Okay, what have you done? Have you graphed the function?
Also there is a lack of information. The region enclosed by y= x^2+ x, x= 0, and x= 2 is not bounded and does not have a finite area. You are missing one boundary. Is y= 0 also a boundary line?
If so, to do this you only need to know:
1) The area is $\displaystyle \int_0^2 f(x)- g(x) dx$ where f(x) and g(x) are the equations of the upper and lower boundaries, respectively. (And, here, g(x)= 0.)
2) The anti-derivative of $\displaystyle x^n$ is $\displaystyle \frac{1}{n+1}x^{n+1}$.
well sketch the graph first of all. complete the square and you can see that its just a shifted parabola. now they want you to find the area enclosed by the graph and the vertical lines x=0 and x=2. i also assume y=0 is a boundary otherwise the area would not be finite. draw a rectangle in the region with height f(x) and width dx. the area of that rectangle would be (x^2 + x)dx so when you sum up a lot of those rectangles and take the limit as the width of those rectangles go to zero, you get the integral from 0 to 2 of (x^2 + x)dx. that is an easy integral to compute and you can complete the problem from there.