Starting from the definition, calculate the following

Starting from the definition, calculate the following integral: $$ \iint_{\substack{a \leqslant x \leqslant b \\ c \leqslant y \leqslant d}} e^{x+y} \, dx \, dy. $$

Step 1 · Understanding the definition

$$ \iint_R f(x,y)\,dA = \lim_{n,m\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{m} f(x_i^*,y_j^*)\,\Delta x\,\Delta y, \qquad R=[a,b]\times[c,d],\ f(x,y)=e^{x+y}=e^x\cdot e^y. $$

Step 2 · Recognize the separable form

$$ e^{x+y}=e^x\cdot e^y. $$

This lets us write $f$ as a product of a function of $x$ alone and a function of $y$ alone.

Step 3 · Set up the partition

$$ \Delta x=\frac{b-a}{n}, \qquad \Delta y=\frac{d-c}{m}, $$

with $x_i=a+i\frac{b-a}{n}$ ($i=1,\dots,n$) and $y_j=c+j\frac{d-c}{m}$ ($j=1,\dots,m$).

Step 4 · Write the Riemann sum

$$ S_{n,m}=\sum_{i=1}^{n}\sum_{j=1}^{m} e^{x_i+y_j}\,\Delta x\,\Delta y = \sum_{i=1}^{n}\sum_{j=1}^{m} e^{x_i}\cdot e^{y_j}\cdot\Delta x\,\Delta y. $$

Step 5 · Separate the sums

$$ S_{n,m}= \sum_{i=1}^{n} e^{x_i}\Delta x \sum_{j=1}^{m} e^{y_j}\Delta y = \left(\sum_{i=1}^{n} e^{x_i}\Delta x\right)\left(\sum_{j=1}^{m} e^{y_j}\Delta y\right). $$

Step 6 · Take the limit

$$ \lim_{n\to\infty}\sum_{i=1}^{n} e^{x_i}\Delta x = \int_a^b e^x\,dx, \qquad \lim_{m\to\infty}\sum_{j=1}^{m} e^{y_j}\Delta y = \int_c^d e^y\,dy. $$

Therefore $\displaystyle \iint_{\substack{a\leqslant x\leqslant b\\ c\leqslant y\leqslant d}} e^{x+y}\,dx\,dy = \int_a^b e^x\,dx \cdot \int_c^d e^y\,dy$.

Step 7 · Evaluate the integrals

$$ \int_a^b e^x\,dx=\left[e^x\right]_a^b=e^b-e^a, \qquad \int_c^d e^y\,dy=\left[e^y\right]_c^d=e^d-e^c. $$

Step 8 · Multiply the results

$$ \iint_{\substack{a\leqslant x\leqslant b\\ c\leqslant y\leqslant d}} e^{x+y}\,dx\,dy = (e^b-e^a)(e^d-e^c) = e^{b+d}-e^{b+c}-e^{a+d}+e^{a+c}. $$

Step 9 · Verification using iterated integrals

$$ \int_a^b\int_c^d e^{x+y}\,dy\,dx = \int_a^b e^x(e^d-e^c)\,dx = (e^d-e^c)(e^b-e^a). $$

This confirms our result. ✓

Step 10 · Alternative order of integration

$$ \int_c^d\int_a^b e^{x+y}\,dx\,dy = \int_c^d e^y(e^b-e^a)\,dy = (e^b-e^a)(e^d-e^c). $$

Again we get the same result. ✓

Step 11 · Special case verification

For $a=0,b=1,c=0,d=1$:

$$ \iint_{[0,1]\times[0,1]} e^{x+y}\,dx\,dy = (e-1)(e-1)=(e-1)^2=e^2-2e+1. $$

Directly: $\int_0^1 e^x\,dx\cdot\int_0^1 e^y\,dy=(e-1)(e-1)=(e-1)^2$. Perfect match. ✓

Step 12 · Geometric interpretation

$(e^b-e^a)(e^d-e^c)$ is the product of the areas under $y=e^x$ from $a$ to $b$ and under $y=e^y$ from $c$ to $d$; equivalently, the volume under $z=e^{x+y}$ over $[a,b]\times[c,d]$.

Step 13 · Numerical example

For $a=0,b=1,c=0,d=2$:

$$ \iint_{[0,1]\times[0,2]} e^{x+y}\,dx\,dy = (e-1)(e^2-1) \approx (1.71828)(6.38906) \approx 10.977. $$

Summary

Key steps: recognize the separable form $e^{x+y}=e^x\cdot e^y$; build the Riemann sum and separate it as $\left(\sum e^{x_i}\Delta x\right)\left(\sum e^{y_j}\Delta y\right)$; take limits to get $\int_a^b e^x\,dx\cdot\int_c^d e^y\,dy$; evaluate to $(e^b-e^a)(e^d-e^c)$; verify with iterated integrals in both orders.

Additional Notes

• For any separable function $f(x,y)=g(x)h(y)$, $\iint_{[a,b]\times[c,d]} f\,dA = \left(\int_a^b g\,dx\right)\left(\int_c^d h\,dy\right)$.
• The factored form $(e^b-e^a)(e^d-e^c)$ and the expanded form $e^{b+d}-e^{b+c}-e^{a+d}+e^{a+c}$ are equal representations of the same volume.