General Contents
Detailed Contents
Index
Programmed tutorial: Integral of the Absolute Value of a Quadratic
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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 lets look at a related integral involving the same linear quantity, but without the absolute value signs:
Plot the function (
x
2
+ 2
x
3) and show the area involved in this integral.
Then check your work by clicking Next.
alt="Graph of a quadratic factor. Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!
For this integral, is all of the area positive?
No.
What part is negative?
The part below the
x
-axis, shown in red, or the part for which
.
Next plot the function |
x
2
+
2
x
3| and show the area involved in the original integral
.
Then check your work by clicking Next.
alt="Graph of the absolute value of a quadratic factor. Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!
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 three parts, separated by the
x
-intercepts.
Determine the
x
-intercepts.
We solve for the
x
value which results in a
y
-coordinate of 0:
x
2
+ 2
x
3 = 0 = (
x
+ 3)*(
x
1),
or
x
= 3 or
x
= 1.
Set up the integral for the part to the left of
x
= 3.
Here |
x
2
+ 2
x
3| = +
(
x
2
+ 2
x
3), so we get
Do the integration.
Substitute the limits.
We get
What does this represent?
The area of the left shape.
Set up the integral for the shape between the intercepts.
Here |
x
2
+ 2
x
3| =
(
x
2
+ 2
x
3), so we get
Do the integration.
Substitute the limits.
We get
What does this represent?
The area of the hump between the intercepts.
Set up the integral for the part to the right of
x
= 1.
Here |
x
2
+ 2
x
3| = +
(
x
2
+ 2
x
3), so we get
Do the integration.
Substitute the limits.
We get
What does this represent?
The area of the right-hand shape.
Have we solved the original problem?
Not quite. We needed the integral from -4 to 2.
How can we find that now?
We can add the three integrals 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 three 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.
In the center, we see that the triangle has less area than that under the hump.
The other triangles each have slightly more area than that under the curve.
alt="Geometric approximation to the area of the absolute value of a quadratic. Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!
As a review, state the area of a triangle.
Area = ½ * 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 2.5 to compare with 2.3.
These are close and the area of the triangle is slightly larger, as expected from the diagram.
Now find the area of the center triangle.
We found an area of
in our integration.
How do the decimal approximations to these fractions compare?
We get a value of 8 to compare with 10.7.
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.
The area of the right-hand triangle is the same as that of the left-hand one, and also confirms our integration.
In summary, we see that by calculating areas geometrically, we can independently estimate our answers from integration.
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.
.