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Definite Integrals of Negative Functions

Author: Sophia
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what's covered
In this lesson, you will connect the ideas from Riemann sums, integrals, and regions that are below the x-axis. Specifically, this lesson will cover:

Table of Contents

1. Riemann Sums of Functions That Are Below the x-Axis

Up until now, we only integrated functions that were above the x-axis on open square brackets a comma space b close square brackets. We’ll use this example to see what happens when that is not the case.

Consider the function f open parentheses x close parentheses equals 1 minus x on the interval open square brackets 1 comma space 5 close square brackets.

The graph on the left is f open parentheses x close parentheses on the interval open square brackets 1 comma space 5 close square brackets comma and the graph on the right shows the rectangles that could be used in a Riemann sum. Remember, the rectangles have a base on the x-axis and extend out to the graph of f open parentheses x close parentheses.

A graph with an x-axis ranging from 0 to 6 and a y-axis ranging from −6 to 0. A line slants downward from the point (1, 0) on the x-axis to the point (5, –4). A graph with an x-axis ranging from 0 to 6 and a y-axis ranging from −6 to 0. Four rectangular bars correspond to subintervals 1–2, 2–3, 3–4, 4–5, and 5–6 and extend downward to −0.5, −1.5, −2.5, and −3.5 units from the x-axis. The bars increase in height as x increases. A line slants downward from the point (1, 0) on the x-axis to the point (5, –4), touching the top midpoint of each rectangular bar.

Now consider a Riemann sum, sum from k equals 1 to n of f open parentheses c subscript k close parentheses increment x.

  • When f open parentheses c subscript k close parentheses is positive, we know the quantity f open parentheses c subscript k close parentheses times increment x is the area of one rectangle.
  • As a result, when f open parentheses c subscript k close parentheses is negative, the quantity f open parentheses c subscript k close parentheses times increment x is the negative of the area of that one rectangle.
Now picture adding these quantities, all of which are negative. The Riemann sum would be an estimate for the negative of the area.

Now, consider the definite integral of this function on the interval open square brackets 1 comma space 5 close square brackets comma written integral subscript 1 superscript 5 open parentheses 1 minus x close parentheses d x.

We know the value of this integral is the limit of the Riemann sums as the number of rectangles gets larger and larger open parentheses n rightwards arrow infinity close parentheses.

Note that the area of the region between f open parentheses x close parentheses and the x-axis is 1 half open parentheses 4 close parentheses open parentheses 4 close parentheses equals 8 space units squared.

Then, integral subscript 1 superscript 5 open parentheses 1 minus x close parentheses d x equals short dash 8 comma the negative of the area of the region.

big idea
If the graph of f open parentheses x close parentheses is below the x-axis on open square brackets a comma space b close square brackets comma then integral subscript a superscript b f open parentheses x close parentheses d x is the negative of the area of the region between f open parentheses x close parentheses and the x-axis on open square brackets a comma space b close square brackets.

2. Evaluating Definite Integrals When f (x) ≤ 0 on [a, b]

EXAMPLE

Evaluate integral subscript short dash 4 end subscript superscript 2 open parentheses open vertical bar x close vertical bar minus 5 close parentheses d x.

The graph of f open parentheses x close parentheses is shown on the left and the region is shown on the right.

A graph with an x-axis ranging from −6 to 6 and a y-axis ranging from −5 to 0. A line slants downward from the marked point at (−5, 0) on the x-axis to the marked point at (0, −5) on the y-axis and then slants upward to the marked point at (5, 0) on the x-axis. A graph with an x-axis ranging from −6 to 6 and a y-axis ranging from −5 to 0. A line starts from the point (−5, 0) on the x-axis and slants downward, passing through a marked point at (−4, −1) up to a marked point at (0, −5) on the y-axis. The line then slants upward up to the point (5, 0) by passing through the marked point at (2, −3). The area above the slanted line, between x equals −4 and x equals 2, is shaded up to the x-axis, forming a trapezoid.


