Evaluate $\iint y e^{x y} d A$ for the region $R = {(x, y) ∣ 0 \leq x \leq 1, 1 \leq y \leq 2}$ .
A. $e^{2} + 2 e - 1$
B. $e^{2} + e - 1$
C. $e^{2} - e - 1$
D. $e^{2} + e + 1$

Answer

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Answer

Thus, the final answer is $e^{2} + e - 1$ .

Steps

Step 1 ：We are given the double integral $\iint y e^{x y} d A$ over the region $R = {(x, y) ∣ 0 \leq x \leq 1, 1 \leq y \leq 2}$ .

Step 2 ：To solve this, we first integrate with respect to x while keeping y constant. This gives us $\int_{0}^{1} y e^{x y} d x = e^{y} - 1$ .

Step 3 ：We then integrate this result with respect to y. This gives us $\int_{1}^{2} (e^{y} - 1) d y = - e - 1 + e^{2}$ .

Step 4 ：Thus, the final answer is $e^{2} + e - 1$ .