Tag: fundamental theorem of integral calculus

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Questions Related to fundamental theorem of integral calculus

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $

$=\cfrac { 1 }{ 2 } \int _{ 0 }^{ \pi /2 }{ \cfrac { z }{ 2\sin { \cfrac { z }{ 2 }  } \cos { \cfrac { z }{ 2 }  }  }  } dz$

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

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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] $ 

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

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

$=\dfrac{3}{\pi}+\dfrac{1}{\pi^2}$  

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

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

  1. Is equal to zero

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

  4. Can not be evaluated
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $"A"$

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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,

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

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

$I = \dfrac{1-\sqrt{3}}{2}$
Multiple choice physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

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

  1. $\cfrac { { e }^{ \pi }+1 }{ 2 } $
  2. $\cfrac { { e }^{ \pi }-1 }{ 2 } $
  3. ${ e }^{ \pi }+1$
  4. $\cfrac { { e }^{ \pi }+1 }{ 4 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
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:-
$\Rightarrow \int _{0}^{\pi}e^xsinxdx=(0\times e^\pi-0\times 1)-\int _{0}^{\pi}\left (cosx e^xdx  \right )$

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

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

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

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

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