Question

Approximate the area under the curve. Use right side approximation.

y=.4x^2+1, over [0,1] and the width of the rectangles are 1/4 each.

Answer 1

You have four intervals [0,1/4][1/4,1/2][1/2,3/4][3/4,1]
The right sides are x = 1/4, 1/2, 3/4, and 1
Evaluate for the four values of y.

You're almost done!

Answer 2

\[y=.4x^2+1\]
So I got the other question correct, but I've worked through this one twice and got the same answer, yet it's wrong.

So I started off by graphing it using a graphing calculator and I set the window and everything correctly. I wound up getting y values of 1, 1.025, 1.1, 1.225, 1.4, and I multiplied the entire thing by .25 because that's the width of each triangle. I got the second question correct by doing the same thing.

Answer 3

I added the y values, that is.

Answer 4

each triangle? I'm tired. ._. rectangle*

Answer 5

Why do you care about x = 0 giving y = 1? If you were doing the left side, that would be needed. Right side? No. Just do the four right sides you have enumerated.

Answer 6

I'm so sorry man. I wasn't there yesterday when they did this and I am just going off of a single question I watched the teacher do. It doesn't help that I don't have a graphing calculator around.

Answer 7

So the area approximation is derived from the area formula itself? Because area=x*y where x is length and y is height, and the rectangle width would be .25 in this case and the y would be the length?

Answer 8

I just used Desmos graphing calculator or whatever. It helps being able to visualize it. x=0 is irrelevant, it's only the .25, .5, .75, and 1 that matter, right?

Answer 9

It's x = 0.25, 0.50, 0.75, and 1.00
This gives you four y-values to pair up with the common width of 0.25

0.25*1.025 + 0.25*1.1 + 0.25*1.225 + 0.25*1.4

Answer 10

Thank you sir, I've got it now. :3