Answer to Question #245474 in Calculus for Jummy

Question

Integrate f(x,y)=x+y over the region bounded by y=x^2-1, x=2 and x=0.

Answer 1

The sketch of the region is as shown in the figure below:

The integration of the function is evaluated as,

$\iint f(x,y)dA=\int_{1}^{2}\int_{0}^{x^2-1}(x+y)dydx$

$=\int_{1}^{2}[xy+(y^2/2)]_{0}^{x^2-1}dx$

$=\int_{1}^{2}(x^3-x+\frac {1}{2}(x^4-2x^2+1))dx$

$=\frac{1}{2}\int_{1}^{2}(x^4+2x^3-2x^2-2x+1)dx$

$=\frac{1}{2}[\frac{x^5}{5}+\frac{2}{4}x^4-\frac{2}{3}x^3-x^2+x]_{1}^{2}$

$=\frac{211}{60}$

Therefore, the double integral is $\iint f(x,y)dA=\frac{211}{60}$