Lesson 6.7: Applications of Integration

Introduction

Welcome to Lesson 6.7 of Foundation Further Mathematics! This lesson will focus on the applications of integration, which is a fundamental concept in calculus. By the end of this lesson, you, students, will be equipped with the skills to:

• Find the area under a curve and between curves.
• Determine the volumes of revolution about the x- and y-axes.
• Calculate the mean value of a function over an interval.
• Set up an integral from a worded context.
• Compute areas between curves using correct limits.

Hook

Have you ever wondered how engineers design roller coasters? 🏗️ Or how we can calculate the volume of complex shapes? Integration plays a critical role in these real-world applications! In this lesson, we'll explore how to apply integration techniques to solve practical problems.

Area Under a Curve

One of the most common applications of integration is finding the area under a curve defined by a function $f(x)$ over an interval $[a, b]$. The area can be found using the definite integral:

$$ \text{Area} = \int_a^b f(x) \, dx $$

Example 1: Area Under a Straight Line

Let’s consider the function $f(x) = 2x + 1$ between $x = 1$ and $x = 3$. To find the area under the curve, we will calculate:

$$ \text{Area} = \int_1^3 (2x + 1) \, dx $$

Calculating this integral involves:

1. Finding the antiderivative:

• The antiderivative of $2x + 1$ is $x^2 + x$.

1. Evaluating the definite integral:

• $$ \text{Area} = \left[ x^2 + x

ight]_1^3 = (3^2 + 3) - (1^2 + 1) = (9 + 3) - (1 + 1) = 12 - 2 = 10 $$

Thus, the area under the line from $x=1$ to $x=3$ is 10 square units!

Area Between Curves

Next, we will learn how to find the area between two curves. Suppose we have two functions, $f(x)$ and $g(x)$, with $f(x) \geq g(x)$ over an interval $[a, b]$. The area between the curves is given by:

$$ \text{Area} = \int_a^b (f(x) - g(x)) \, dx $$

Example 2: Area Between Two Functions

Consider the functions $f(x) = x^2$ and $g(x) = x$. We need to find the area between these two curves from $x = 0$ to $x = 1$. First, we determine which function is above the other within this interval. Since $x^2$ is below $x$ at these points, we have:

$$ \text{Area} = \int_0^1 (x - x^2) \, dx $$

Calculating this integral:

1. Find the antiderivative:

• The antiderivative of $x - x^2$ is $ \frac{x^2}{2} - \frac{x^3}{3} $.

1. Evaluate the definite integral:

• $$ \text{Area} = \left[ \frac{x^2}{2} - \frac{x^3}{3}

$ight]_0^1 = \left( \frac{1^2}{2} - \frac{1^3}{3} $

$ight) - \left( 0 $

ight) = $\frac{1}{2}$ - $\frac{1}{3}$ = $\frac{3}{6}$ - $\frac{2}{6}$ = $\frac{1}{6}$ $$

So, the area between the curves from $x = 0$ to $x = 1$ is $\frac{1}{6}$ square units.

Volume of Revolution

Another fascinating application of integration is calculating the volume of a solid obtained by rotating a function around an axis. The method of washers or disks can be used depending on the axis of rotation.

Example 3: Volume Around the x-Axis

Let’s find the volume of the solid obtained by rotating the area between the curves $f(x) = x^2$ and $g(x) = 0$ from $x=0$ to $x=1$ around the x-axis. The volume is given by:

$$ V = \pi \int_0^1 (f(x))^2 \, dx $$

Substituting for $f(x)$, we have:

$$ V = \pi \int_0^1 (x^2)^2 \, dx = \pi \int_0^1 x^4 \, dx $$

Calculating this:

1. Find the antiderivative of $x^4$, which is $\frac{x^5}{5}$.
2. Evaluate the integral:

• $$ V = \pi \left[ \frac{x^5}{5}

ight]_0^1 = $\pi$ $\left($ $\frac{1^5}{5}$ - 0

ight) = $\frac{\pi}{5}$ $$

Thus, the volume of the solid is $\frac{\pi}{5}$ cubic units!

Mean Value of a Function

The mean value of a function $f(x)$ over the interval $[a, b]$ is defined as:

$$ \text{Mean Value} = \frac{1}{b - a} \int_a^b f(x) \, dx $$

Example 4: Mean Value of $f(x) = x^2$ from 0 to 2

To find the mean value of the function $f(x) = x^2$ over the interval $[0, 2]$:

1. Calculate the definite integral:

• $$ \int_0^2 x^2 \, dx = \left[ \frac{x^3}{3}

ight]_0^2 = $\frac{8}{3}$ $$

1. Calculate the mean value:

• $$ \text{Mean Value} = \frac{1}{2 - 0} \cdot \frac{8}{3} = \frac{8}{6} = \frac{4}{3} $$

So, the mean value of the function over this interval is $\frac{4}{3}$.

Conclusion

In this lesson, we explored the applications of integration, focusing on areas under curves, between curves, volumes of revolution, and mean values of functions. Integration is a powerful tool that allows us to calculate quantities that are not always easy to measure directly.

Study Notes

• The area under a curve can be found using the definite integral.
• The area between curves requires the difference of the two functions under the integral.
• Volumes of revolution can be calculated using the method of disks or washers.
• The mean value of a function is the average output over an interval.
• Always ensure to establish the correct limits for integration based on the problem.