Note that the graph of f open parentheses x close parentheses is below the x-axis on the interval open square brackets short dash 4 comma space 2 close square brackets. The region itself is not a standard shape, so let’s split the region at x equals 0.

On open square brackets short dash 4 comma space 0 close square brackets comma the region is a trapezoid with parallel (vertical) bases 1 and 5, and (horizontal) height 4. The area is 1 half open parentheses 4 close parentheses open parentheses 1 plus 5 close parentheses equals 12 space units squared.

On open square brackets 0 comma space 2 close square brackets comma the region is a trapezoid with parallel (vertical) bases 5 and 3, and (horizontal) height 2. The area is 1 half open parentheses 2 close parentheses open parentheses 5 plus 3 close parentheses equals 8 space units squared.

Then, the total area is 20 space units squared.

Since the region is completely below the x-axis on open square brackets short dash 4 comma space 2 close square brackets comma integral subscript short dash 4 end subscript superscript 2 open parentheses open vertical bar x close vertical bar minus 5 close parentheses d x equals short dash 20.

3. Evaluating Definite Integrals When f (x) Is Both Negative and Positive on [a, b]

Suppose we wish to evaluate integral subscript a superscript b f open parentheses x close parentheses d x for the function whose graph is shown in the figure.

A graph with an x-axis and a y-axis has a curve that oscillates above and below the x-axis, creating three distinct regions. The curve rises from the third quadrant, passing through a marked point labeled ‘(a, f(a))’ on the negative x-axis. It then extends into the second quadrant, reaches a small peak, and then dips into the fourth quadrant after passing through the origin. The curve then extends downward, reaches a minimum point, rises upward in the first quadrant after crossing the x-axis, continues to extend in the first quadrant until it reaches a larger peak, and then falls sharply downward in the fourth quadrant after crossing the x-axis at the marked point labeled ‘(b, f(b))’. The region above the negative x-axis is labeled ‘A1’, the region below the positive x-axis is labeled ‘A2’, and the region above the positive x-axis is labeled ‘A3’. All three regions are shaded.

Notice how this region is broken into 3 smaller regions with areas A subscript 1 comma A subscript 2 comma and A subscript 3.

Now, consider the definite integral on open square brackets a comma space b close square brackets.

  • For the region with area A subscript 1 comma the definite integral is equal to A subscript 1 since the region is above the x-axis.
  • For the region with area A subscript 2 comma the definite integral is equal to short dash A subscript 2 since the region is below the x-axis.
  • For the region with area A subscript 3 comma the definite integral is equal to A subscript 3 since the region is above the x-axis.
Thus, the definite integral over open square brackets a comma space b close square brackets is equal to the sum of the three definite integrals, or A subscript 1 minus A subscript 2 plus A subscript 3. In general, we would add any area above the x-axis and subtract any area below the x-axis.

big idea
If f left parenthesis x right parenthesis takes on both positive and negative values on open square brackets a comma space b close square brackets comma then:
integral subscript a superscript b f open parentheses x close parentheses d x equals (sum of all areas above the x-axis) - (sum of all areas below the x-axis)

Let’s look at an example.

EXAMPLE

Evaluate integral subscript 0 superscript 10 open parentheses x minus 4 close parentheses d x.

Consider the graph of f open parentheses x close parentheses equals x minus 4 on the interval open square brackets 0 comma space 10 close square brackets comma as shown in the figure on the left. The figure on the right shows the graph with the relevant regions.

A graph with an x-axis ranging from −4 to 12 at intervals of 2 and a y-axis ranging from −4 to 6 at intervals of 2. A line slants upward from the point (0, −4) on the y-axis to the point (10, 6) after crossing through the x-axis at the point (4, 0). A graph with an x-axis ranging from −4 to 12 at intervals of 2 and a y-axis ranging from −4 to 6 at intervals of 2. A line slants upward from the marked point (0, −4) on the y-axis to the marked point (10, 6) after crossing through the x-axis at the marked point (4, 0), forming two right-angled triangles. The triangle below the x-axis has vertices at (0, 0), (0, −4), and (4, 0), and the triangle above the x-axis has vertices at (4, 0), (10, 0), and (10, 6). Both the triangles are shaded.

