Tag: fundamental theorem of integral calculus

Questions Related to fundamental theorem of integral calculus

The value of $\displaystyle \int _0^1\tan^{-1}\left (\frac {2x-1}{1+x-x^2}\right )dx$ is

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $1$

  2. $0$

  3. $-1$

  4. $\dfrac {\pi}{4}$

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Correct Option: B
Explanation:

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.

$\int _{0}^{\pi /2}sin2xtan^{-1}\left ( sinx \right )dx=$

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1.  $\dfrac{\pi }{2}$-1

  2.  $\dfrac{\pi }{2}$+1

  3.  $\dfrac{3\pi }{2}$+1

  4.  $\dfrac{3\pi }{2}$-1

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Correct Option: A
Explanation:

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: $\displaystyle \int _{0}^{\sqrt{3}}[x^{3} -1] dx$

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $\dfrac{1}{4}-\sqrt3$

  2. $\dfrac{1}{4}-\sqrt2$

  3. $\dfrac{9}{4}-\sqrt3$

  4. $\dfrac{9}{4}-\sqrt2$

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Correct Option: C
Explanation:

Consider, $\displaystyle I= \int _{0}^{\sqrt{3}}[x^{3} -1] dx$


$\Rightarrow I=\left [\dfrac{x^4}{4}-x\right]^{\sqrt3} _{0}$

$I=\dfrac94-\sqrt3$

$\displaystyle\int^{100} _0[\tan^{-1}x]dx$.

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $100+\tan 1$

  2. $100-\tan 1$

  3. $\tan 1$

  4. $99+\tan 1$

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Correct Option: B

Solve $\displaystyle\int^{100} _0e^{x-[x]}dx=?$ where $[x]$ is greatest integer function.

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $100e$

  2. $100(e-1)$

  3. $100(e+1)$

  4. $100(1-e)$

Show answer
Correct Option: B
Explanation:
Consider, $I=\displaystyle\int^{100} _0e^{x-[x]}dx$

$I=100\displaystyle\int^{1} _0e^{x}dx$

$I=100(e^x) _0^1$

$I=100(e^1-e^0)$

$I=100(e-1)$

$\int _0^\pi  {{x^2}\,g\left( x \right)\,dx\, = } $

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. 0

  2. $\frac{\pi }{8}$

  3. $\frac{8}{{{\pi ^2}}}$

  4. $\frac{16}{{{\pi ^2}}}$

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Correct Option: B

$\int _0^\pi  {f\left( x \right)\,dx\, = } $

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. 0

  2. $\frac{8}{\pi }$

  3. $\frac{8}{{{\pi ^2}}}$

  4. $\frac{16}{{{\pi ^2}}}$

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Correct Option: B

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 

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $\dfrac{1}{3}(\sqrt{32}-\sqrt{27})$

  2. $\dfrac{1}{3}(\sqrt{32}-\sqrt{25})$

  3. $\dfrac{1}{3}(\sqrt{27}-\sqrt{25})$

  4. None

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Correct Option: A

The value of the definite integral, $\displaystyle \int _0^{\pi/2} \dfrac{sin5x}{sinx}dx$ is 

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. 0

  2. $\dfrac{\pi}{2}$

  3. $\pi$

  4. $2\pi$

Show answer
Correct Option: B
Explanation:
$\displaystyle = \int _{0}^{\dfrac{\pi}{2}} \dfrac{\sin 5x}{\sin x} dx$
We are going to use a important property if.
$\displaystyle \int _{0}^{\dfrac{\pi}{2}} \dfrac{\sin nx}{\sin x} = \begin{cases} \dfrac{\pi}{2} & ,\ if\ n\ is\ odd \\ o & ,\ if\ is\ even \end{cases}$
So, less $n=s (odd)$
$\displaystyle \int _{0}^{\dfrac{\pi}{2}} \dfrac{\sin 5x}{\sin x} =\dfrac{\pi}{2}$

The value of the definite integral $\int _{ 0 }^{ \pi /2 }{ \sin { x } \sin { 2x } \sin { 3x } dx } $ is equal to:

business maths definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $\cfrac{1}{3}$

  2. $-\cfrac{2}{3}$

  3. $-\cfrac{1}{3}$

  4. $\cfrac{1}{6}$

Show answer
Correct Option: D
Explanation:

$\int _0^{\pi/2}\sin x\sin 2x\sin 3x dx$


$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2}2\sin x\sin 2x\sin 3x dx$

$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2}2\sin x\sin 3x\sin 2x dx$

We know that       $2\sin A \sin B=\cos(A-B)-\cos (A+B)$

$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2}(\cos (x-3x)-\cos(x+3x))\sin 2x dx$

$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2}(\cos 2x-\cos 4x)\sin 2x dx$

$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2} \sin 2x \cos 2x dx-\dfrac{1}{2}\int _0^{\pi /2}\sin 2x \cos 4x dx$

$\Rightarrow \dfrac{1}{4}\int _0^{\pi/2} 2\sin 2x \cos 2x dx-\dfrac{1}{4}\int _0^{\pi /2}2\sin 2x \cos 4x dx$

We know that   $2\sin A \cos B=\sin(A+B)+\sin (A-B)$

$\Rightarrow \dfrac{1}{4}\int _0^{\pi/2} (\sin (2x+2x)+\sin (2x-2x))dx-\dfrac{1}{4}\int _0^{\pi /2}(\sin (2x+4x)+\sin (2x-4x)) dx$

$\Rightarrow \dfrac{1}{4}\int _0^{\pi/2} \sin 4xdx-\dfrac{1}{4}\int _0^{\pi /2}(\sin 6x-\sin 2x) dx$

$\Rightarrow \dfrac{1}{4}\int _0^{\pi/2} \sin 4xdx-\dfrac{1}{4}\int _0^{\pi /2}\sin 6x dx+\dfrac{1}{4}\int _{0}^{\pi/4}\sin 2x dx$

$\Rightarrow \dfrac{1}{4}[\dfrac{\sin 4x}{4}] _0^{\pi/4}-\dfrac{1}{4}[\dfrac{-\sin 6x}{6}] _0^{\pi/4}+\dfrac{1}{4}[\dfrac{-\cos 2x}{2}] _0^{\pi/4}$

$\Rightarrow \dfrac{-1}{12}+\dfrac{1}{4}=\dfrac{1}{6}$