12. \nevaluate the integral: (iint_{r} e^{2 x+3 y} d a), where (r=0,5 \times0,9)\nanswer :\n

Explanation:

Step1: Rewrite the double - integral as an iterated integral

Since (R = [0,5]\times[0,9]) and (e^{2x + 3y}=e^{2x}\cdot e^{3y}), we can write (\iint_{R}e^{2x+3y}dA=\int_{0}^{9}\int_{0}^{5}e^{2x}\cdot e^{3y}dxdy). By the property of double - integrals (\int_{a}^{b}\int_{c}^{d}f(x)g(y)dxdy=\int_{a}^{b}g(y)dy\int_{c}^{d}f(x)dx) (when the limits of integration are constants), we have (\int_{0}^{9}e^{3y}dy\int_{0}^{5}e^{2x}dx).

Step2: Integrate (e^{2x}) with respect to (x)

Using the formula (\int e^{kx}dx=\frac{1}{k}e^{kx}+C) ((k\neq0)), for (k = 2), (\int_{0}^{5}e^{2x}dx=\left[\frac{1}{2}e^{2x}\right]{0}^{5}). [ \begin{align*} \left[\frac{1}{2}e^{2x}\right]{0}^{5}&=\frac{1}{2}(e^{2\times5}-e^{2\times0})\ &=\frac{1}{2}(e^{10}-1) \end{align*} ]

Step3: Integrate (e^{3y}) with respect to (y)

Using the formula (\int e^{ky}dy=\frac{1}{k}e^{ky}+C) ((k\neq0)), for (k = 3), (\int_{0}^{9}e^{3y}dy=\left[\frac{1}{3}e^{3y}\right]{0}^{9}). [ \begin{align*} \left[\frac{1}{3}e^{3y}\right]{0}^{9}&=\frac{1}{3}(e^{3\times9}-e^{3\times0})\ &=\frac{1}{3}(e^{27}-1) \end{align*} ]

Step4: Multiply the two results

[ \begin{align*} \int_{0}^{9}e^{3y}dy\int_{0}^{5}e^{2x}dx&=\frac{1}{2}(e^{10}-1)\times\frac{1}{3}(e^{27}-1)\ &=\frac{(e^{10}-1)(e^{27}-1)}{6} \end{align*} ]

Answer:

(\frac{(e^{10}-1)(e^{27}-1)}{6})