1. The volume under the surface [tex]f(x,y)=e^{x+y}[/tex] is given by the double integral,
[tex]\displaystyle\int_0^1\int_0^1e^{x+y}\,\mathrm dx\,\mathrm dy[/tex]
We split up the integration region into a 2x3 grid of rectangles whose upper right corners are determined by the right endpoints of the partition along either axis. That is, we split up the [tex]x[/tex] interval [0, 1] into 2 subintervals,
[0, 1/2], [1/2, 1]
with right endpoints given by the arithmetic sequence,
[tex]r_i=0+\dfrac{i(1-0)}2=\dfrac i2[/tex]
for [tex]i\in\{1,2\}[/tex], and the [tex]y[/tex] interval [0, 1] into 3 subintervals,
[0, 1/3], [1/3, 2/3], [2/3, 1]
with right endpoints
[tex]r_j=0+\dfrac{j(1-0)}3=\dfrac j3[/tex]
for [tex]j\in\{1,2,3\}[/tex].
Then the upper right corners of the 6 rectangles are the points
(1/2, 1/3), (1/2, 2/3), (1/2, 1), (1, 1/3), (1, 2/3), (1, 1)
generated by the sequence [tex](r_i,r_j)[/tex].
The integral is thus approximated by the sum
[tex]\displaystyle\sum_{j=1}^3\sum_{i=1}^2f(r_i,r_j)\dfrac{1-0}m\dfrac{1-0}n=\dfrac16\sum_{j=1}^3\sum_{i=1}^2f(r_i,r_j)=\frac{e^{5/6}+e^{7/6}+e^{4/3}+e^{5/3}}6[/tex]
or approximately 2.4334. (Compare to the actual value of the integral, which is close to 2.952.)
For the midpoint rule estimate, we replace the sampling points [tex](r_i,r_j)[/tex] with [tex](m_i,m_j)[/tex], i.e. the midpoints of each subinterval, so the set of sampling points is
(1/4, 1/6), (3/4, 1/6), (1/4, 1/2), (3/4, 1/2), (1/4, 5/6), (3/4, 5/6)
and the integral is approximately
[tex]\displaystyle\sum_{j=1}^3\sum_{i=1}^2f(m_i,m_j)\dfrac{1-0}m\dfrac{1-0}n=\frac{e^{5/12}+e^{3/4}+e^{11/12}+e^{13/12}+e^{5/4}+e^{19/12}}6[/tex]
or about 2.908.
2. We approach the second integral the same way. Split up the [tex]x[/tex] interval into 8 subintervals with left and right endpoints given respectively by
[tex]\ell_i=-2+\dfrac{(i-1)(2-(-2))}8=\dfrac{i-5}2[/tex]
[tex]r_i=-2+\dfrac{i(2-(-2))}8=\dfrac{i-4}2[/tex]
for [tex]i\in\{1,2,\ldots,8\}[/tex], and the [tex]y[/tex] interval into 2 subintervals with
[tex]\ell_j=0+\dfrac{(j-1)(2-0)}2=j-1[/tex]
[tex]r_j=0+\dfrac{j(2-0)}2=j[/tex]
for [tex]j\in\{1,2\}[/tex].
The upper left corners of the rectangles in this grid are given by the sequence [tex](\ell_i,r_j)[/tex]. So the integral is approximately
[tex]\displaystyle\sum_{j=1}^2\sum_{i=1}^8f(\ell_i,r_j)\frac{2-(-2)}m\frac{2-0}n=51[/tex]
(Compare to the actual value, 32.)
