Approximate the integral $$ \int_{R} \int f(x, y) d A $$ by dividing the rectangle $$ R $$ with vertices $$ (0 $$, of the th square. Evaluate the iterated integral and compare it with the approximation. $$ \frac{1}{5} \int_{0}^{4} \int_{0}^{2} x^{2} y d y d x $$

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We want to evaluate the iterated integral

$$
\frac{1}{5} \int_{0}^{4} \int_{0}^{2} x^{2} y \, dy \, dx
$$

and then compare this exact value with an approximation of the corresponding double integral over the rectangle $$ R $$ (which, from context, is taken as $$ [0,4] \times [0,2] $$). We proceed step by step.

### Step 1. Evaluate the Inner Integral

Consider the inner integral

$$
\int_{0}^{2} x^{2} y \, dy.
$$

Since $$ x^{2} $$ is treated as a constant with respect to $$ y $$, we have

$$
\int_{0}^{2} x^{2} y \, dy = x^{2} \int_{0}^{2} y \, dy.
$$

Evaluating the $$ y $$-integral,

$$
\int_{0}^{2} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} = 2.
$$

Thus, the inner integral becomes

$$
\int_{0}^{2} x^{2} y \, dy = 2x^{2}.
$$

### Step 2. Evaluate the Outer Integral

Now substitute the result of the inner integral back into the iterated integral:

$$
\frac{1}{5} \int_{0}^{4} 2x^{2} \, dx.
$$

Simplify the constant factor:

$$
\frac{1}{5} \cdot 2 \int_{0}^{4} x^{2} \, dx = \frac{2}{5} \int_{0}^{4} x^{2} \, dx.
$$

Evaluate the outer integral:

$$
\int_{0}^{4} x^{2} \, dx = \left[ \frac{x^3}{3} \right]_{0}^{4} = \frac{4^3}{3} - 0 = \frac{64}{3}.
$$

Thus, we have

$$
\frac{2}{5} \int_{0}^{4} x^{2} \, dx = \frac{2}{5} \cdot \frac{64}{3} = \frac{128}{15}.
$$

### Step 3. Comparison with the Approximation

If we approximate the double integral

$$
\iint_R f(x,y) \, dA
$$

by dividing the rectangle $$ R = [0,4] \times [0,2] $$ into several subrectangles and using an appropriate sample point (for example, the center of each subrectangle) the approximation process might involve summing the contributions $$ f(x_i^*, y_j^*) \Delta A $$ over the divided regions. In our case the approximation is given as

$$
\frac{1}{5} \int_{0}^{4} \int_{0}^{2} x^{2} y \, dy \, dx,
$$

which we have just evaluated exactly. Thus the exact value of the iterated (and hence the double) integral is

$$
\frac{128}{15} \approx 8.5333.
$$

This shows that the approximation procedure (when combined with a factor of $$ \frac{1}{5} $$ as described) correctly yields the value $$ \frac{128}{15} $$ for the integral.

### Final Answer

$$
\frac{128}{15}
$$
The exact value of the integral is $$ \frac{128}{15} $$, which is approximately 8.5333.

Approximate the integral by dividing the rectangle with vertices ,
of the th square. Evaluate the iterated integral and compare it with the approximation.

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Approximate the integral by dividing the rectangle with vertices , of the th square. Evaluate the iterated integral and compare it with the approximation.