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8.12: Area Under the Curve

Difficulty Level: At Grade Created by: CK-12

You probably remember Becky and her track meet from earlier lessons. She won her race after pulling away from the pack in a hard push at the finish, and her boyfriend got a great picture of her just as she began pulling away. We have learned that by using derivatives, she could actually calculate her speed at the moment the picture was taken, and then with the second derivative she could calculate her acceleration similarly.

In this lesson we will discuss the integral , which is the process which would allow Becky to calculate the distance she actually covered during a given interval of the race, even though her speed was not constant!

Watch This

Embedded Video:

- Khan Academy: Introduction to Definite Integrals

Guidance

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.

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 ( x 1 ), the second f ( x 2 ), and the last is f ( x n ). 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 ( x 1 )∆ x , The second is f ( x 2 )∆ x , and so on. Thus the total area A n 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 and a curve over a defined interval.

The Definite Integral (The Limit Method)
The area between a curve f ( x ) and the x -axis over the interval [ a , b ] can be calculated by
A = \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 A

Calculate the area between the curve y = x 2 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 x i .
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 B

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 x i .
\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 x i .
\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.

Example C

Approximate the definite integral between x = 0 and x = 40 by calculating the areas of rectangles which fill the area in the image below. Use at least 3 successively narrower sizes of rectangles.

The equation of the curve in the image is: y = 60 -40(1 - \frac{9}{10}^x)

Solution

\therefore the closest approximated area is 1173.44 units (The actual calculated area is 1174.0373)

Vocabulary

The integral is used to calculate the area under a curve. It has many applications in science, including finding distance traveled by an object moving at inconstant speeds.

The symbol "∆" or " \delta " is used to denote "change in", as in "the change in velocity over time" = \frac{\delta v}{t}

The symbol \int_{x}^{y} is used to denote "the integral" or "the area under the curve between x and y "

Guided Practice

Questions

1) Approximate the area under y = x + 3 on the interval [5,6] using the middle Riemann Sum with 5 subintervals.

Sketch of graph:

2) Approximate the area between y = 3x 2 + x + 5 and the x-axis on the interval between x = 2 and x = 5 using the right Riemann Sum with 2 subintervals.

3) Use a definite integral to find the area under the curve y = 5x^2 + 2x + 4 on the interval [0, 3].

Solutions

1) First, we divide the interval [5,6] into pieces:

Between x = 5 and x = 5.2, the middle value is 5.1 + 3 = 8.1
Between x = 5.2 and x = 5.4, the middle value is 5.3 + 3 = 8.3
Between x = 5.4 and x = 5.6, the middle value is 5.5 + 3 = 8.5
Between x = 5.6 and x = 5.8, the middle value is 5.7 + 3 = 8.7
Between x = 5.8 and x = 6, the middle value is 5.9 + 3 = 8.9
Adding these, we get 42.5.
To get the Riemann sum, take this answer and multiply by the width of each segment: .2

\therefore 8.5 is our approximated area.

2) First, we divide the interval [2,5] into subintervals: Between x = 2 and x = 3.5, the right value is 3(3.5) 2 + (2.5) + 5 = 45.25 Between x = 3.5 and x = 5, the right value is 3(5) 2 + (5) + 5 = 85 Adding these, we get 130.25. Take this answer and multiply by the width of each segment: 1.5.

\therefore \approx 195.38 is the area

3) \int_{0}^{3} 5x^2 + 2x + 4dx = \frac{5}{3} x^3 + x^2 +4x |_{4}^{5}

Practice

  1. Use the limit method to find the area under the curve of f ( x ) = x 2 in the interval [0, 2].

Find the area between the curve and the x-axis:

  1. Curve y = x on the interval x = 1 to x = 3.
  2. Curve y = - x from x = 1 to x = 3.
  3. Curve y = x from x = -3 to x = 3
  4. Approximate the area under y = 2x + 3 on the interval [0,3] using the middle Riemann Sum for y with 6 subintervals.

Find the Area Under the Curve

  1. y = 3 on [4, 5]
  2. y = 3x + 1 on [1, 5]
  3. y = \frac{1}{x} on [3, 4]
  4. y = 2x + 4 on [5, 6]
  5. y = 5x^3 + 4x^2 + x + 2 on [2, 5]
  6. y = \frac {1}{x} on [3, 7]
  7. y = 3x^2 + 2x on [5, 6]
  8. y = 4 on [2, 6]
  9. y = 2x^2 + 4x + 5 on [1, 5]
  10. Sketch y = x 2 and y = x on the same coordinate system and then find the area of the region enclosed between them.

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Difficulty Level:

At Grade

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Date Created:

Nov 01, 2012

Last Modified:

Aug 04, 2014

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