Area using the midpoint rule

[skatru]

Use the midpoint rule with n = 4 to approximate the area of the region bounded by th given curves.

y = sin[sup:2k6qi55h]2[/sup:2k6qi55h](xpi/4)
y = cos[sup:2k6qi55h]2[/sup:2k6qi55h](xpi/4)

0 is equal to or greater than x is equal to or less than 1

So I need to find the integral at least I believe I do.

Integral cos[sup:2k6qi55h]2[/sup:2k6qi55h](xpi/4) - sin[sup:2k6qi55h]2[/sup:2k6qi55h](xpi/4) =?

the answer is approximately .64

I'm just confused... Any help is appreciated. Thanks.

 

[galactus]

One thing we could do is to rewrite:

\(\displaystyle cos^{2}(\frac{{\pi}x}{4})-sin^{2}(\frac{{\pi}x}{4})=2cos^{2}(\frac{{\pi}x}{4})-1\)

Now, the widths of the rectangles are \(\displaystyle \frac{1-0}{4}=\frac{1}{4}\)

For the first step we go to the midpoint, which is 1/8.

Plug that into the given function and get .980785280403

Now, go to the next interval by adding 1/4, which is 3/8, then 5/8, then 7/8.

We get

  x                        f(x)
-------------------------------
1/8                    .9878528
3/8                    .8314696
5/8                    .5555702
7/8                    .1950903

Notice that is 1/8 to the the end of the interval.

Add them up and we get 2.562915

Multiply by .25, the width of our interval and get .64072889

Check this against the actual which is .6366198

It's a little more than the actual because the corners of the rectangles stick out of the curve. The more we add, the better it'll become.