<meta http-equiv="refresh" content="1; url=/nojavascript/"> Integration: The Area Under the Curve | CK-12 Foundation
Dismiss
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Math Analysis Go to the latest version.

8.6: Integration: The Area Under the Curve

Created by: CK-12

Learning objectives

  • Demonstrate an understanding of the notion of the definite integral.
  • Learn how to compute the area under a curve by the Limit Method.

Introduction

To understand integration, consider the area under the curve y = f(x) for the interval from x = a to x = b in the figure below.

As discussed in Lesson 8.1, one way to calculate the area is to fill the region with rectangles. If the region is curved, the rectangles will not fit exactly but we can improve the approximation by using rectangles of thinner width. If we continue to make the rectangles thinner and thinner, the area under the curve would reach the exact area under the curve. This is the limiting process that we discussed. In other words, the area under the curve is the limit of the total area of the rectangles as the widths of the rectangles approach zero.

The Area Under the Curve

Consider again the figure above. The interval from x = a to x = b is subdivided into n equal subintervals. Rectangles are drawn in each subinterval. Each rectangle touches the curve at its upper right corner. The height of the first rectangle is f(x1), the second f(x2), and the last is f(xn). Since the length of the entire interval from a to b is b - a, then the width of each subinterval is \frac{b-a}{b}. We will refer to this width as ∆x. (The Greek letter ∆ is Delta and thus “delta x”.) That is,

\Delta x = \frac{b - a} {n}

is defined as the width of each subinterval. The area of the first rectangle is f(x1)∆x, The second is f(x2)∆x, and so on. Thus the total area An of the n rectangles, is the sum of all areas:

A_n = f(x_1) \Delta x + f(x_2) \Delta x + . . . + f(x_n) \Delta x
= \sum_{i = 1}^n f(x_i) \Delta x

To make use of the concept of limit, we make the width of each rectangle approach 0 , which is equivalent to making the number of rectangles, n, approach infinity. By doing so, we find the exact area under the curve,

\lim_{n \rightarrow \infty} A_n = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} f(x_i) \Delta x.

This limit is defined as the definite integral and it is denoted by

\int_{a}^{b} f(x) dx.

The Definite Integral

A definite integral gives us the area between the x-axis a curve over a defined interval.

The Definite Integral (The Limit Method)

The area a curve f(x) and the x-axis over the interval [a, b] can be calculated byA = \int_{a}^{b} f(x) dx = \lim_{n \rightarrow \infty} \sum_{i = 1}^n f(x_i) \Delta x

where

\Delta x = \frac{b - a} {n}

is the width of the subintervals.

It is important to keep in mind that the area under the curve can assume positive and negative values. It is more appropriate to call it “the net signed area”. Example 2 below illustrates this point.

Example 1: Calculate the area between the curve y = x2 and the x-axis from x = 0 to x = 1.

Solution: We divide the region into n number of subintervals, each of width ∆x (see figure below).

First find ∆x.
\Delta x = \frac{b - a} {n}
= \frac{1 - 0} {n}
= \frac{1} {n}
The next step is to find xi.
x_i = a + i \Delta x
= 0 + i \cdot \frac{1} {n}
= \frac{i} {n}
Therefore,
f(x_i) = x^2_i = \left (\frac{i} {n}\right )^2
Using the integration formula
A = \int_{a}^{b} f(x) dx = \lim_{n \rightarrow \infty} \sum_{i = 1}^n f(x_i) \Delta x
= \int_{0}^{1} x^2 dx= \lim_{n \rightarrow \infty} \sum_{i = 1}^n \left (\frac{i} {n}\right )^2 \left (\frac{1} {n}\right )
= \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \frac{i^2} {n^3}
Since we are summing over i, not n, the summation becomes,
A = \lim_{n \rightarrow \infty} \frac{1} {n^3} \sum_{i = 1}^{n} i^2
= \lim_{n \rightarrow \infty} \frac{1} {n^3} (1^2 + 2^2 + 3^2 + . . . + n^2)
But since
\sum_{i = 1}^{n} i^2 = \frac{n(n + 1) (2n + 1)} {6}
then
A = \lim_{n \rightarrow \infty} \frac{1} {n^3} \frac{n(n + 1)(2n + 1)} {6}
\lim_{n \rightarrow \infty} \frac{1} {6} \left (2 + \frac{3} {n} + \frac{1} {n^2}\right )
Taking the limit,
A = \frac{1} {6} (2 + 3(0) + (0))
= \frac{1} {3}
Thus the area under the curve is (1/3).

