Mathematics

Integral Calculus Applications

51 Questions

Applications of integral calculus focus on calculating geometric properties like the area under curves and the volume of revolving solids. Questions also involve numerical integration techniques using rectangles to approximate complex regions. These spatial mathematics problems are essential for advanced competitive examinations.

Area bounded by curvesVolume of revolving solidsNumerical integration rectanglesCurve approximation methodsIntegration theorems

Integral Calculus Applications Questions

Multiple choice

What is the area enclosed by line 2|x| + 3|y| < = 6?

  1. 12

  2. 3

  3. 6

  4. 9

  5. 24
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From this equation, there are four possible equations of line: 2x+3y<=6 -2x+3y<=6 -2x-3y<=6 2x-3y<=6 All these lines will enclose an area of a triangle = (1/2*2*3) = 3 units square  Since there are four such triangles and figure is symmetric, total area enclosed will be = 3*4 = 12 square units (Correct Answer)

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 conjugate hyperbola hyperbola conic section maths

The area of quadrilateral formed by focil hyperbola $\dfrac{x^2}{4}-\dfrac{y^2}{3}=1$ & its conjugate hyperbola is

  1. $14$
  2. $24$
  3. $12$
  4. $10$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For x^2/4 - y^2/3 = 1, a=2, b=sqrt(3). Foci are (+-ae, 0). e = sqrt(1 + 3/4) = sqrt(7)/2. Foci are (+-sqrt(7), 0). For the conjugate hyperbola y^2/3 - x^2/4 = 1, foci are (0, +-be') where e' = sqrt(1 + 4/3) = sqrt(7)/sqrt(3). Foci are (0, +-sqrt(7)). The quadrilateral vertices are (+-sqrt(7), 0) and (0, +-sqrt(7)). Area = 1/2 * d1 * d2 = 1/2 * (2*sqrt(7)) * (2*sqrt(7)) = 14.

Multiple choice construction : division of a line segment dividing a line segment into three or five equal parts divsion of line segmet in given ratio constructions maths

If a straight line $y-x=2$ divides the region ${x}^{2}+{y}^{2}\le 4$ into two parts, then the ratio of the area of the smaller part to the area of the greater part is 

  1. $\pi-2 : 3\pi+2$
  2. $3\pi-4 : \pi+4$
  3. $\pi-3 : 3\pi+3$
  4. $3\pi-8 : \pi+8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line divides the circle into two segments. The area of the circle is 4*pi. The line distance from the center is sqrt(2), which allows calculating the sector and triangle areas to find the ratio.

Multiple choice addition and subtraction vedic methods of multiplication vedic mathematics history of mathematics maths

If $A, {A} _{1}, {A} _{2}, {A} _{3}$ be the area of the in-circle and ex-circles, then $\dfrac {1}{\sqrt {{A} _{1}}}+\dfrac {1}{\sqrt {{A} _{2}}}+\dfrac {1}{\sqrt {{A} _{3}}}$ is equal to

  1. $\dfrac {1}{\sqrt {{A}}}$
  2. $\dfrac {2}{\sqrt {{A}}}$
  3. $\dfrac {3}{\sqrt {{A}}}$
  4. $None$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$A _1={\pi}{r _1}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s-a)^{2}}$

$A _2={\pi}{r _2}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s-b)^{2}}$
$A _3={\pi}{r _3}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s-c)^{2}}$
$A={\pi}{r}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s)^{2}}$
$\dfrac{1}{\sqrt{A _1}}+\dfrac{1}{\sqrt{A _2}}+\dfrac{1}{\sqrt{A _3}}=\dfrac{1}{\sqrt{\pi}}\bigg[\dfrac{s-a}{\Delta}+\dfrac{s-b}{\Delta}+\dfrac{s-c}{\Delta}\bigg]=\dfrac{1}{\sqrt{\pi}\Delta}[3{s}-(a+b+c)]=\dfrac{s}{\sqrt{\pi}\Delta}=\dfrac{1}{\sqrt{A}}$

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

The area of the triangle formed y a tangents to the curve $2xy=a^{2}$ and the coordinates axes is

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

The equation of the curve is xy = a^2 / 2. The tangent at any point (x1, y1) forms a triangle with the coordinate axes whose area is a constant independent of the point, specifically equal to 2 * (x1 * y1) = a^2, making option D correct.

