find the area of the rgion bounded by the graphs of y=x^3-2x and g(x)=-x

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

3*-81=0

3*3-81=0

Yes, you are correct in breaking the region into two parts based on the regions above and below the x-axis. The first integral represents the area of the region above the x-axis, and the second integral represents the area of the region below the x-axis.

Upon evaluating the integrals, the first integral gives the value of 1/4, and the second integral also gives the value of 1/4.

Since the graph is symmetrical about the x-axis, the area above the x-axis is equal to the area below the x-axis. Therefore, the total area bounded by the graphs of y = x^3 - 2x and g(x) = -x is 1/4 + 1/4 = 1/2.

So, the correct answer to the question is indeed 1/2.

Yes, you have correctly set up the integrals to find the area of the region bounded by the graphs of y=x^3-2x and g(x)=-x. Since the region is symmetrical above and below the x-axis, you only need to evaluate one of the integrals and then double the result.

For the first integral, you have correctly set up the limits (-1 to 0) and the integrand [x^3-2x-(-x)]. Evaluating this integral will give you the area of the region above the x-axis.

For the second integral, you have also correctly set up the limits (0 to 1) and the integrand [-x-(x^3-2x)]. This integral will give you the area of the region below the x-axis.

However, since the region is symmetrical, you can choose to evaluate either of the integrals and then simply double the result to get the total area.

By evaluating both integrals, you obtained 1/4 for each integral. Therefore, the total area of the region bounded by the graphs is 1/2, which is the sum of the areas obtained from each integral.

It's important to note that although you mentioned the answer given is "none of the above," the correct answer is indeed 1/2.