okay, so i'm not at home and hence cannot use all the tools i usually use to make the graphs look pretty, by shading in the area that you want to find and such. so you have to look at the basic graph and figure it out. we want to find the area enclosed by the red, blue and green curves. the point of sketching was to figure out which graph is above which. why, because the area will be given by: int{higher graph - lower graph}dx

now we need to find our limits of integration. the last one is easy, it's x = 1, they told us that. the first one is where the graphs intersect, so let's find that by equating the functions.

y = x^3 - 2x + 1

y = -2x

=> for intersection: x^3 - 2x + 1 = -2x

=> x^3 + 1 = 0 ...............this is the sum of two cubes, there's a formula for that (the alternative is doing long division or synthetic division..yuck!)

=> (x + 1)(x^2 - x + 1) = 0

=> x = -1 since x^2 - x + 1 has no real roots.

so our limits of integration is from -1 to 1

so, Area = int{x^3 - 2x + 1 - (-2x)}dx

............= int{x^2 + 1}dx

............= (1/3)x^3 + x evaluated between -1 and 1, we use the fundamental theorem of calculus, i hope you know it

=> Area = (1/3)(1)^3 + 1 - (1/3)(-1)^3 - 1 = 2/3