Tag: fundamental theorem of calculus

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Questions Related to fundamental theorem of calculus

Multiple choice 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

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

  1. $1$
  2. $0$
  3. $-1$
  4. $\dfrac {\pi}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice 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

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

  1. $\dfrac{\pi }{2}$-1
  2. $\dfrac{\pi }{2}$+1
  3. $\dfrac{3\pi }{2}$+1
  4. $\dfrac{3\pi }{2}$-1
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice 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

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

  1. $\dfrac{1}{4}-\sqrt3$
  2. $\dfrac{1}{4}-\sqrt2$
  3. $\dfrac{9}{4}-\sqrt3$
  4. $\dfrac{9}{4}-\sqrt2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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


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

$I=\dfrac94-\sqrt3$

Multiple choice 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

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

  1. $100+\tan 1$
  2. $100-\tan 1$
  3. $\tan 1$
  4. $99+\tan 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The integral is the sum of integrals from n to n+1. For x in [tan(n), tan(n+1)], [tan^-1(x)] = n. The sum is 0*integral(0 to tan(1)) + 1*integral(tan(1) to tan(2)) + ... + 99*integral(tan(99) to tan(100)). This simplifies to 100 - tan(1).

Multiple choice 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

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

  1. $100e$
  2. $100(e-1)$
  3. $100(e+1)$
  4. $100(1-e)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Consider, $I=\displaystyle\int^{100} _0e^{x-[x]}dx$

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

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

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

$I=100(e-1)$
Multiple choice 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

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 

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice 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

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

  1. 0
  2. $\dfrac{\pi}{2}$
  3. $\pi$
  4. $2\pi$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $n=s (odd)$
$\displaystyle \int _{0}^{\dfrac{\pi}{2}} \dfrac{\sin 5x}{\sin x} =\dfrac{\pi}{2}$

Multiple choice 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

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

  1. $\cfrac{1}{3}$
  2. $-\cfrac{2}{3}$
  3. $-\cfrac{1}{3}$
  4. $\cfrac{1}{6}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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   $2\sin A \cos B=\sin(A+B)+\sin (A-B)$

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

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

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

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

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