Answer:
Iterated integral = 1.8
Approximation = 1.7
Step-by-step explanation:
The iterated integral will be:
[tex]\int\limits^2_0 \int\limits^4_0 \frac{1}{(x+1)(y+1)} dxdy = \int\limits^2_0 (\int\limits^4_0 \frac{1}{(x+1)(y+1)} dy)dx = \int\limits^2_0 \frac{ln(5)}{x+1} dx =[/tex]
[tex]= ln(3)ln(5) = 1.7682[/tex] = 1.8 (since rounding one decimal place is asked)
The coordinates of each 8 squares are as following:
[tex](0.5,0.5),(1.5,0.5),(2.5,0.5),(3.5,0.5),(0.5,1.5),(1.5,1.5),(2.5,1.5),(3.5,1.5)[/tex]
So, the approximation will be:
[tex]\sum\limits^8_{i=1} f(x_i,y_i)\Delta x_i\Delta y_i = \sum\limits^{3.5}_{x=0.5}\sum\limits^{1.5}_{y=0.5}\frac{1}{(x+1)(y+1)}=[/tex]
[tex]= \frac{4}{9}+ \frac{4}{15}+ \frac{4}{21}+ \frac{4}{27}+\frac{4}{15} +\frac{4}{25} +\frac{4}{35} +\frac{4}{45} = 1.6796[/tex] = 1.7 (since rounding one decimal place is asked)
Thus, the result from iterated integral and approximation is slightly different.