Example 2: Find the between the curve y = x and the x-axis from x = -1 to x = 1.

Solution: As you can see in figure a, the integral represents the total areas of all the rectangles above and below the x-axis. First, we divide the region into two regions, one above x-axis and one below the x-axis. Then we divide each region into n subintervals, each of width ∆x (figure b).

Region I: Find ∆x and xi.
\Delta x = \frac{1 - 0} {n} = \frac{1} {n}
x_{i} = a + i \Delta x
= 0 + i \left (\frac{1} {n}\right ) = \frac{i} {n}
f(x_i) = \frac{i} {n}
Region II: Again, find ∆x and xi.
\Delta x = \frac{-1 -0} {n} = \frac{-1} {n}
x_i = b + i \Delta x
= -1 + i \left (\frac{-1} {n}\right )
= -1 - \frac{i} {n}
f(x_i) = -1 - \frac{i} {n}
The integral represents the net area of the two regions I and II:
A = A_1 - A_2 = (\text{area above x-axis in }[a, b]) - (\text{area below x-axis in }[a, b])
Thus,
A = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} f(x_i) \Delta x - \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} f(x_i) \Delta x
= \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \left (\frac{i} {n}\right ) \left (\frac{1} {n}\right ) - \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \left (-1 - \frac{i} {n}\right ) \left (\frac{1} {n}\right )
= \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \left (\frac{i} {n^2}\right ) - \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \left (\frac{-1} {n} - \frac{i} {n^2}\right )
\mathit = \lim_{n \rightarrow \infty} \frac{1} {n^2} \sum_{i = 1}^{n} i - \left [\lim_{n \rightarrow \infty} \frac{-1} {n} + \lim_{n \rightarrow \infty} \frac{1} {n^2} \sum_{i = 1}^{n} i\right ]
= \lim_{n \rightarrow \infty} \frac{1} {n^2} \frac{n(n+1)} {2} - \left [0 + \lim_{n \rightarrow \infty} \frac{1} {n^2} \frac{n(n + 1)} {2}\right ]
= \frac{1} {2} - \left [\frac{1} {2}\right ]
= 0
We conclude that the net area is zero.

Lesson Summary

In this lesson we once again looked at the second fundamental problem in calculus, that of finding the area between a curve and the x-axis. Using limits, we derived the definite integral which gives us the area between a curve and the x-axis. In the final lesson of this chapter we will look at the connection between the integral and the derivative.

Points to Consider

  1. How might you have solved the problem in Example 2 without using calculus?
  2. Can you make a conjecture at the area between the curve y=x3 and the x-axis over the interval [-2,2]?

Review Questions

  1. Use the limit method to find the area under the curve of f(x) = x2 in the interval [0, 2].
  2. Use the limit method to find the area under the curve of f(x) = x3 in the interval [1, 4].
  3. Find the area between the curve y = x3 and the x-axis from -1 to 2.
  4. Find the area between the curve y = x and the x-axis from x = 1 to x = 3.
  5. Find the area between the curve y = -x and the x-axis from x = 1 to x = 3.
  6. Find the area between the curve y = x and the x-axis from x = -3 to x = 3
  7. Find the area between the curve y = x2 - 3x + 5 from x = 0 to x = 3.
  8. Sketch y = x2 and y = x on the same coordinate system and then find the area of the region enclosed between them.

Review Answers

  1. 8/3
  2. 63 \frac{3} {4}
  3. \frac{15} {4}
  4. 4
  5. -4
  6. 0
  7. 10.5
  8. Area is \frac{1} {6}.

Image Attributions

Files can only be attached to the latest version of None

Reviews

Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
CK.MAT.ENG.SE.1.Math-Analysis.8.6

Original text