Mathematics

Definite Integration

26 Questions

Definite integration involves calculating the value of an integral within specific bounds, often using trigonometric and logarithmic functions. This set of questions covers evaluating limits and solving complex integrands. It is a highly scoring area in the mathematics sections of various exams.

Trigonometric integralsIntegration limitsLogarithmic integralsIntegral properties

Definite Integration Questions

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

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?

  1. $4I$
  2. $2I$
  3. $I$
  4. $\dfrac{I}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$I = \displaystyle \int _{0}^{\pi} {ln(\sin x)dx}$

 using property,
$I = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {(ln(\sin (2\pi -x) +ln(\sin x)) dx}$

we know that $\sin(x) = \sin(2\pi -x)$ 

$I = 2\displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin x)dx}$

$\dfrac{I}{2} = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin x)dx}$

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

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?

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

$I = \displaystyle \int _{0}^{\pi} {ln(\sin x)dx}$

 using property,
$I = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {(ln(\sin (2\pi -x) +ln(\sin x)) dx}$

we know that $\sin(x) = \sin(2\pi -x)$ 
$I = 2\displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin x)dx}$

$\dfrac{I}{2} = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin x)dx}$
 by property,

$\dfrac{I}{2} = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin (\dfrac{\pi}{2} - x))dx}$

$\dfrac{I}{2} = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\cos x)dx}$

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

$\displaystyle\int _{ 0 }^{ 1 }{ \cfrac { \tan ^{ -1 }{ x }  }{ x }  } dx$ 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 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:-
$\Rightarrow \int _{0}^{\pi}e^xsinxdx=(0\times e^\pi-0\times 1)-\int _{0}^{\pi}\left (cosx e^xdx  \right )$

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

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

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

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

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

Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes
Evaluate the following definite integral:
$\displaystyle \int _{0}^1 \dfrac {1-x}{1+x}dx$
  1. $2\log 2+1$
  2. $2\log 2$
  3. $2\log 2-1$
  4. $\log 4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$I=\displaystyle \int _{0}^1 \dfrac {1-x}{1+x}dx$ 

$=\displaystyle \int _{0}^1 \dfrac {2-(1+x)}{1+x} dx$ 

$=\displaystyle \int _{0}^1 \left (\dfrac {2}{1+x} -1 \right)dx $

$=\left [2\log (x+1)-x \right] _0^1$

$\Rightarrow \ I=(2\log 2-1)- (2\log 1-0)$ 

$=2\log 2-1$

Multiple choice position of point wrt ellipse ellipse maths

Evaluate $\displaystyle \int x^2+3x+5\ dx$ 

  1. $\dfrac {x^3}3+3\dfrac {x^2}3+\dfrac 53+c$
  2. $\dfrac {x^2}2+ 3x+5+c $
  3. $ \dfrac {x^3}3+3\dfrac {x^2}2+5x+c $

  4. None of the above.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle \int x^2+3x+5\ dx$ 


$=\displaystyle \int x^2 dx+\int 3x dx+\int 5 dx$


$=\dfrac {x^3}{3}+\dfrac {3x^2}{2}+5x+C$


Multiple choice

What is the volume of the solid generated by revolving the region bounded by the curves y = x^2 and y = 4 - x^2 about the x-axis?

  1. 32π cubic units

  2. 48π cubic units

  3. 64π cubic units

  4. 80π cubic units
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We can use the method of cylindrical shells to find the volume of the solid. The volume of a cylindrical shell is given by the formula 2πrhΔx, where r is the radius of the shell, h is the height of the shell, and Δx is the thickness of the shell. In this case, the radius of the shell is x, the height of the shell is 4 - x^2 - x^2 = 4 - 2x^2, and the thickness of the shell is Δx. Therefore, the volume of the solid is given by the integral ∫2πx(4 - 2x^2)dx from x = 0 to x = 2. Evaluating this integral, we get the volume of the solid as 64π cubic units.

Multiple choice

What is the value of the Gamma function at (\frac{3}{2})?

  1. $\frac{\sqrt{\pi}}{2}$
  2. $\frac{2}{\sqrt{\pi}}$
  3. $\frac{\pi}{2}$
  4. $\frac{3\sqrt{\pi}}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Gamma function at (\frac{3}{2}) is equal to (\frac{\sqrt{\pi}}{2}).