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 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 the nth roots of unity complex numbers maths

Value of $\displaystyle sin \frac{\pi}{2n + 1} sin \frac{2 \pi}{2n + 1} sin \frac{3 \pi}{2n + 1} ..... sin \frac{n\pi}{2n + 1}$.

  1. $\dfrac{\sqrt{2n+1}}{2^n}$

  2. 1
  3. $\dfrac{n(n+1)}{2}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The roots of the equation $x^{2n + 1}- 1 = 0$ are
$1, \displaystyle cos \frac{2 \pi}{2n + 1} + i  sin  \frac{2 \pi}{2n + 1}, cos \frac{4 \pi}{2n + 1} + i  sin \frac{4 \pi}{2n + 1}, ........, cos \frac{4 n \pi}{2n + 1} + i  sin  \frac{4n  \pi}{2n + 1}$
Therefore $\displaystyle x^{2n+1} - 1 = (x - 1) \left ( x - cos \frac{2 \pi}{2n + 1} - i  sin \frac{2 \pi}{2n + 1} \right ) \left ( x - cos \frac{4 \pi}{2n + 1} - i  sin \frac{4 \pi}{2n + 1} \right ) ....... \left ( x - cos \frac{4 n\pi}{2n + 1} - i  sin \frac{4n \pi}{2n + 1} \right )$
Further since
$\displaystyle cos \left ( \frac{(2n + 1) - r}{2n + 1} \right ) 2\pi = cos \frac{2 r \pi}{2n + 1}$
and $\displaystyle sin \left ( \frac{(2n + 1) - r}{2n + 1} \right ) 2\pi = -sin \frac{2 r \pi}{2n + 1}$
it follows that
$\displaystyle \left ( x - cos \frac{2 \pi}{2n + 1} - i  sin \frac{2 \pi}{2n + 1}\right ) \left ( x - cos \frac{4 \pi}{2n + 1} - i  sin \frac{4 \pi}{2n + 1}\right )$
$= x^2 - 2x  cos \displaystyle \frac{2 \pi}{2n + 1} + 1$
$\left ( x - cos \frac{4 \pi}{2n + 1} - i  sin \frac{4 \pi}{2n + 1}\right )\left ( x - cos \frac{(4n - 2) \pi}{2n + 1} - i  sin \frac{(4n - 2) \pi}{2n + 1}\right )$
$=x^2 - 2x  cos \displaystyle \frac{4 \pi}{2n + 1} + 1$
$\left ( x - cos \frac{2 n\pi}{2n + 1} - i  sin \frac{2 n\pi}{2n + 1}\right )\left ( x - cos \frac{(2n + 2) \pi}{2n + 1} - i  sin \frac{(2n + 2)}{(2n + 1)} \pi\right )$
$= x^2 - 2x   cos \frac{2 n \pi}{2n + 1} + 1$
Thus the polynomial $x^{2n + 1} - 1$ can be rewritten thus
$x^{2n + 1} - 1 = (x - 1) \displaystyle \left ( x^2 - 2x  cos  \frac{2 \pi}{2n + 1} + 1\right ) \left ( x^2 - 2x  cos  \frac{4 \pi}{2n + 1} + 1\right )........ \left ( x^2 - 2x  cos  \frac{2 n\pi}{2n + 1} + 1\right ) $
or $\displaystyle \frac{x^{2n + 1} - 1}{x - 1} = \left ( x^2 - 2x  cos  \frac{2 \pi}{2n + 1} + 1\right ) \left ( x^2 - 2x  cos  \frac{4 \pi}{2n + 1} + 1\right ) ......... \left ( x^2 - 2x  cos  \frac{2 n\pi}{2n + 1} + 1\right ) $
Taking $\displaystyle \lim _{x \rightarrow 1}$ on both sides
$(2n + 1) = 2^{2n} sin^2 \displaystyle \frac{\pi}{2n + 1} sin^2 \frac{2 \pi}{2n + 1} ..... sin^2 \frac{n \pi}{2n + 1}$
Hence, $\displaystyle sin \frac{\pi}{2n + 1} sin \frac{2 \pi}{2n + 1} ...... sin \frac{n \pi}{2n + 1} = \frac{\sqrt{(2n + 1)}}{2^n}$


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

$\displaystyle \int x^2+3x+5\ dx$ 


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


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