The triangle between x equals 0 and x equals 4 has area 1 half open parentheses 4 close parentheses open parentheses 4 close parentheses equals 8 comma and is below the x-axis.

The triangle between x equals 4 and x equals 10 has area 1 half open parentheses 6 close parentheses open parentheses 6 close parentheses equals 18 comma and is above the x-axis.

Then, integral subscript 0 superscript 10 open parentheses x minus 4 close parentheses d x equals short dash 8 plus 18 equals 10.

watch
Given the graph of y equals f open parentheses x close parentheses comma we’ll find integral subscript 0 superscript 6 f open parentheses x close parentheses d x.

big idea
In this context, the definite integral can be thought of as a “net area.” If integral subscript a superscript b f open parentheses x close parentheses d x greater than 0 comma there is more area above the x-axis than below the x-axis on open square brackets a comma space b close square brackets.

If integral subscript a superscript b f open parentheses x close parentheses d x less than 0 comma there is more area below the x-axis than above the x-axis on open square brackets a comma space b close square brackets.

If integral subscript a superscript b f open parentheses x close parentheses d x equals 0 comma there is as much area above the x-axis as there is below the x-axis on open square brackets a comma space b close square brackets.

try it
Consider the graph of f open parentheses x close parentheses as shown in the figure below that can be used to evaluate integral subscript 0 superscript 12 f open parentheses x close parentheses d x.

A graph with an x-axis ranging from −4 to 12 at intervals of 2 and a y-axis ranging from −6 to 10 at intervals of 2. A line starts from a marked point at (0, −4) on the y-axis, continues horizontally until the marked point at (4, −4), then slants upward to the marked point at (6, 0) and then up to the marked point at (10, 8). From this point, the line continues horizontally until the marked point at (12, 8).

Evaluate this integral.
To evaluate this integral, we must pay attention to the graph’s location relative to the x-axis. Here is the graph, but with the regions shaded.

A graph with an x-axis ranging from −4 to 12 and a y-axis ranging from −6 to 10. A line starts from the marked point at (0, −4), passes through the marked points at (4, −4), (6, 0), (10, 8), and ends at the marked point at (12, 8). A vertical dashed line extends downward from the marked point (12, 8) up to the point (12, 0). The area above the line between the points (0, 0), (0, −4), (4, −4), and (6, 0) and the area below the line between the points (10, 8), (6, 0), (12, 0), and (12, 8) are shaded.

Consider the region between x equals 0 and x equals 6. This region is in the shape of a trapezoid with bases 4 and 6 and height 4. Its area is A subscript 1 equals 4 over 2 open parentheses 4 plus 6 close parentheses equals 20 space units squared.

Now consider the region between x equals 6 and x equals 12. This is also a trapezoid, but with bases 2 and 6, and with height 8. Its area is A subscript 2 equals 8 over 2 open parentheses 2 plus 6 close parentheses equals 32 space units squared.

Since the first region is below the x-axis and the second region is above the x-axis, the value of the definite integral is short dash A subscript 1 plus A subscript 2 equals short dash 20 plus 32 equals 12.

summary
In this lesson, you learned about the Riemann sums of functions that are below the x-axis, understanding how to evaluate integral subscript a superscript b f open parentheses x close parentheses d x when the graph of f(x) is below the x-axis. You applied this knowledge in evaluating definite integrals when f (x) ≤ 0 on [a, b]. Lastly, you learned that when evaluating definite integrals when f (x) is both negative and positive on [a, b], you can interpret the value of the definite integral as “net area,” considering regions that are above and below the x-axis. This will be very useful when investigating applications in the next tutorial.

Source: THIS TUTORIAL HAS BEEN ADAPTED FROM CHAPTER 4 OF "CONTEMPORARY CALCULUS" BY DALE HOFFMAN. ACCESS FOR FREE AT WWW.CONTEMPORARYCALCULUS.COM. LICENSE: CREATIVE COMMONS ATTRIBUTION 3.0 UNITED STATES.

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