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 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.

Multiple choice

What is the name of the theorem that states that the area under a curve can be calculated by dividing it into an infinite number of rectangles?

  1. Fundamental Theorem of Calculus

  2. Mean Value Theorem

  3. Rolle's Theorem

  4. Cauchy's Integral Theorem
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Fundamental Theorem of Calculus is a central theorem in calculus that establishes a connection between differentiation and integration, allowing for the calculation of areas under curves.

Multiple choice

What is the name of the theorem that states that the area under a curve can be approximated by a sum of rectangles?

  1. Fundamental theorem of calculus

  2. Mean value theorem

  3. Rolle's theorem

  4. Cauchy's mean value theorem
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 approximated by a sum of rectangles.

Multiple choice

What is the volume of the solid generated by revolving the region bounded by the curves y = x^2 and y = 4 - x^2 about the x-axis?

  1. 32π/3

  2. 64π/3

  3. 96π/3

  4. 128π/3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The volume of the solid generated by revolving the region bounded by the curves y = f(x) and y = g(x) about the x-axis is given by the formula V = π∫[f(x)]^2 - [g(x)]^2 dx. Substituting f(x) = x^2 and g(x) = 4 - x^2, we get V = π∫(x^2)^2 - (4 - x^2)^2 dx = π∫x^4 - 16 + 8x^2 dx = π[(1/5)x^5 - 16x + (8/3)x^3] + C, where C is a constant. Evaluating the integral from x = -2 to x = 2, we get V = π[(1/5)(2)^5 - 16(2) + (8/3)(2)^3 - (1/5)(-2)^5 + 16(-2) - (8/3)(-2)^3] = π[(32/5) - 32 + (64/3) - (32/5) + 32 - (64/3)] = π(64/3).

Multiple choice

Which mathematical concept is used to describe the area under the curve of a function?

  1. Integral

  2. Derivative

  3. Limit

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

The integral of a function represents the area under the curve of the function over a specified interval.

Multiple choice

Which theorem states that the integral of a function (f(x)) over an interval ([a, b]) is equal to the area under the curve of (f(x)) between (a) and (b)?

  1. Pythagorean Theorem

  2. Triangle Inequality Theorem

  3. Angle Sum Theorem

  4. Fundamental Theorem of Calculus
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The Fundamental Theorem of Calculus states that the integral of a function (f(x)) over an interval ([a, b]) is equal to the area under the curve of (f(x)) between (a) and (b).

Multiple choice

Which theorem states that the integral of a function (f(x)) over an interval ([a, b]) is equal to the area under the curve of (f(x)) between (a) and (b)?

  1. Pythagorean Theorem

  2. Triangle Inequality Theorem

  3. Angle Sum Theorem

  4. Fundamental Theorem of Calculus
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The Fundamental Theorem of Calculus states that the integral of a function (f(x)) over an interval ([a, b]) is equal to the area under the curve of (f(x)) between (a) and (b).

Multiple choice

Which theorem states that the integral of a function (f(x)) over an interval ([a, b]) is equal to the area under the curve of (f(x)) between (a) and (b)?

  1. Pythagorean Theorem

  2. Triangle Inequality Theorem

  3. Angle Sum Theorem

  4. Fundamental Theorem of Calculus
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The Fundamental Theorem of Calculus states that the integral of a function (f(x)) over an interval ([a, b]) is equal to the area under the curve of (f(x)) between (a) and (b).

Multiple choice

What is the name of the theorem that states that the integral of a function is equal to the area under the curve of the function?

  1. Fundamental theorem of calculus

  2. Second fundamental theorem of calculus

  3. Mean value theorem

  4. Cauchy's mean value theorem
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The fundamental theorem of calculus states that the integral of a function is equal to the area under the curve of the function.

Multiple choice

What is the name of the mathematical method used to approximate the area under a curve?

  1. Integration

  2. Differentiation

  3. Exponentiation

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

Integration is a mathematical method used to approximate the area under a curve, which is essential in calculus and various applications.

Multiple choice

Which numerical integration algorithm is based on approximating the area under a curve using a series of rectangles?

  1. Trapezoidal rule

  2. Simpson's rule

  3. Gaussian quadrature

  4. Monte Carlo integration
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The trapezoidal rule approximates the area under a curve by dividing it into a series of trapezoids and summing their areas.

Multiple choice

What is Aryabhata's formula for the area of a parabola?

  1. $$A = \frac{1}{2}bh$$
  2. $$A = \pi r^2$$
  3. $$A = \frac{1}{3}Bh$$
  4. $$A = \frac{1}{4}Bh$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Aryabhata's formula for the area of a parabola is given by $$A = \frac{1}{2}bh$$, where $$b$$ is the length of the base of the parabola and $$h$$ is its height.

Multiple choice

Find the area of the region bounded by the curves (y = x^2) and (y = 2x + 1).

  1. \(\frac{1}{3}\) square units
  2. \(\frac{2}{3}\) square units
  3. \(1\) square unit
  4. \(\frac{3}{2}\) square units
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To find the area of the region, we need to find the points of intersection of the two curves. Setting (y = x^2) and (y = 2x + 1) equal to each other, we get: (x^2 = 2x + 1) which gives (x^2 - 2x - 1 = 0). Solving for (x), we get (x = 1) and (x = -1). Therefore, the area of the region is: (Area = \int_{-1}^{1} (2x + 1 - x^2) dx = \left[x^2 + x - \frac{x^3}{3}\right]_{-1}^{1} = \left[(1)^2 + (1) - \frac{(1)^3}{3}\right] - \left[(-1)^2 + (-1) - \frac{(-1)^3}{3}\right] = \frac{2}{3}) square units.

Multiple choice

Find the volume of the solid generated by revolving the region bounded by the curves (y = x^2) and (y = 2 - x) about the (x)-axis.

  1. \(\frac{8\pi}{3}\) cubic units
  2. \(\frac{16\pi}{3}\) cubic units
  3. \(\frac{24\pi}{3}\) cubic units
  4. \(\frac{32\pi}{3}\) cubic units
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To find the volume of the solid, we can use the formula: (Volume = \pi \int_{a}^{b} [f(x)]^2 dx), where (f(x)) is the function that defines the upper boundary of the region and ([a, b]) are the limits of integration. Plugging in the values of (f(x) = 2 - x) and ([a, b] = [0, 2]), we get: (Volume = \pi \int_{0}^{2} (2 - x)^2 dx = \pi \left[2x - \frac{x^3}{3}\right]_{0}^{2} = \pi \left[(2)(2) - \frac{(2)^3}{3} - (2)(0) - \frac{(0)^3}{3}\right] = \frac{16\pi}{3}) cubic units.