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

The area of the triangle formed by the lines  $x ^ { 2 } - 3 x y + y ^ { 2 } = 0$  and  $x + y + 1 = 0$  is square units. is

  1. $\dfrac {1}{12}$
  2. $\dfrac { 1 } { 2 \sqrt { 5 } }$
  3. $\dfrac { 2 } { \sqrt { 3 } }$
  4. $\dfrac { \sqrt { 3 } } { 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
${ x }^{ 2 }-3xy+2{ y }^{ 2 }=0$
$\Rightarrow \left( x-y \right) \left( x-2y \right) =0$
Hence three sides are
$x-y=0\quad \longrightarrow \left( i \right) $
$x-2y=0\quad \longrightarrow \left( ii \right) $
$x+y+1=0\quad \longrightarrow \left( iii \right) $
Solving $(i)$, $(ii)$ & $(iii)$
three vertices are $A\left( 0,0 \right) ,\quad B\left( \dfrac { -2 }{ 3 } ,\dfrac { -1 }{ 3 }  \right) ,\quad C\left( \dfrac { -1 }{ 2 } ,\dfrac { -1 }{ 2 }  \right) $
$AB=\sqrt { \dfrac { 1 }{ 9 } +\dfrac { 4 }{ 9 }  } =\sqrt { \dfrac { 5 }{ 9 }  } =\dfrac { \sqrt { 5 }  }{ 3 } $
$BC=\sqrt { \dfrac { 1 }{ 36 } +\dfrac { 1 }{ 36 }  } =\dfrac { \sqrt { 2 }  }{ 6 } $
$AC=\sqrt { \dfrac { 1 }{ 4 } +\dfrac { 1 }{ 4 }  } =\dfrac { 1 }{ \sqrt { 2 }  } =\dfrac { \sqrt { 2 }  }{ 2 } $
$\therefore$   area using heron's formula
$\Delta =\sqrt { S\left( S-AB \right) \left( S-BC \right) \left( S-CA \right)  } $
    $=\dfrac { 1 }{ 12 } { unit }^{ 2 }$
Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

Area of the triangle formed by the lines $x-y=0, x+y=0$ and ant tangent to the hyparabola $x^{2}-y^{2}=a^{2}$ is 

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

Equation of line

$ x-y=0\,\,......\,\,\left( 1 \right) $

$ x+y=0\,\,.......\,\,\left( 2 \right) $


Equation of hyperbola is

${{x}^{2}}-{{y}^{2}}={{a}^{2}}\,\,......\,\,\left( 3 \right)$


Let the point $P(a\sec \theta, a\tan \theta)$ on the hyperbola.


Equation of tangent is,

$ x{{x} _{1}}-y{{y} _{1}}={{a}^{2}} $

$\Rightarrow a(x\sec \theta-y\tan\theta)=a^2$

$ \Rightarrow x\sec \theta -y\tan \theta =a\,\,......\,\,\left( 4 \right) $


Now,

Area of $\Delta AOB$ $=\dfrac{1}{2}$

$=\dfrac{1}{2}\left|a^2(\tan^2\theta-\sec^2\theta)-a^2(\sec^2\theta-\tan^2\theta)\right|$

$ =\dfrac{1}{2}\left| {{a}^{2}}\left( -1 \right)-{{a}^{2}}\left( 1 \right) \right| $

$ =\dfrac{1}{2}\left| -2{{a}^{2}} \right| $

$ =\left| -{{a}^{2}} \right| $

$ =\left| {{a}^{2}} \right| $


Hence, this is the answer.
Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

The area of the sector of circle ${x}^{2}+{y}^{2}=16$ and the line $y=x$ in the first quadrant is 

  1. $8\pi sq.units$
  2. $\pi sq.units$
  3. $4\pi sq.units$
  4. $2\pi sq.units$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The circle x^2 + y^2 = 16 has a radius of 4. The line y = x makes an angle of 45 degrees (pi/4 radians) with the positive x-axis. The area of the sector in the first quadrant bounded by this line is (pi/4) / (2pi) of the total area, or 1/8 of the circle's area, which is (1/8) * pi * 4^2 = 2pi square units.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The graph $y^2 + 2xy + 40 |x| = 400$ divides the plane into regions. Then the area of bounded region is

  1. $200$ sq. units
  2. $400$ sq. units
  3. $800$ sq. units
  4. $500$ sq. units
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For $x < 0$, the equation is
$(y - 20) (y+2x + 20) =0$
Hence, the area is $20 \times 40 = 800$ sq. units.
For $x\geq0$ the equation simplifies to  $y^2-400+2xy+40x=(y-20)(y+20)+2x(y+20)=(y+20)(y+2x-20)=0$
Thus, obtained lines make a quadrilateral $ABCD,$
Area of quadrilateral $ABCD =2\times$ Area of triangle $BCD $$=2\times\dfrac{1}{2}\times$ base $\times $height$ = 20\times 40=800$ sq.units

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

Let $L$ be an end of the latus rectum of $y^2 = 4x$. The normal at $L$ meets the curve again at $M$. The normal at $M$ meets the curve again at $N$. The area of $\Delta LMN$ is

  1. $\dfrac{1280}{9} sq.$ units
  2. $\dfrac{640}{9} sq.$ units
  3. $\dfrac{320}{9} sq.$ units
  4. $\dfrac{160}{9} sq.$ units
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the parabola y^2=4x, the latus rectum end L is (1, 2). The normal at (1, 2) is y+2x=4, which meets the parabola at M(9, -6). The normal at M is y+x/2 = -3, which meets the parabola at N(1/4, -1). The area of triangle LMN with vertices (1, 2), (9, -6), and (0.25, -1) is 1280/9.