Mathematics

Integration and Definite Integrals

51 Questions

Integration and definite integrals measure the accumulation of quantities and the area under curves. This topic evaluates limits of integration, exponential functions, and numerical methods like the trapezoidal rule. These advanced mathematical concepts are crucial for high level quantitative aptitude tests.

Definite integral limitsNumerical integration trapezoidal ruleExponential function integralsAverage value functionsUnbounded integrals

Integration and Definite Integrals Questions

Multiple choice

Changing the order of the integration in the double integral $I=\int\limits_0^8\int\limits^2_{\frac{x}{4}}f(x,y)dydx$ leads to $I=\int\limits_r^s\int\limits^q_pf(x,y)dydx$. What is the value of q?

  1. 4y

  2. 16y2

  3. x

  4. 8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\text{Given}\hspace{3cm}I=\int\limits_0^8\int\limits^2_{\pi/4}f(x,y)dydx \\ \text{Here we can draw the graoh from the limits of the integration, the limit of y is from$ y=\frac{x}{4}to$ y=2}\\ \text{For x the limit is$\hspace{1cm}$ x=0 to x=8} $

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

$\int _{0}^{\pi /2}sin2xtan^{-1}\left ( sinx \right )dx=$

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

We have,

$I=\int _{0}^{\dfrac{\pi }{2}}{\sin 2x{{\tan }^{-1}}\left( \sin x \right)dx}$

$=\int _{0}^{\dfrac{\pi }{2}}{2\sin x\cos x{{\tan }^{-1}}\left( \sin x \right)dx}$

Let

$ \sin x=t $

$ \cos xdx=dt $

Change limit

$ \sin 0=t $

$ t=0 $

And,

$ \sin \dfrac{\pi }{2}=t $

$ t=1 $

Then,

$ \int _{0}^{1}{2t{{\tan }^{-1}}t\cos xdx} $

$ =\int _{0}^{1}{2t{{\tan }^{-1}}tdt} $

$ =2\int _{0}^{1}{t{{\tan }^{-1}}tdt} $

On integrating and we get,

$ 2\left[ {{\tan }^{-1}}t\int _{0}^{1}{t}dt-\int _{0}^{1}{\left( \dfrac{d\left( {{\tan }^{-1}}t \right)}{dt}\int _{0}^{1}{tdt} \right)}dt \right] $

$ =2\left[ {{\tan }^{-1}}t\left[ {{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1} \right]-\int _{0}^{1}{\dfrac{1}{1+{{t}^{2}}}}\dfrac{{{t}^{2}}}{2}dt \right] $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{\dfrac{{{t}^{2}}}{1+{{t}^{2}}}}dt $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{\dfrac{{{t}^{2}}+1-1}{1+{{t}^{2}}}}dt $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{\dfrac{{{t}^{2}}+1}{1+{{t}^{2}}}}dt+\int _{0}^{1}{\dfrac{1}{1+{{t}^{2}}}}dt $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-\int _{0}^{1}{1}dt+\int _{0}^{1}{\dfrac{1}{1+{{t}^{2}}}}dt $

$ =2{{\tan }^{-1}}t{{\left( \dfrac{{{t}^{2}}}{2} \right)} _{0}}^{1}-{{\left[ t \right]} _{0}}^{1}+{{\left[ {{\tan }^{-1}}t \right]} _{0}}^{1}+C $

$ =2\left[ {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right]\left[ \dfrac{{{1}^{2}}}{2}-\dfrac{{{0}^{2}}}{2} \right]-\left[ 1-0 \right]+\left[ {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right]+C $

$ =2\left[ \dfrac{\pi }{4}-0 \right]\left[ \dfrac{1}{2} \right]-1+\left[ \dfrac{\pi }{4}-0 \right] $

$ =\dfrac{\pi }{4}+\dfrac{\pi }{4}-1 $

$ =\dfrac{2\pi }{4}-1 $

$ =\dfrac{\pi }{2}-1 $

Hence, this is the 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 $\int _{ 0 }^{ \pi /2 }{ \sin { x } \sin { 2x } \sin { 3x } dx } $ is equal to:

  1. $\cfrac{1}{3}$
  2. $-\cfrac{2}{3}$
  3. $-\cfrac{1}{3}$
  4. $\cfrac{1}{6}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\int _0^{\pi/2}\sin x\sin 2x\sin 3x dx$


$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2}2\sin x\sin 2x\sin 3x dx$

