Answer
Using a computer algebra system, we obtain:
1. for $N=25$: ${S_{25,25}} \simeq 1.07308$
2. for $N=50$: ${S_{50,50}} \simeq 1.07827$
3. for $N=100$: ${S_{100,100}} \simeq 1.08092$
Work Step by Step
The Riemann sum ${S_{N,N}}$ to estimate $\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^1 {{\rm{e}}^{{x^3} - {y^3}}}{\rm{d}}y{\rm{d}}x$ is given by
${S_{N,N}} = \mathop \sum \limits_{i = 1}^N \mathop \sum \limits_{j = 1}^N f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j}$,
where $f\left( {x,y} \right) = {{\rm{e}}^{{x^3} - {y^3}}}$.
The domain is ${\cal R} = \left[ {0,1} \right] \times \left[ {0,1} \right]$. We evaluate ${S_{N,N}}$ using lower left-hand vertices as sample points.
Case 1. For $N=25$
Using the regular partition, dimensions of the subrectangles are
$\Delta x = \frac{{1 - 0}}{{25}} = \frac{1}{{25}}$, ${\ \ \ \ }$ $\Delta y = \frac{{1 - 0}}{{25}} = \frac{1}{{25}}$
The Riemann sum becomes
${S_{25,25}} = \mathop \sum \limits_{i = 1}^{25} \mathop \sum \limits_{j = 1}^{25} f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j}$
${S_{25,25}} = \frac{1}{{625}}\left( {f\left( {0,0} \right) + f\left( {\frac{1}{{25}},0} \right) + \cdot\cdot\cdot + f\left( {\frac{{23}}{{25}},\frac{{24}}{{25}}} \right) + f\left( {\frac{{24}}{{25}},\frac{{24}}{{25}}} \right)} \right)$
Using a computer algebra system, we obtain ${S_{25,25}} \simeq 1.07308$.
Case 2. For $N=50$
Using the regular partition, dimensions of the subrectangles are
$\Delta x = \frac{{1 - 0}}{{50}} = \frac{1}{{50}}$, ${\ \ \ \ }$ $\Delta y = \frac{{1 - 0}}{{50}} = \frac{1}{{50}}$
The Riemann sum becomes
${S_{50,50}} = \mathop \sum \limits_{i = 1}^{50} \mathop \sum \limits_{j = 1}^{50} f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j}$
${S_{50,50}} = \frac{1}{{2500}}\left( {f\left( {0,0} \right) + f\left( {\frac{1}{{50}},0} \right) + \cdot\cdot\cdot + f\left( {\frac{{24}}{{25}},\frac{{49}}{{50}}} \right) + f\left( {\frac{{49}}{{50}},\frac{{49}}{{50}}} \right)} \right)$
Using a computer algebra system, we obtain ${S_{50,50}} \simeq 1.07827$.
Case 3. For $N=100$
Using the regular partition, dimensions of the subrectangles are
$\Delta x = \frac{{1 - 0}}{{100}} = \frac{1}{{100}}$, ${\ \ \ \ }$ $\Delta y = \frac{{1 - 0}}{{100}} = \frac{1}{{100}}$
The Riemann sum becomes
${S_{100,100}} = \mathop \sum \limits_{i = 1}^{100} \mathop \sum \limits_{j = 1}^{100} f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j}$
${S_{100,100}} = \frac{1}{{10000}}\left( {f\left( {0,0} \right) + f\left( {\frac{1}{{100}},0} \right) + \cdot\cdot\cdot + f\left( {\frac{{49}}{{50}},\frac{{99}}{{100}}} \right) + f\left( {\frac{{99}}{{100}},\frac{{99}}{{100}}} \right)} \right)$
Using a computer algebra system, we obtain ${S_{100,100}} \simeq 1.08092$.
