Evaluate the following Integral $$ \iint(4 x+y)^{3} d A, R_{7}(0,2),(0,8) $$.

Question

UpStudy AI · Accepted Answer

To evaluate the double integral

$$
\iint (4x + y)^3 \, dA
$$

over the region $$ R $$ defined by $$ x $$ from 0 to 2 and $$ y $$ from 0 to 8, we proceed as follows:

### Step 1: Set Up the Integral

The region $$ R $$ is a rectangle with:
- $$ x $$ ranging from 0 to 2.
- $$ y $$ ranging from 0 to 8.

Thus, the double integral becomes:

$$
\int_{0}^{2} \int_{0}^{8} (4x + y)^3 \, dy \, dx
$$

### Step 2: Integrate with Respect to $$ y $$

First, integrate the inner integral with respect to $$ y $$:

$$
\int_{0}^{8} (4x + y)^3 \, dy
$$

Let $$ u = 4x + y $$. Then $$ du = dy $$.

$$
\int (4x + y)^3 \, dy = \int u^3 \, du = \frac{u^4}{4} \Big|_{0}^{8} = \frac{(4x + 8)^4 - (4x)^4}{4}
$$

### Step 3: Substitute Back and Integrate with Respect to $$ x $$

Now, the double integral becomes:

$$
\frac{1}{4} \int_{0}^{2} \left[ (4x + 8)^4 - (4x)^4 \right] \, dx
$$

Factor out $$ 4^4 = 256 $$:

$$
\frac{256}{4} \int_{0}^{2} \left[ (x + 2)^4 - x^4 \right] \, dx = 64 \int_{0}^{2} \left[ (x + 2)^4 - x^4 \right] \, dx
$$

### Step 4: Compute the Integral

Evaluate the integral:

$$
\int_{0}^{2} (x + 2)^4 \, dx - \int_{0}^{2} x^4 \, dx
$$

First integral:

$$
\int_{0}^{2} (x + 2)^4 \, dx = \int_{2}^{4} u^4 \, du = \frac{u^5}{5} \Big|_{2}^{4} = \frac{4^5 - 2^5}{5} = \frac{1024 - 32}{5} = \frac{992}{5}
$$

Second integral:

$$
\int_{0}^{2} x^4 \, dx = \frac{x^5}{5} \Big|_{0}^{2} = \frac{32}{5}
$$

Subtract the two results:

$$
\frac{992}{5} - \frac{32}{5} = \frac{960}{5} = 192
$$

### Step 5: Multiply by the Constant

Finally, multiply by 64:

$$
64 \times 192 = 12,288
$$

### Conclusion

The value of the double integral is:

$$
\boxed{12288}
$$
The value of the integral is 12,288.

Evaluate the following Integral \( \iint(4 x+y)^{3} d A, R_{7}(0,2),(0,8) \).