Application of Integrals 

Integrals are one of the most important concepts in calculus. The application of integrals class 12 includes calculating quantities that accumulate continuously, such as areas, volumes, work, energy, and probabilities. They help solve virtually all practical problems in varying fields, such as physics, economics, and biology.

1.0Application of Integrals in Maths

1. Finding the Area Under a Curve

The most common application of integrals is calculating the area under a curve. For a given function f(x), the area between the graph of the function and the x-axis over the interval [a,b] is determined using a definite integral:

$\text{ Area }={\int }_a^{b}f(x)dx$

This concept is used for finding the total area in many applications, such as in finding the distance traveled by an object moving along a path or to calculate the area between curves in geometry.

Example: Find the area under the curve

$\mathbf{y}=x^2\text{ from }x=0\text{ to }x=2\text{. }$

Solution:

$\text{ Area }={\int }_0^2{x}^{2}dx={\left[\frac{x^3}{3}\right]}_0^2=\frac{8}{3}\text{ square units }$

2. Calculating the Area Between Two Curves

Many times, we need to calculate the area between two curves

$y=f(x)\text{ and }y=g(x)$, over an interval [a, b].  The area between the curves is found by subtracting the area under the lower curve from the area under the upper curve:

Area between curves =

${\int }_a^b[f(x)-g(x)]dx$

Example: Find the area between the curves

$y=x^2\text{ and }y=x\text{ from }x=0\text{ to }x=1$

Solution:

$\text{ Area }={\int }_0^1\left(x-x^2\right)dx=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$

3. Finding the Average Value of a Function

The average value of a function f(x) over an interval [a, b][a, b][a, b] can be calculated using the formula:

Average value =

$\frac{1}{b-a}{\int }_a^{b}f(x)dx$

This concept is used in many fields, like physics and economics, to determine the mean value of a quantity over time or space.

Example: Find the average value

$f(x)=x^2$ over the interval [0,2].

Solution: Average value =

$\frac{1}{2-0}{\int }_0^2{x}^{2}dx=\frac{1}{2}\left[\frac{x^3}{3}\right]=\frac{1}{2}\times \frac{8}{3}$

4. Applications in Probability

In probability theory, integrals are used to find probabilities for continuous random variables. It f(x) is the PDF or Probability Density Function of any random variable; the probability (P) that X lies in the interval [a, b] is : 

$P(a\le X\le b)={\int }_a^{b}f(x)dx$

Example: For a continuous random variable with the PDF,

$f(x)=3x^2,0\le x\le 1$, the probability that X lies between 0.5 and 1 is.

Solution:

$P(0.5\le X\le 1)={\int }_{0.5}^{1}3x^{2}dx=\left[x^3\right]=1-0.125=0.875$

2.0Application of integrals in Real life

1. Work Done by a Variable Force

Work is defined in physics as the force over a distance, and if the force is changing, then we make use of integrals to determine how much work gets done. The work formula is

$W={\int }_a^{b}F(x)dx$

Where function F(x) is the force applied at position x, and a and b are the limits of the distance over which the force acts?

Example: If a force

$F(x)=2x(\text{ in newtons })$is applied along a straight line from

$x=0\text{ to }x=3$, the work done.

Solution:

$W={\int }_0^{3}2xdx={\left[x^2\right]}_0^3=9\text{ joules }$

2. Finding the Volume of Solids

Integrals are also used to compute the volume of solids when the solids are of curved or irregular shapes. The volume of solids generated by revolving a region about an axis can be easily determined through techniques such as the disk/washer method or the shell method.

Disk Method

If a region is rotated around the x-axis, the volume of the solid formed is given by the formula:

$\mathrm{V}=2\pi {\int }_a^{b}x\cdot f(x)dx$

Example (Disk Method): Find the volume (V) of the solid formed by rotating the specific region bounded by the

$curve\mathbf{y}=x^{2}andthex-axisbetweenx=0\text{ and }x=1$.

Solution:

$\mathrm{v}=\pi {\int }_0^1{\left(x^2\right)}^{2}dx=\pi {\int }_0^1{x}^{4}dx=\pi \left[\frac{x^5}{5}\right]=\frac{\pi }{5}$