Question

[15 points] Compute ffR2(x + 1)y2 dA, R = [ 0, 1] x [0,3), by Riemann sum definition. You must use the Riemann sum definition to receive credit.

Answer 1

Looks like the integral is

$\displaystyle\iint_R2(x+1)y^2\,\mathrm dA$

where $R=[0,1]\times[0,3]$. (The inclusion of $y=3$ will have no effect on the value of the integral.)

Let's split up $R$ into $mn$ equally-sized rectangular subintervals, and use the bottom-left vertices of each rectangle to approximate the integral. The intervals will be partitioned as

$[0,1]=\left[0,\dfrac1m\right]\cup\left[\dfrac1m,\dfrac2m\right]\cup\cdots\cup\left[\dfrac{m-1}m,1\right]$

and

$[0,3]=\left[0,\dfrac3n\right]\cup\left[\dfrac3n,\dfrac6n\right]\cup\cdots\cup\left[\dfrac{3(n-1)}n,3\right]$

where the bottom-left vertices of each rectangle are given by the sequence

$v_{i,j}=\left(\dfrac{i-1}n,\dfrac{3(j-1)}n\right)$

with $1\le i\le m$ and $1\le j\le n$. Then the Riemann sum is

$\displaystyle\lim_{m\to\infty,n\to\infty}\sum_{i=1}^m\sum_{j=1}^nf(v_{i,j})\frac{1-0}m\frac{3-0}n$

$\displaystyle=\lim_{m\to\infty,n\to\infty}\frac3{mn}\sum_{i=1}^m\sum_{j=1}^n\frac{18}{mn^2}(j-1)^2(i-1+m)$

$\displaystyle=\lim_{m\to\infty,n\to\infty}\frac{54}{m^2n^3}\sum_{i=0}^{m-1}\sum_{j=0}^{n-1}j^2(i+m)$

$\displaystyle=\frac92\lim_{m\to\infty,n\to\infty}\frac{(3m-1)(2n^3-3n^2+n)}{mn^3}$

$\displaystyle=\frac92\left(\lim_{m\to\infty}\frac{3m-1}m\right)\left(\lim_{n\to\infty}\frac{2n^3-3n^2+n}{n^3}\right)=\boxed{27}$