{1}{2}
\left [ \left ( \dfrac{3}{\sqrt{2}} \right )-\dfrac{\sqrt{3}}{\sqrt{2}} \right
]$
$=\dfrac{3}{\sqrt{2}}$
The value of $\displaystyle \int _{0}^{2}(x-\log _{2}a)dx=2\log _{2}(\frac{2}{a})$ for which of the following conditions?
-
$\mathrm{a}>0$
-
$\mathrm{a}>2$
-
$\mathrm{a}=4$
-
$\mathrm{a}=8$
$\int _{ 0 }^{ 2 }{ (x-\log _{ 2 }a } )dx=2\log _{ 2 }(\cfrac { 2 }{ a } )$
Consider the integral $I=\displaystyle\int^{\pi} _0 ln(\sin x)dx$.What is $\displaystyle\int^{\dfrac{\pi}{2}} _{0}$ ln $(\sin x)dx$ equal to?
-
$4I$
-
$2I$
-
$I$
-
$\dfrac{I}{2}$
$I = \displaystyle \int _{0}^{\pi} {ln(\sin x)dx}$
Consider the integral $I=\displaystyle\int^{\pi} _0 ln(\sin x)dx$.What is $\displaystyle\int^{\frac{\pi} {2}} _0 ln(\cos x)dx$ equal to?
-
$\dfrac{I}{2}$
-
$I$
-
$2I$
-
$4I$
$I = \displaystyle \int _{0}^{\pi} {ln(\sin x)dx}$
$ \int _{\sin x}^1 t^2 f(t) dt = 1 - \sin x \forall x \epsilon (0, \pi / 2 ) $ then $ f \left( \dfrac {1}{\sqrt3} \right) $ is :
-
$3$
-
$\sqrt3$
-
$1/3$
-
None of these
$ \int _{\sin x}^1 t^2 f(t) dt = 1 - \sin x \forall x \epsilon (0, \pi/2 ) $
Differentiating both sides we get
$ \dfrac {d}{dx} (1) [ 1 \cdot f (1)] - \cos x ( \sin^2 x) f ( \sin x) = -\cos x $
$ \Rightarrow f ( \sin x) = \dfrac {1}{ \sin^2 x} $
$ \therefore f \left( \dfrac {1}{\sqrt3} \right) = f \left( \sin \left( \sin^{-1} \dfrac {1}{\sqrt3} \right) \right) $
$ = \left[ \dfrac {1}{ \sin \left( \sin^{-1} \dfrac {1}{\sqrt3} \right)} \right]^2 = 3 $
Consider the integrals ${I _1} = \int _0^1 {{e^{ - x}}{{\cos }^2}xdx,} {I _2} = \int _0^1 {{e^{ - {x^2}}}{{\cos }^2}xdx,} {I _3} = \int _0^1 {{e^{ - x}}dx} $ and ${I _4} = \int _0^1 {{e^{ - (1/2){x^2}}}} dx$. The greatest of these integrals is
-
$I _1$
-
$I _2$
-
$I _3$
-
$I _4$
$I _1=\int _{0}^{4}e^{-x} cos^2x dx$
$I _2=\int _{0}^{1}e^{-x^2}cos^2x dx$
Both have $cos^2x$ so value get restricted more in (0, 1)
Now in (0, 1) $e^{-\frac {x^2}{2}}>e^{-x}$
$\therefore \int _{0}^{1}e^{-\frac {x^2}{2}}>\int _{0}^{1}e^{-x}$
$\therefore I _4>I _3>I _1>I _2$
Let $ f\left( a,b \right) =\int _{ a }^{ b }{ \left( { x }^{ 2 }-4x+3 \right) dx,\left( b>a \right) }$ then
-
$ f\left( a,3 \right)$ is least when $a=1$
-
$f\left( 4,b \right)$ is an increasing function $ \forall b\ge 4$
-
$ f\left( 0,b \right)$ is least for $b=2$
-
$ \min { \left{ f\left( a,b \right) \right} =-\dfrac { 4 }{ 3 } } \forall a,b\in R$
First of all integrate the function.
