Mathematics

Integral Calculus Applications

45 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 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 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 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.

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

Area of the triangle formed by the ${x}$ axis, the tangent and normal at $(3,2)$ to the ellipse $\displaystyle \frac{x^{2}}{18}+\frac{y^{2}}{8}=1$ is 

  1. $5$
  2. $\dfrac{13}{3}$
  3. $\displaystyle \frac{15}{2}$
  4. $\displaystyle \frac{9}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given equation of ellipse $\displaystyle \frac{x^{2}}{18}+\frac{y^{2}}{8}=1$
$\displaystyle \frac {dy}{dx}=\displaystyle \frac {-4x}{9y}$
Slope of tangent to ellipse at $(3,2)$ is 
$m=\displaystyle \frac{-2}{3}$

Equation of tangent to ellipse is 
$y-2=-\displaystyle \frac{2}{3}(x-3)$
$\Rightarrow 2x+3y=12$
Since , the tangent intersect x-axis i.e. $y=0$
$\Rightarrow x=6$
So, tangent intersects x-axis at $(6,0)$

Equation of normal to ellipse is 
$y-2=\displaystyle \frac{3}{2}(x-3)$
$\Rightarrow 3x-2y=5$
Since , the tangent intersect x-axis i.e. $y=0$
$\Rightarrow x=\displaystyle \frac{5}{3}$
So, normal intersects x-axis at $\left(\displaystyle \frac{5}{3} ,0\right)$

So, area of triangle $=\displaystyle \frac { 1 }{ 2 } \begin{vmatrix} 3 & 2 & 1 \ 6 & 0 & 1 \ \frac { 5 }{ 3 }  & 0 & 1 \end{vmatrix}$
$=\displaystyle \frac{13}{3}$ sq.units

Multiple choice

What is the name of the theorem that states that the area under a curve can be found by integrating the function that defines the curve?

  1. Fundamental theorem of calculus

  2. Mean value theorem

  3. Chain rule

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

The fundamental theorem of calculus states that the area under a curve can be found by integrating the function that defines the curve.