evaluate $\\int_{2}^{5}\\int_{2}^{8}xye^{x + y}dydx$. (hint: at some point you will need to use integration…

Explanation:

Step1: Separate the variables

Since (e^{x + y}=e^x\cdot e^y), the double - integral (\int_{2}^{5}\int_{2}^{8}xye^{x + y}dydx=\int_{2}^{5}xe^{x}dx\int_{2}^{8}ye^{y}dy).

Step2: Use integration by parts for (\int ye^{y}dy)

The formula for integration by parts is (\int u;dv=uv-\int v;du). Let (u = y), (dv=e^{y}dy). Then (du = dy), (v = e^{y}). So (\int ye^{y}dy=ye^{y}-\int e^{y}dy=ye^{y}-e^{y}+C=(y - 1)e^{y}+C).

Step3: Evaluate (\int_{2}^{8}ye^{y}dy)

[ \begin{align*} \int_{2}^{8}ye^{y}dy&=\left[(y - 1)e^{y}\right]_{2}^{8}\ &=(8 - 1)e^{8}-(2 - 1)e^{2}\ &=7e^{8}-e^{2} \end{align*} ]

Step4: Use integration by parts for (\int xe^{x}dx)

Let (u = x), (dv=e^{x}dx). Then (du = dx), (v = e^{x}). So (\int xe^{x}dx=xe^{x}-\int e^{x}dx=xe^{x}-e^{x}+C=(x - 1)e^{x}+C).

Step5: Evaluate (\int_{2}^{5}xe^{x}dx)

[ \begin{align*} \int_{2}^{5}xe^{x}dx&=\left[(x - 1)e^{x}\right]_{2}^{5}\ &=(5 - 1)e^{5}-(2 - 1)e^{2}\ &=4e^{5}-e^{2} \end{align*} ]

Step6: Calculate the product of the two integrals

[ \begin{align*} \int_{2}^{5}xe^{x}dx\int_{2}^{8}ye^{y}dy&=(4e^{5}-e^{2})(7e^{8}-e^{2})\ &=4e^{5}\times7e^{8}-4e^{5}\times e^{2}-e^{2}\times7e^{8}+e^{2}\times e^{2}\ &=28e^{13}-4e^{7}-7e^{10}+e^{4} \end{align*} ]

Answer:

(28e^{13}-4e^{7}-7e^{10}+e^{4})