Acproximate the integral $$ \int_{R} \int R(x, y) d A $$ by dividing the rectangle $$ R $$ with vertices $$ (0,0),(4,0),(4,2) $$, and ( 0,2 ) into elaht equal squares and finding the sumn $$ \sum_{i=1}^{n} R\left(x_{i}, y_{j}\right) \Delta A_{i} $$, where $$ \left(x_{i}, y_{j}\right) $$ is the center of th th square. Evaluate the iterated integral and compare it with the approximation. (Round your answers to one decimal place.) $$ \frac{1}{2} \int_{0}^{4} \int_{0}^{2} x^{2} y d y \text { dx } $$

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We wish to approximate the integral
$$
\frac{1}{2}\int_{0}^{4}\int_{0}^{2}x^2y\;dy\,dx
$$
by subdividing the rectangle
$$
R:\; (0,0),\,(4,0),\,(4,2),\,(0,2)
$$
into eight equal squares and using the midpoint method. Then we will evaluate the integral exactly and compare.

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**1. Exact Evaluation**

We start with
$$
I=\frac{1}{2}\int_{0}^{4}\int_{0}^{2}x^2y\;dy\,dx.
$$

*Step 1: Integrate with respect to $$y$$:*

For a fixed $$x$$,
$$
\int_{0}^{2}x^2y\;dy = x^2\int_{0}^{2}y\;dy = x^2\left[\frac{y^2}{2}\right]_{0}^{2} = x^2\left(\frac{4}{2}\right) = 2x^2.
$$

*Step 2: Integrate with respect to $$x$$:*
$$
I=\frac{1}{2}\int_{0}^{4}2x^2\;dx = \frac{1}{2}\cdot 2\int_{0}^{4}x^2\;dx = \int_{0}^{4}x^2\;dx.
$$
$$
\int_{0}^{4}x^2\;dx = \left[\frac{x^3}{3}\right]_{0}^{4} = \frac{4^3}{3} = \frac{64}{3}\approx 21.3.
$$

Thus, the exact value is $$\frac{64}{3}\approx 21.3$$.

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**2. Approximation by the Midpoint Rule**

The rectangle $$R$$ has dimensions $$4$$ (in $$x$$) and $$2$$ (in $$y$$), so its area is $$8$$. We divide it into 8 equal squares, each of area
$$
\Delta A=1.
$$
Because the total area is 8, having eight squares means each square has side length $$1$$. The squares are arranged in 4 columns ($$x$$-direction) and 2 rows ($$y$$-direction).

The centers of these $$1\times1$$ squares have coordinates:
- In the $$x$$-direction: $$0.5,\, 1.5,\, 2.5,\, 3.5$$
- In the $$y$$-direction: $$0.5,\, 1.5$$

Since the integrand is $$\frac{1}{2}x^2y$$, our function is
$$
f(x,y)=\frac{1}{2}x^2y.
$$
For each center, we compute $$f(x_i,y_j)$$ and then sum:
$$
\text{Approximation} \approx \sum f(x_i,y_j) \Delta A.
$$

*For the row at $$y=0.5$$:*  
At $$x=0.5$$,
$$
f(0.5,0.5)=\frac{1}{2}(0.5)^2(0.5)=\frac{1}{2}(0.25)(0.5)=0.0625.
$$
At $$x=1.5$$,
$$
f(1.5,0.5)=\frac{1}{2}(1.5)^2(0.5)=\frac{1}{2}(2.25)(0.5)=0.5625.
$$
At $$x=2.5$$,
$$
f(2.5,0.5)=\frac{1}{2}(2.5)^2(0.5)=\frac{1}{2}(6.25)(0.5)=1.5625.
$$
At $$x=3.5$$,
$$
f(3.5,0.5)=\frac{1}{2}(3.5)^2(0.5)=\frac{1}{2}(12.25)(0.5)=3.0625.
$$
Sum for $$y=0.5$$ row:
$$
S_1=0.0625+0.5625+1.5625+3.0625=5.25.
$$

*For the row at $$y=1.5$$:*
At $$x=0.5$$,
$$
f(0.5,1.5)=\frac{1}{2}(0.5)^2(1.5)=\frac{1}{2}(0.25)(1.5)=0.1875.
$$
At $$x=1.5$$,
$$
f(1.5,1.5)=\frac{1}{2}(1.5)^2(1.5)=\frac{1}{2}(2.25)(1.5)=1.6875.
$$
At $$x=2.5$$,
$$
f(2.5,1.5)=\frac{1}{2}(2.5)^2(1.5)=\frac{1}{2}(6.25)(1.5)=4.6875.
$$
At $$x=3.5$$,
$$
f(3.5,1.5)=\frac{1}{2}(3.5)^2(1.5)=\frac{1}{2}(12.25)(1.5)=9.1875.
$$
Sum for $$y=1.5$$ row:
$$
S_2=0.1875+1.6875+4.6875+9.1875=15.75.
$$

*Total approximate sum:*
$$
S=S_1+S_2=5.25+15.75=21.0.
$$
Since $$\Delta A=1$$ for each square, this is our approximation.

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**3. Comparison**

The exact value is $$\frac{64}{3}\approx 21.3$$ and the midpoint approximation gives $$21.0$$ (both rounded to one decimal place). The two values are very close.
The exact value of the integral is approximately 21.3. Using the midpoint rule with 8 equal squares, the approximation is 21.0.

Acproximate the integral by dividing the rectangle with vertices , and ( 0,2 ) into elaht equal squares and finding the sumn , where is the center of th
th square. Evaluate the iterated integral and compare it with the approximation. (Round your answers to one decimal place.)

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Acproximate the integral by dividing the rectangle with vertices , and ( 0,2 ) into elaht equal squares and finding the sumn , where is the center of th th square. Evaluate the iterated integral and compare it with the approximation. (Round your answers to one decimal place.)