Definite Integral: Reciprocal of (a Radical times a Power)

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Find .

First let's look at this problem graphically.
Draw the graph of this function
in the interval .

alt=" Graph of reciprocal of (a radical times a power). Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag! . .

Shade the area involved in the integral.

alt=" Graph of reciprocal of (a radical times a power) with area shaded. 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 area positive or negative?

Since it is above the x-axis, it is positive.

Now let's start the integration. How do we need to rewrite the problem in order to integrate?

We need to convert it to powers in the numerator and use the Power Rule of integration.

What powers will we get?

We'll get and the negative third power.

Set up the problem with these powers.

We get

Notice that the exponent inside the parentheses doesn't change.
In order to integrate, we need to try a substitution.
What looks like a good candidate to be replaced by a new variable, u?

Let

With this choice, what is du?

, or

Use these results to set up the indefinite integral in terms of u and du.

We get

Determine the result of the indefinite integration in terms of u.

We get

Simplify.

Rewrite this result in terms of x.

Use this result to set up the definite integral.

Substitute the limits.

Simplify.

Now let's estimate the area graphically. Add a horizontal line to the diagram which omits some of the desired green area and includes about the same amount of the white area not involved in the integral.

alt=" Graph of reciprocal of (a radical times a power) showing the average value. 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 the rectangle under this line?

It is 3 units wide and 0.016 units high.

Determine its area.

The area is 3 * 0.016 = 0.048.

This area is in approximate agreement with the area in our integration, indicating that our integration may be correct.

The end. If you found this helpful and would recommend that I create more pages like this one, please let me know: Email to John Taylor

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