Mathematics

Definite Integration

20 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 absolute value real numbers (rational and irrational numbers) real numbers basic algebra maths

The absolute value of $\dfrac { \displaystyle\int _{ 0 }^{ \pi /2 }{ \left( x\cos { x+1 }  \right) { e }^{ \sin { x }  }dx }  }{ \displaystyle\int _{ 0 }^{ \pi /2 }{ \left( x\sin { x-1 }  \right) { e }^{ \cos { x }  }dx }  } $ is equal to 

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

$\begin{array}{l} We\, have \ I=\dfrac { { \int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ { e^{ \sin  x } }\left( { x\cos  x+1 } \right) dx }  } }{ { \int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ { e^{ \cos  x } }\left( { x\sin  x-1 } \right) dx }  } }  \ =\dfrac { { \left[ { x{ e^{ \sin  x } } } \right] _{ 0 }^{ \frac { \pi  }{ 2 }  } } }{ { \int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ { e^{ \cos  x } }\left( { 1-x\sin  x } \right) dx }  } } =\dfrac { { \frac { \pi  }{ 2 } \times e } }{ { \left[ { { e^{ \cos  x } }x } \right] _{ \frac { \pi  }{ 2 }  }^{ 0 } } }  \ =\dfrac { { \frac { \pi  }{ 2 } e } }{ { 0-\frac { \pi  }{ 2 }  } } =-e \ Hence,\, absolute\, value\, =e \ Hence,\, option\, A\; is\, the\, correct\, answer. \end{array}$

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

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 integral $\displaystyle\int{\sin{x}{\cos}^{4}{x}dx}$ where $x\in\left[-1,\,1\right]$ is 

  1. 1

  2. 1\2

  3. 0

  4. 4
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$f\left(x\right)=\sin{x}{\cos}^{4}{x},$

$f\left(-x\right)=\sin{\left(-x\right)}{\cos}^{4}{\left(-x\right)}=-f\left(x\right)$

Since $f\left(x\right)$ is an odd function, $\displaystyle\int _{-1}^{1}\sin{x}{\cos}^{4}{x}dx=0$
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


$\displaystyle \int _{1}^{4}\frac{\mathrm{x}\mathrm{d}\mathrm{x}}{\sqrt{2+4\mathrm{x}}}=$

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

$\int _{1}^{4}\dfrac{x   dx}{\sqrt{2 + 4x}}=\dfrac{1}{2}\int _{1}^{4}\dfrac{x   dx}{\sqrt{x+\dfrac{1}{2}}}$
$=\dfrac{1}{2}\left [ \int _{1}^{4}\dfrac{(x+\dfrac{1}{2})dx}{\sqrt{x+\dfrac{1}{2}}}-\int _{1}^{4}\dfrac{\dfrac{1}{2}dx}{\sqrt{x+\dfrac{1}{2}}} \right ]$
$=\dfrac{1}{2}\left [ \int _{1}^{4} \sqrt{x+\dfrac{1}{2}} dx-\int _{1}^{4}(x+\dfrac{1}{2})^{\dfrac{1}{2}} \int _{1}^{4} \right ]$
$=\dfrac{1}{2} \left [ \dfrac{2}{3} (x+\dfrac{1}{2})^{\dfrac{3}{2}} \int _{1}^{4}-(x+\dfrac{1}{2})^{\dfrac{4}{2}} \int _{1}^{4} \right ]$
$=\dfrac{1}{3} \left [ \left ( \dfrac{9}{2} \right )^{\dfrac{3}{2}}-\left ( \dfrac{3}{2} \right )^{\dfrac{3}{2}} \right ] -\dfrac{1}{2} \left [ \left ( \dfrac{9}{2} \right )^{\dfrac{1}{2}}-\left ( \dfrac{3}{2} \right )^{\dfrac{1}{2}} \right ]$
$=\dfrac{1}{3} \left [ \left ( \dfrac{9}{2} \right )\left ( \dfrac{9}{2} \right )^{\dfrac{1}{2}}-\left ( \dfrac{3}{2} \right )\left ( \dfrac{3}{2} \right )^{\dfrac{1}{2}} \right ] - \dfrac{1}{2} \left [ \left ( \dfrac{3}{\sqrt{2}} \right )-\dfrac{\sqrt{3}}{\sqrt{2}} \right ]$
$=\dfrac{3}{\sqrt{2}}$

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 $\displaystyle \int _{0}^{2}(x-\log _{2}a)dx=2\log _{2}(\frac{2}{a})$ for which of the following conditions?

  1. $\mathrm{a}>0$
  2. $\mathrm{a}>2$
  3. $\mathrm{a}=4$
  4. $\mathrm{a}=8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\int _{ 0 }^{ 2 }{ (x-\log _{ 2 }a } )dx=2\log _{ 2 }(\cfrac { 2 }{ a } )$

The solution exists only if function is defined.
$ x-\log _{ 2 }a\longrightarrow$ defined
$ x\longrightarrow$ is defined for all values 
But $\log _{ 2 }a\longrightarrow$ defined for all values
But $ \log _{ 2 }a\longrightarrow$ defined for only a>0$
$\therefore \log (0)$ and $\log \text {(negative values)} )\longrightarrow$ not defined
Hence, required condition is $a>0$.

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] $ 

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

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

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