General Contents
Detailed Contents
Index
Programmed tutorial: Integral of the Absolute Value of a Cubic
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If you want to see all of the following steps at once, click the "All Steps" button. Otherwise, use the "Next" button.
Find the integral
General Contents
Detailed Contents
Index
First let’s look at a related integral involving the same quantity, but without the absolute value signs:
In order to plot this integrand, we need the factors and x-intercepts.
Factor out one power of
x
.
becomes
Now factor the quadratic.
We get
State the x-intercepts.
Setting each factor to zero, we get
x-intercepts at –1, 0, and 2.
Plot the function
.
Then check your work by clicking “Next”.
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For this graph, is all of the area positive?
No.
What part is negative?
The part below the x-axis, or the part for which
. It is shown in red.
Next let’s plot the absolute value,
,
and show the area involved in the original integral
.
Then check your work by clicking “Next”.
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Is any of the area negative?
No, because of the absolute value signs.
How do we set up the integral?
We handle the integral in two parts, separated by the x-intercept at
x
= 0.
Set up the integral for the part to the left of
x
= 0.
Here
,
so we get
Do the integration.
Substitute the limits.
We get
What does this represent?
The area of the left “hump”.
Set up the integral for the right-hand “hump”.
Here
, so we get
Do the integration.
Substitute the limits.
We get
What does this represent?
The area of the right-hand “hump”.
Have we solved the original problem?
Not quite. We needed the integral from –1 to 2.
How can we find that now?
We can add the two areas just calculated.
Do that.
We get
for the value of the original definite integral.
How can we check our result?
We can approximate the area of each of the two shapes geometrically. In the figure below, the curves representing the function we are integrating are shown (magnified on the right) in red. In blue are shown triangles which have approximately the same area. On the left, we see that the triangle has more area than that under the hump. On the right, the triangle has less area than that under the hump.
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As a review, state the area of a triangle.
Area = ˝ * base * height
Apply that to the left-hand triangle.
We found an area of
in our integration. How do the decimal approximations to these fractions compare?
We get a value of 0.5 to compare with 0.42. These are close and the area of the triangle is slightly larger, as expected from the diagram.
Now find the area of the right triangle.
We found an area of
in our integration. How do the decimal approximations to these fractions compare?
We get a value of 2 to compare with 2.7. These are not as close as on the left, which we should expect from the diagram. The area of the triangle is the smaller, as expected
In summary, we see that by calculating areas geometrically, we can independently estimate our answers from integration.
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