Answer
The approximation value $52$ and the integration value $\dfrac{160}{3} \approx 53.3$.
Work Step by Step
The approximation value can be found as:
$\Sigma_{i=1}^{8} f(x_i, y_i) \triangle x_i \triangle y_i=\dfrac{2}{4}+\dfrac{10}{4}+\dfrac{26}{4}+\dfrac{50}{4}+\dfrac{18}{4}+\dfrac{34}{4}+\dfrac{58}{4} =52 $
We need to set up the integration in the iterated form.
$I=\int_{0}^4 \int_{0}^{2} (x^2+y^2) \ dy \ dx$
Next, we will solve the integral as follows:
$I=\int_0^4 [x^2 y+\dfrac{y^3}{3}]_0^2 \ dx $
or, $=\int_0^4 [2x^2+\dfrac{(2)^3}{3}] \ dx$
or, $=[\dfrac{2x^3}{3}+\dfrac{8x}{3}]_0^4$
or, $=\dfrac{2(4)^3}{3}+\dfrac{8(4)}{3}$
or, $=\dfrac{160}{3} \approx 53.3$
So, we have the approximation value $52$ and the integration value $\dfrac{160}{3} \approx 53.3$.
