Suggested languages for you:

Americas

Europe

Q25E

Expert-verified
Found in: Page 699

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Calculate the double integral$$\int {\int\limits_R {y{e^{ - xy}}dA,R = \left( {0,2} \right)X\left( {0,3} \right)} }$$

Therefore, integral of the given function is$$\frac{1}{2}\left( {{e^{ - 6}} + 5} \right)$$.

See the step by step solution

## Step(1):- Fubini’s theorem

Recollect the fubini’s theorem

If f is continuous on the rectangle R,

$$\begin{array}{l}R = \{ (x,y)|a \le x \le b,c \le y \le d|\} then,\\\int {\int\limits_R {f\left( {x,y} \right)dA = \int\limits_a^b {\int\limits_c^d {f\left( {x,y} \right)dydx = } } } } \int\limits_a^b {\int\limits_c^d {f\left( {x,y} \right)dxdy} } \end{array}$$

## Step (2):- Determining integral value

Notice that $$0 \le x \le 2$$ which means that x=0 and x=2 are the lower and upper limits of integration of x respectively, and $$0 \le y \le 3$$, which means that y=0 and y=3 are the lower and upper limits of integration of y respectively.

Integrating the function f(x,y)=$$y{e^{ - xy}}$$ with respect to x from 0 to 2 and holding y as a constant.

$$\begin{array}{l}\int {\int\limits_R {y{e^{ - xy}}dA = \int\limits_0^3 {\left( {\int\limits_0^2 {y{e^{ - xy}}dx} } \right)dy} } } \\\int\limits_0^3 {\left( {y\frac{{{e^{ - xy}}}}{{\left( { - y} \right)}}} \right)_{x = 0}^2dy} \\\int\limits_0^3 {\left( { - {e^{ - xy}}} \right)} _{x = 0}^2dy\\\int\limits_0^3 {\left( { - {e^{ - xy}} + 1} \right)dy} \\\left( {\frac{{ - {e^{ - 2y}}}}{{ - 2}} + y} \right)_0^3\\\left( {\frac{1}{2}\frac{{{e^{ - 2y}}}}{{ - 2}} + y} \right)_0^3\\\frac{1}{2}{e^{ - 6}} + 3 - \frac{1}{2}\\\frac{1}{2}\left( {{e^{ - 6}} + 5} \right)\end{array}$$

Therefore, integral of the given function is,

$$\frac{1}{2}\left( {{e^{ - 6}} + 5} \right)$$.