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Definite Integral: Fractional Power
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Find
.
First, let's look at the graph of this function. Draw that graph.
alt=" Graph of a fractional power. Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!
..
Is this an even or odd function?
Because the numerator and denominator of the fractional power are odd, this is an odd function.
When we have an integral of an odd function between symmetric limits (–1 and 1 here), we can do some prediction about the result of the integration. What do we expect for this case?
We expect a result of zero.
How does the amount of area between the curve and the
x
-axis in the third quadrant compare to the amount of area in the first quadrant?
They are equal.
Do they cancel?
Yes, the area in the third quadrant is negative and that in the first quadrant is positive.
Show this graphically.
alt=" Graph of a fractional power. Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!
. .
Which color represents the negative area?
The red area is negative.
Let's integrate in two stages. Consider
Should we expect to get a positive or negative value?
Negative, because the area is below the
x
-axis.
Let's do the integral
in stages.
Do the indefinite integral
Substitute the limits to get
I
1.
How do we evaluate
?
We can think of it as
Evaluate the sixth power of (–1).
It becomes 1.
Take the one fifth root of that.
We get 1.
Now use these results to evaluate I1.
Now let's estimate the area in the third quadrant graphically. Add a horizontal line to the diagram which omits some of the desired red area and includes a compensating amount of the white area not involved in the third quadrant.
alt=" Graph of a fractional power. Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!
What are the dimensions of this rectangle?
It is one unit wide and 0.8 units high.
Determine its area.
The area is 1.0 * 0.8 = 0.8.
How does this compare with the result of our integration?
It is close to the value
.
so we can trust our work so far.
Now let's do the integral in the first quadrant
Is this what we expect from the diagram?
Yes, the area is above the
x
-axis so it must be positive. It is the same in size as the red area, so the value has to be
.
How do we use these results to solve the original problem?
We add I3 and I1.
What is the result?
We get
, as expected.
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General Contents
Detailed Contents
Index