$\displaystyle \int _0^1 \dfrac{xe^x}{(x + 1)^2} dx =$
-
$\dfrac{e}{2}$
-
$\dfrac{e - 1}{2}$
-
$\dfrac{3e}{2} -1$
-
$\dfrac{e - 3}{2}$
$\displaystyle\int _{ 0 }^{ 1 }{ \cfrac { \tan ^{ -1 }{ x } }{ x } } dx$ equals
-
$\displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { x }{ \sin { x } } dx } $
-
$\cfrac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { x }{ \sin { x } } dx } $
-
$\displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { \sin { x } }{ x } dx } $
-
None of the above
Let $I=\int _{ 0 }^{ 1 }{ \cfrac { \tan ^{ -1 }{ x } }{ x } } dx$
Put $\tan ^{ -1 }{ x } =\cfrac { z }{ 2 } \Rightarrow x=\tan { \cfrac { z }{ 2 } } $
$\Rightarrow dx=\cfrac { 1 }{ 2 } \sec ^{ 2 } \cfrac { z }{ 2 } dz$
$\therefore \quad I=\displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { \cfrac { z }{ 2 } \left( \cfrac { 1 }{ 2 } \sec ^{ 2 } \cfrac { z }{ 2 } \right) }{ \tan { \cfrac { z }{ 2 } } } } dz$
$I=\cfrac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { z }{ \cfrac { \sin { \cfrac { z }{ 2 } } }{ \cos { \cfrac { z }{ 2 } } } } .\cfrac { 1 }{ 2\cos ^{ 2 }{ \cfrac { z }{ 2 } } } dz } $
$=\cfrac { 1 }{ 2 } \int _{ 0 }^{ \pi /2 }{ \cfrac { z }{ 2\sin { \cfrac { z }{ 2 } } \cos { \cfrac { z }{ 2 } } } } dz$
$=\cfrac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { z }{ \sin { z } } } dz=\cfrac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { x }{ \sin { x } } } dx$
Evaluate $\displaystyle\int^{\frac{3}{2}} _{-1}|x\sin(\pi x)|dx$.
-
$\dfrac {3}{\pi} +\dfrac {1}{\pi^2}$
-
$3\pi +\pi^2$
-
$\dfrac { 2 }{ \pi } +\dfrac { 1 }{ { \pi }^{ 2 } }$
-
none of the above
$|x\sin(\pi x)|=\begin{cases}x\sin \pi x ,\,\,\,\,x\in(-1,1)\ -x\sin\pi x,x\in (1,\dfrac{3}{2}) \end{cases}$
$\int _{ 0 }^{ \infty }{ f\left( x+\cfrac { 1 }{ x } \right) .\cfrac { \ln { x } }{ x } } dx$
-
Is equal to zero
-
Is equal to one
-
Is equal to $\cfrac { 1 }{ 2 } $
-
Can not be evaluated
Evaluate $I = \displaystyle \int _{\pi /6}^{\pi /3}\sin x:dx$
-
$\displaystyle \frac{1-\sqrt{3}}{2}$
-
$\displaystyle \frac{\sqrt{3}+1}{2}$
-
$\displaystyle \frac{\sqrt{3}-1}{2\sqrt{3}}$
-
None of these
What is $\displaystyle \int _{ 0 }^{ \pi }{ { e }^{ x } } \sin { x } dx$ equal to?
-
$\cfrac { { e }^{ \pi }+1 }{ 2 } $
-
$\cfrac { { e }^{ \pi }-1 }{ 2 } $
-
${ e }^{ \pi }+1$
-
$\cfrac { { e }^{ \pi }+1 }{ 4 } $
Using Integration by part$:-$