Area under Cosine

If you find this page helpful and would recommend that I create more pages like this one, please let me know. Please let me know your opinion of the video.
Email to John Taylor

If you want to see all of the following steps at once, click the "All Steps" button. Otherwise, use the "Next" button.
Scroll to the end of the page for a helpful video.

Find the area between Area bounded by the cosine

First, let’s draw a graph on paper. Draw the cosine curve..
Then check your work by clicking “Next”.

alt="Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!

Show the left boundary of the area.. Then check your work by clicking “Next”.

alt="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. Then check your work by clicking “Next”.

alt="Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!

Show a typical infinitesimal rectangle to be used to find the area. Then check by clicking “Next”.

alt="Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!

What is the width of this rectangle?

dx.

Describe the top of the rectangle.

The cosine curve determines the top.

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.

Set up the indefinite integral for the integration.

Substitute the expressions for the top and the bottom

Add the limits of integration.

Do the integration.

Substitute the limits.

We get

Is this a reasonable area based on a visible estimate on the diagram?

Yes. For example the area to the right of the y-axis is larger than a triangle with a base of 1.57 and a height of 1 unit, so that area is about one square unit. The area to the left of the vertical axis is about three fourths of that. Consequently 1.71 looks like a reasonable value.

The end. If you found this helpful and would recommend that I create more pages like this one, please let me know. Please let me know your opinion of the video. Email to John Taylor

General Contents

Detailed Contents

Index

This video is brought to you by Brightstorm, where you can see over 2,000 free math videos. See more Definite Integrals and Area videos at Brightstorm.