Programmed tutorial: Integral of the Absolute Value of an Offset Linear

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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 .

First let’s look at a series of related functions. Plot the function . Then check your work by clicking “Next”.

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Plot the absolute value function . Then check your work by clicking “Next”.

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Since the negative, , is involved in our integral, plot it and then check.

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How will the graph of differ?

It will be translated upward.

Plot it and then check by clicking “Next”.

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Indicate limits of integration on the diagram.

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The preceding diagram also shows a typical infinitesimal rectangle used to find the area. What is the width of this rectangle?

dx.

Describe the top of the rectangle.

The lines determine the top of the rectangle.

Which line determines the bottom?

The x-axis.

How can we use such rectangles to find the area?

We will describe the area of such rectangles and add them up over the whole area by integration.

Is our above description of the top and bottom applicable over the whole area?

Yes, in general. Mathematically, we’ll have a different equation for the top on either side of the peak.

Add shading to your diagram to indicate the two regions of area.

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Determine the equation of the line to be used for the top of the infinitesimal rectangle for the left portion.

Determine the equation of the line to be used for the top of the infinitesimal rectangle for the right portion.

Set up the indefinite integral for the integration for x < 5 in general terms.

Substitute values and limits.

Do the integration.

Substitute the limits.

Set up the indefinite integral for the integration for x > 5 in general terms.

Substitute values and limits.

Do the integration.

Substitute the limits and simplify.

Have we solved the original problem?

Not quite. We need the integral from 3 to 8.

How can we find that now?

We can add the two areas.

Do that.

We get for the value of the original definite integral.

How can we check our result?

We can estimate an average height over the area and multiply it by the full width of 5.

Do that with an average height of 4 (which is too large).

Roughly, the area then would be 4 * 5 = 20.

How does this compare with our result?

It is slightly larger, as we would expect from an average height that is too large.

Try an average height of 3 (which is too low).

We get 3 * 5 = 15.

Again our result from integration is consistent.

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

General Contents

Detailed Contents

Index