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Applications of Integration

  • Understand the concept of integration and its applications.
  • Learn how to compute areas under curves using definite integrals.
  • Explore the concept of finding the area between two curves.

Integration has a wide range of applications in mathematics, physics, engineering, and economics. It helps us compute areas, volumes, work, and even probabilities. In this section, we explore some fundamental applications of integration.

1. Area Under a Curve

The definite integral can be used to find the area enclosed between a function \( f(x) \) and the x-axis over an interval \([a, b]\):

\[ A = \int_a^b f(x) \, dx \]

 

If \( f(x) \) is below the x-axis in some region, its integral will be negative. To find the total area, we take the absolute value of such integrals.

Example 1: Finding the Area Between a Function and the X-Axis

Find the area enclosed by \( f(x) = x^2 - 4 \) from \( x = -2 \) to \( x = 2 \).

Step 1: Find where the function crosses the x-axis by solving \( x^2 - 4 = 0 \):

\[ x^2 = 4 \quad \Rightarrow \quad x = \pm 2 \]

Step 2: The function is below the x-axis for \( x \in (-2,2) \), so we integrate the absolute value:

\[ A = \int_{-2}^{2} |x^2 - 4| \, dx \]

Since \( x^2 - 4 \) is negative in this interval, we rewrite it as:

\[ A = \int_{-2}^{2} (4 - x^2) \, dx \]

Step 3: Compute the integral:

\[ \int (4 - x^2) \, dx = 4x - \frac{x^3}{3} \]

Step 4: Evaluate at \( x = 2 \) and \( x = -2 \):

\[ \left[ 4(2) - \frac{(2)^3}{3} \right] - \left[ 4(-2) - \frac{(-2)^3}{3} \right] \] \[ = \left[ 8 - \frac{8}{3} \right] - \left[ -8 + \frac{8}{3} \right] \] \[ = \left( \frac{24}{3} - \frac{8}{3} \right) - \left( \frac{-24}{3} + \frac{8}{3} \right) \] \[ = \frac{16}{3} - \left(-\frac{16}{3} \right) = \frac{32}{3} \]

Thus, the total area is \( \frac{32}{3} \).

1. Area Between Two Curves

The area between two functions \( f(x) \) and \( g(x) \) over an interval \([a, b]\) is given by:

\[ A = \int_a^b \big( f(x) - g(x) \big) \, dx \]

Example 2: Finding the Area Between Two Curves

Find the area enclosed between \( f(x) = x^2 \) and \( g(x) = x+2 \) from \( x = -1 \) to \( x = 2 \).

Step 1: Determine which function is on top.

Since \( x^2 \) is a parabola and \( x+2 \) is a straight line, \( g(x) = x+2 \) is greater over the given interval.

Step 2: Compute the integral:

\[ A = \int_{-1}^{2} \big( (x+2) - x^2 \big) \, dx \]

Step 3: Integrate:

\[ \int (x+2 - x^2) \, dx = \int (x - x^2 + 2) \, dx \] \[ = \frac{x^2}{2} - \frac{x^3}{3} + 2x \]

Step 4: Evaluate at \( x = 2 \) and \( x = -1 \):

\[ \left[ \frac{(2)^2}{2} - \frac{(2)^3}{3} + 2(2) \right] - \left[ \frac{(-1)^2}{2} - \frac{(-1)^3}{3} + 2(-1) \right] \] \[ = \left[ \frac{4}{2} - \frac{8}{3} + 4 \right] - \left[ \frac{1}{2} + \frac{1}{3} - 2 \right] \] \[ = \left[ 2 - \frac{8}{3} + 4 \right] - \left[ \frac{1}{2} + \frac{1}{3} - 2 \right] \] \[ = \frac{18}{3} - \frac{8}{3} + 4 - \left(\frac{3}{6} + \frac{2}{6} - 2 \right) \] \[ = \frac{10}{3} + 4 - \left(\frac{5}{6} - 2 \right) \] \[ = \frac{10}{3} + 4 - \frac{5}{6} + 2 \] \[ = \frac{10}{3} + 6 - \frac{5}{6} \] \[ = \frac{10}{3} + \frac{36}{6} - \frac{5}{6} \] \[ = \frac{20}{6} + \frac{31}{6} = \frac{51}{6} = \frac{17}{2} \]

Thus, the total area enclosed between the curves is \( \frac{17}{2} \).

Practice Problems

Problem 1: Find the area enclosed by \( f(x) = x^3 - 3x \) from \( x = -2 \) to \( x = 2 \).

Solution:

Find where \( x^3 - 3x = 0 \):

\[ x(x^2 - 3) = 0 \quad \Rightarrow \quad x = 0, \pm\sqrt{3} \]

Set up the integral and compute:

\[ A = \int_{-2}^{2} |x^3 - 3x| \, dx \]

(Complete solution follows the same process as Example 1.)

Problem 2: Find the area between \( f(x) = x^2 - 2x \) and \( g(x) = 3x - x^2 \) from \( x = 0 \) to \( x = 3 \).

Solution:

Find where \( x^2 - 2x = 3x - x^2 \) and set up the integral.

Compute \( \int_0^3 ( (3x - x^2) - (x^2 - 2x) ) \, dx \).

(Complete solution follows the same process as Example 2.)


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