Midpoint Rule With Area Between Curves

[InterserveVB]

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

y = (1+x^3)^1/2, y = 1 - 8x, x = 2

How would I go about solving this?

 

[stapel]

First, find the intersection points. (A graph might help for this.) Clearly, x = 2 will be the right-hand interval endpoint "b". What is the left-hand interval endpoint "a"? So how long is the interval |b - a|?

If n = 4, how many subintervals do you have? How long are they? What are their midpoints x_i?

Which function is the upper limit of the area? Which is the lower? So what is the difference that you'll be integrating?

Once you've determined all this, then just plug into the formulation they gave you.

Eliz.

 

[InterserveVB]

Ok, they intersect when x = 0 so I guess 2 and 0 are my endpoints.

The midpoint forumla is delta x = b-a / n

Would I just solve each of the parts seperatly to get b and a then plug them in. Also, I am not sure which equation to put first. 2 is the upper limit and 0 is the lower. the diffrence between the two is 2. Correct?

 

[stapel]

I am not sure which equation to put first.

Which function is the upper limit of the area? Which is the lower?

Eliz.

 

[InterserveVB]

Ok I got it. 17.22 thx again.

 

[stapel]

Ok I got it. 17.22

Excellent! Good work!

Eliz.

P.S. If you have a TI-83 or TI-84, send me a private message if you like, reminding me of this thread and providing your e-mail address, and I can send you a "Riemann sums" program I wrote that does left-endpoint, right-endpoint, and midpoint computations.