Double Integral over a region

**M.R** · July 7th 2012, 04:50 PM

Question: Let $R$ be the region in the plane defined by:

$R = \{(x,y) \in R^2 : 1\leq x \leq 2, x\leq y \leq x^3 \}$

Calculate:

$\int \int_R e^{\frac{y}{x}} dA$

My attempt:

$\int_1^2 \int_x^x^3 e^{\frac{y}{x}} dy dx$ $=$ $\int_1^2 x e^x^2 - xe dx$

$=$ $\frac{1}{2}[e^x^2 - x^2e]_1^2$ $=$ $\frac{1}{2}e^4 - 2e$

Can someone please verify if my answer is correct? Thank you

**Prove It** · July 7th 2012, 08:18 PM

Originally Posted by M.R

Question: Let $R$ be the region in the plane defined by:

$R = \{(x,y) \in R^2 : 1\leq x \leq 2, x\leq y \leq x^3 \}$

Calculate:

$\int \int_R e^{\frac{y}{x}} dA$

My attempt:

$\int_1^2 \int_x^x^3 e^{\frac{y}{x}} dy dx$ $=$ $\int_1^2 x e^x^2 - xe dx$

$=$ $\frac{1}{2}[e^x^2 - x^2e]_1^2$ $=$ $\frac{1}{2}e^4 - 2e$

Can someone please verify if my answer is correct? Thank you

Yes it's correct

**M.R** · July 7th 2012, 08:27 PM

Thank you

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