$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2}2\sin x\sin 3x\sin 2x dx$

We know that       $2\sin A \sin B=\cos(A-B)-\cos (A+B)$

$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2}(\cos (x-3x)-\cos(x+3x))\sin 2x dx$

$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2}(\cos 2x-\cos 4x)\sin 2x dx$

$\Rightarrow \dfrac{1}{2}\int _0^{\pi/2} \sin 2x \cos 2x dx-\dfrac{1}{2}\int _0^{\pi /2}\sin 2x \cos 4x dx$

$\Rightarrow \dfrac{1}{4}\int _0^{\pi/2} 2\sin 2x \cos 2x dx-\dfrac{1}{4}\int _0^{\pi /2}2\sin 2x \cos 4x dx$

We know that   $2\sin A \cos B=\sin(A+B)+\sin (A-B)$

$\Rightarrow \dfrac{1}{4}\int _0^{\pi/2} (\sin (2x+2x)+\sin (2x-2x))dx-\dfrac{1}{4}\int _0^{\pi /2}(\sin (2x+4x)+\sin (2x-4x)) dx$

$\Rightarrow \dfrac{1}{4}\int _0^{\pi/2} \sin 4xdx-\dfrac{1}{4}\int _0^{\pi /2}(\sin 6x-\sin 2x) dx$

$\Rightarrow \dfrac{1}{4}\int _0^{\pi/2} \sin 4xdx-\dfrac{1}{4}\int _0^{\pi /2}\sin 6x dx+\dfrac{1}{4}\int _{0}^{\pi/4}\sin 2x dx$

$\Rightarrow \dfrac{1}{4}[\dfrac{\sin 4x}{4}] _0^{\pi/4}-\dfrac{1}{4}[\dfrac{-\sin 6x}{6}] _0^{\pi/4}+\dfrac{1}{4}[\dfrac{-\cos 2x}{2}] _0^{\pi/4}$

$\Rightarrow \dfrac{-1}{12}+\dfrac{1}{4}=\dfrac{1}{6}$

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 _{\sin x}^1 t^2 f(t) dt = 1 - \sin x \forall x \epsilon (0, \pi / 2 ) $ then $ f \left( \dfrac {1}{\sqrt3} \right) $ is :

  1. $3$
  2. $\sqrt3$
  3. $1/3$

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

$ \int _{\sin x}^1 t^2 f(t) dt = 1 - \sin x \forall x \epsilon (0, \pi/2 ) $
Differentiating both sides we get 
$ \dfrac {d}{dx} (1) [ 1 \cdot f (1)] - \cos x ( \sin^2 x) f ( \sin x) = -\cos x  $
$ \Rightarrow f ( \sin x) = \dfrac {1}{ \sin^2 x} $
$ \therefore f \left( \dfrac {1}{\sqrt3} \right) = f \left( \sin \left( \sin^{-1} \dfrac {1}{\sqrt3} \right) \right) $
$ = \left[ \dfrac {1}{ \sin \left( \sin^{-1} \dfrac {1}{\sqrt3} \right)} \right]^2 = 3 $

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 integrals ${I _1} = \int _0^1 {{e^{ - x}}{{\cos }^2}xdx,} {I _2} = \int _0^1 {{e^{ - {x^2}}}{{\cos }^2}xdx,} {I _3} = \int _0^1 {{e^{ - x}}dx} $ and ${I _4} = \int _0^1 {{e^{ - (1/2){x^2}}}} dx$. The greatest of these integrals is

  1. $I _1$
  2. $I _2$
  3. $I _3$
  4. $I _4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$I _1=\int _{0}^{4}e^{-x} cos^2x dx$
$I _2=\int _{0}^{1}e^{-x^2}cos^2x dx$
Both have $cos^2x$ so value get restricted more in (0, 1)
Now in (0, 1) $e^{-\frac {x^2}{2}}>e^{-x}$
$\therefore \int _{0}^{1}e^{-\frac {x^2}{2}}>\int _{0}^{1}e^{-x}$
$\therefore I _4>I _3>I _1>I _2$

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

In numerical integration, what is the trapezoidal rule used for?

  1. Approximating the area under a curve

  2. Finding the roots of polynomials

  3. Solving differential equations

  4. Computing eigenvalues and eigenvectors
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The trapezoidal rule is a numerical integration method that approximates the area under a curve by dividing it into trapezoids and summing their areas.