Saturday

April 18, 2015

April 18, 2015

Posted by **david** on Saturday, January 6, 2007 at 9:05pm.

i drew the graph and half of the graph is above the xaxis and the other half is below the axis.

so the integrals i came up with are two because i broke them up and i combined the answers at the end:

integral (-1 to 0)[x^3-2x-(-x)] dx

the second integral is

intg(0 to 1){-x-(x^3-2x)] dx

the second integral is the part that's below the x axis. i was wondering if you can check if i did it right.

this is actually a multiple choice question and the answer is none of the above but i want to know what the answer would be.

Since you drew the graph you should have noticed that the graph is symmetrical, that is the region above is equal to the region below.

So you would have to evaluate only one of them, then double that answer.

I got 1/4 for each of your integrals, so the total area would be 1/2.

Both of integrals are correct

- calculus showed work -
**Siphosakhe**, Tuesday, February 17, 2015 at 1:40pm3*-81=0

- calculus showed work -
**Siphosakhe**, Tuesday, February 17, 2015 at 1:43pm3*3-81=0

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