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

Iterated integral = 1.8

Approximation = 1.7

Explanation:

The iterated integral will be:

`\int\limits^2_0 \int\limits^4_0 (1)/((x+1)(y+1)) dxdy = \int\limits^2_0 (\int\limits^4_0 (1)/((x+1)(y+1)) dy)dx = \int\limits^2_0 (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)(1)/((x+1)(y+1))=`

`= (4)/(9)+ (4)/(15)+ (4)/(21)+ (4)/(27)+(4)/(15) +(4)/(25) +(4)/(35) +(4)/(45) = 1.6796` = 1.7 (since rounding one decimal place is asked)

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