Tag: fundamental theorem of calculus

Questions Related to fundamental theorem of calculus

$\displaystyle \int _0^1 \dfrac{xe^x}{(x + 1)^2} dx =$

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

  1. $\dfrac{e}{2}$

  2. $\dfrac{e - 1}{2}$

  3. $\dfrac{3e}{2} -1$

  4. $\dfrac{e - 3}{2}$

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Correct Option: C

$\displaystyle\int _{ 0 }^{ 1 }{ \cfrac { \tan ^{ -1 }{ x }  }{ x }  } dx$ equals

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

  1. $\displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { x }{ \sin { x } } dx } $

  2. $\cfrac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { x }{ \sin { x } } dx } $

  3. $\displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { \sin { x } }{ x } dx } $

  4. None of the above

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Correct Option: B
Explanation:

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 $\displaystyle\int^{\frac{3}{2}} _{-1}|x\sin(\pi x)|dx$.

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

  1. $\dfrac {3}{\pi} +\dfrac {1}{\pi^2}$

  2. $3\pi +\pi^2$

  3. $\dfrac { 2 }{ \pi } +\dfrac { 1 }{ { \pi }^{ 2 } }$

  4. none of the above

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Correct Option: A
Explanation:

$|x\sin(\pi x)|=\begin{cases}x\sin \pi x ,\,\,\,\,x\in(-1,1)\  -x\sin\pi x,x\in (1,\dfrac{3}{2})  \end{cases}$

$\displaystyle \int _{ 1 }^{ \frac { 3 }{ 2 }  }{ |x\sin(\pi x)| }dx=\int _{ -1 }^{ 1  }{ x\sin(\pi x)dx }+\int _{ 1 }^{ \frac { 3 }{ 2 }  }{ -x\sin(\pi x)dx }$

$=\displaystyle \left|\dfrac{-x\cos\pi x}{\pi}\right|^1 _{-1}-\int _{- 1 }^{ 1  }{ \left(\dfrac{-\cos\pi x}{\pi}dx\right) }-\left[\left|\dfrac{-x\cos (\pi x)}{\pi}\right|^{\dfrac{3}{2}} _1-\int _{ 1 }^{ \frac { 3 }{ 2 }  }{ \dfrac{-\cos x }{\pi}dx } \right] $ 

$=\dfrac{1}{\pi}-\left(\dfrac{-1}{\pi}\right)+0-\left[\dfrac{-1}{\pi}-(\dfrac{1}{\pi^2})\right]$

$=\dfrac{1}{\pi}+\dfrac{1}{\pi}+\dfrac{1}{\pi}+\dfrac{1}{\pi^2}$

$=\dfrac{3}{\pi}+\dfrac{1}{\pi^2}$  

$\int _{ 0 }^{ \infty  }{ f\left( x+\cfrac { 1 }{ x }  \right) .\cfrac { \ln { x }  }{ x }  } dx$

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

  1. Is equal to zero

  2. Is equal to one

  3. Is equal to $\cfrac { 1 }{ 2 } $

  4. Can not be evaluated

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Correct Option: A
Explanation:

Let: $lnx=t \Rightarrow x=e^{t}$
$\Rightarrow \dfrac{1}{x}dx=dt$

As "x" varies from $0$ to $\infty$ "$lnx $  $[t]$" varies $-\infty$ to $\infty$.
Now,
$\int _{0}^{\infty}f(x+\dfrac{1}{x}).\dfrac{lnx}{x}dx$

$\Rightarrow \int _{-\infty}^{\infty}f(e^{t}+e^{-t}).tdt = F(t)$

Now,
Using properties of definite integral:
Here we can see above function is an odd function i.e $F(-t)=-F(t)$
therefore on integrating from $-\infty$ to $\infty$ sum of area of $odd$ $function$ is $zero.$
$\Rightarrow \int _{-\infty}^{\infty}f(e^{t}+e^{-t}).tdt =0$

Thus,
$\int _{0}^{\infty}f(x+\dfrac{1}{x}).\dfrac{lnx}{x}dx=0$
Hence, correct option is $"A"$

Evaluate $I = \displaystyle \int _{\pi /6}^{\pi /3}\sin x:dx$

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. $\displaystyle \frac{1-\sqrt{3}}{2}$

  2. $\displaystyle \frac{\sqrt{3}+1}{2}$

  3. $\displaystyle \frac{\sqrt{3}-1}{2\sqrt{3}}$

  4. None of these

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Correct Option: A
Explanation:
Given : $I = \displaystyle \int _{\pi /6}^{\pi /3}\sin x\:dx$

Integeration of $\sin x dx$ is $-cos x dx + c$

$I = -cos x dx$

Substuting the upper and lower limit values we get,

$I = -cos\dfrac{\pi}{3}+cos\dfrac{\pi}{6}$

$I = \dfrac{-\sqrt{3}}{2} + \dfrac{1}{2}$

$I = \dfrac{1-\sqrt{3}}{2}$

What is $\displaystyle \int _{ 0 }^{ \pi  }{ { e }^{ x } } \sin { x } dx$ equal to?

physics 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 { { e }^{ \pi }+1 }{ 2 } $

  2. $\cfrac { { e }^{ \pi }-1 }{ 2 } $

  3. ${ e }^{ \pi }+1$

  4. $\cfrac { { e }^{ \pi }+1 }{ 4 } $

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Correct Option: A
Explanation:

Using Integration by part$:-$

Considering 1st function $u(x)=sinx$ and 2nd function $v(x)=e^x$
Now using Integration by parts$:-$

$\int _{0}^{\pi}e^xsinxdx=sinx\int _{0}^{\pi}e^xdx-\int _{0}^{\pi}\left (\dfrac{d}{dx}(sinx)\int _{0}^{\pi}e^x  \right )$

$\Rightarrow \int _{0}^{\pi}e^xsinxdx=(sinx\times e^x)| _{0}^{\pi}-\int _{0}^{\pi}\left (cosx e^xdx  \right )$
 
Again using Integration by parts:-
$\Rightarrow \int _{0}^{\pi}e^xsinxdx=(0\times e^\pi-0\times 1)-\int _{0}^{\pi}\left (cosx e^xdx  \right )$

$\Rightarrow \int _{0}^{\pi}e^xsinxdx=(0\times e^\pi-0\times 1)-\left (cosx\int _{0}^{\pi}e^xdx-\int _{0}^{\pi}\dfrac{d}{dx}(cosx)\int _{0}^{\pi}e^xdx \right )$

$\Rightarrow \int _{0}^{\pi}e^xsinxdx=-(cosx\times e^x)| _{0}^{\pi}-\int _{0}^{\pi}\left (sinx e^xdx  \right )         \left \langle \because \dfrac{d}{dx}(cosx)=-sinx  \right \rangle$

$\Rightarrow \int _{0}^{\pi}e^xsinxdx+\int _{0}^{\pi}\left (sinx e^xdx  \right )=-(cosx\times e^x)| _{0}^{\pi}$

$\Rightarrow 2\int _{0}^{\pi}e^xsinxdx=-(-1\times e^\pi-1\times 1)$

$\Rightarrow \int _{0}^{\pi}e^xsinxdx=\dfrac{e^\pi+1}{2}$