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Since this problem tells us that we are using the vertical lines x = -1 and 1, those will be our boundaries for our definite integral:

So we will be doing the integral of ((x^2 + 5 - 1/3x^3))dx which gives us:

1/3x^3 + 5x - 1/12x^4 from -1 to 1, plugging those in we have:

[1/3(1)^3 + 5(1) - 1/12(1)^4] - [1/3(-1)^3 + 5(-1) - 1/12(-1)^4] =

21/4 - (-65/12)

= 32/3 square units

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