Question

Approximate the integral ∫R∫ f(x, y) dA by dividing the rectangle R with vertices (0, 0), (4, 0), (4, 2), and (0, 2) into eight equal squares and finding the sum ∑_(i=1)^8 f(x_i, y_i) ΔA_i, where (x_i, y_i) is the center of the ith square. Evaluate the iterated integral and compare it with the approximation.∫_0^4∫_0^2 1 / (x+ y) (y+1) dy dx. (Round your answers to one decimal place.)

Answer 1

Answer:

Iterated integral = 1.8

Approximation = 1.7

Step-by-step explanation:

The iterated integral will be:

$\int\limits^2_0 \int\limits^4_0 \frac{1}{(x+1)(y+1)} dxdy = \int\limits^2_0 (\int\limits^4_0 \frac{1}{(x+1)(y+1)} dy)dx = \int\limits^2_0 \frac{ln(5)}{x+1} dx =$

$= ln(3)ln(5) = 1.7682$ = 1.8 (since rounding one decimal place is asked)

The coordinates of each 8 squares are as following:

$(0.5,0.5),(1.5,0.5),(2.5,0.5),(3.5,0.5),(0.5,1.5),(1.5,1.5),(2.5,1.5),(3.5,1.5)$

So, the approximation will be:

$\sum\limits^8_{i=1} f(x_i,y_i)\Delta x_i\Delta y_i = \sum\limits^{3.5}_{x=0.5}\sum\limits^{1.5}_{y=0.5}\frac{1}{(x+1)(y+1)}=$

$= \frac{4}{9}+ \frac{4}{15}+ \frac{4}{21}+ \frac{4}{27}+\frac{4}{15} +\frac{4}{25} +\frac{4}{35} +\frac{4}{45} = 1.6796$ = 1.7 (since rounding one decimal place is asked)

Thus, the result from iterated integral and approximation is slightly different.