Integral from 0 to 2 Integral from y squared to 4 y minus y squared 100 left parenthesis y plus 4 right parenthesis dx dy
Part 2
Evaluate the integral in the previous step.

We are tasked with evaluating the double integral:

\[
\int_0^2 \int_{y^2}^4 \left( y - y^2 \right) 100 \left( y + 4 \right) \, dx \, dy
\]

### Step 1: Deal with the inner integral

The inner integral is with respect to \( x \), where the limits of integration are from \( x = y^2 \) to \( x = 4 \). Since the expression \( \left( y - y^2 \right) 100 \left( y + 4 \right) \) does not contain \( x \), it acts as a constant with respect to \( x \). Therefore, we can pull this out of the inner integral:

\[
\int_{y^2}^4 \left( y - y^2 \right) 100 \left( y + 4 \right) \, dx = \left( y - y^2 \right) 100 \left( y + 4 \right) \cdot \left( 4 - y^2 \right)
\]

### Step 2: Expand the expression

Now, expand the expression \( \left( y - y^2 \right) 100 \left( y + 4 \right) \left( 4 - y^2 \right) \).

First, expand \( \left( y - y^2 \right) \left( y + 4 \right) \):

\[
\left( y - y^2 \right) \left( y + 4 \right) = y^2 + 4y - y^3 - 4y^2 = -y^3 + y^2 + 4y
\]

Now multiply this by \( 100 \left( 4 - y^2 \right) \):

\[
100 \left( -y^3 + y^2 + 4y \right) \left( 4 - y^2 \right)
\]

Distribute the terms:

\[
= 100 \left[ (-y^3)(4 - y^2) + (y^2)(4 - y^2) + (4y)(4 - y^2) \right]
\]

Simplify each product:

\[
(-y^3)(4 - y^2) = -4y^3 + y^5
\]
\[
(y^2)(4 - y^2) = 4y^2 - y^4
\]
\[
(4y)(4 - y^2) = 16y - 4y^3
\]

Now, combine all the terms:

\[
= 100 \left( -4y^3 + y^5 + 4y^2 - y^4 + 16y - 4y^3 \right)
\]

Simplify:

\[
= 100 \left( y^5 - y^4 - 8y^3 + 4y^2 + 16y \right)
\]

### Step 3: Outer integral

Now we compute the outer integral:

\[
\int_0^2 \left( y^5 - y^4 - 8y^3 + 4y^2 + 16y \right) \, dy
\]

Compute each term separately:

\[
\int_0^2 y^5 \, dy = \left[ \frac{y^6}{6} \right]_0^2 = \frac{64}{6} = \frac{32}{3}
\]
\[
\int_0^2 y^4 \, dy = \left[ \frac{y^5}{5} \right]_0^2 = \frac{32}{5}
\]
\[
\int_0^2 8y^3 \, dy = 8 \left[ \frac{y^4}{4} \right]_0^2 = 8 \times 4 = 32
\]
\[
\int_0^2 4y^2 \, dy = 4 \left[ \frac{y^3}{3} \right]_0^2 = 4 \times \frac{8}{3} = \frac{32}{3}
\]
\[
\int_0^2 16y \, dy = 16 \left[ \frac{y^2}{2} \right]_0^2 = 16 \times 2 = 32
\]

### Step 4: Combine the results

Now combine the results of the outer integrals:

\[
100 \left( \frac{32}{3} - \frac{32}{5} - 32 + \frac{32}{3} + 32 \right)
\]

First, simplify the terms inside the parentheses:

\[
= 100 \left( 2 \times \frac{32}{3} - \frac{32}{5} \right)
= 100 \left( \frac{64}{3} - \frac{32}{5} \right)
\]

Find a common denominator:

\[
= 100 \left( \frac{320}{15} - \frac{96}{15} \right) = 100 \times \frac{224}{15}
\]

Finally:

\[
= \frac{22400}{15} = 1493.\overline{3}
\]

Thus, the value of the double integral is approximately:

\[
\boxed{1493.\overline{3}}
\]

Would you like further details or have any questions?

### 5 Related Questions:
1. How do you handle double integrals with variable limits?
2. What happens if we swap the order of integration in this type of problem?
3. Can this problem be solved using polar coordinates?
4. How would adding more terms in the integrand affect the solution?
5. What is the significance of each region in a double integral?

### Tip:
For double integrals, it’s often helpful to simplify the inner integral first by considering the variables independently before handling the outer integral.

Integral from 0 to 2 Integral from y squared to 4 y minus y squared 100 (y plus 4) dx dy. Evaluate the integral in the previous step.

Explore the step-by-step solution of a double integral from 0 to 2 and from y^2 to 4 for the function (y - y^2) 100 (y + 4). This example demonstrates the use of polynomial expansion, integral evaluation, and simplification techniques. Ideal for college-level calculus students.

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