To evaluate the double integral $\int \int y \, dx \, dy$ over the region bounded by the line $y = x$ and the parabola $y = 4x - x^2$, let's first visualize and describe the region of integration.

1. Find the points of intersection between the line and the parabola:

We need to determine where the line $y = x$ intersects the parabola $y = 4x - x^2$. Set the equations equal to each other:

$x = 4x - x^2$

Rearrange the equation:

$x^2 - 3x = 0$

Factor:

$x(x - 3) = 0$

So, the solutions are $x = 0$ and $x = 3$. Therefore, the line and parabola intersect at $x = 0$ and $x = 3$.

2. Set up the limits for the double integral:

The region is bounded by:

• The line $y = x$ (lower boundary) and
• The parabola $y = 4x - x^2$ (upper boundary), for $x$ ranging from 0 to 3.

So, we can write the double integral as:

$\int_{x=0}^{3} \int_{y=x}^{4x - x^2} y \, dy \, dx$

3. Evaluate the inner integral:

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

$\int_{y=x}^{4x - x^2} y \, dy$

The antiderivative of $y$ is $\frac{y^2}{2}$, so:

$\left[\frac{y^2}{2}\right]_{y=x}^{y=4x - x^2} = \frac{(4x - x^2)^2}{2} - \frac{x^2}{2}$

4. Simplify the expression:

Now, expand $(4x - x^2)^2$:

$(4x - x^2)^2 = 16x^2 - 8x^3 + x^4$

Thus, the inner integral becomes:

$\frac{16x^2 - 8x^3 + x^4}{2} - \frac{x^2}{2} = \frac{16x^2 - 8x^3 + x^4 - x^2}{2} = \frac{15x^2 - 8x^3 + x^4}{2}$

5. Evaluate the outer integral:

Now integrate with respect to $x$:

$\int_{0}^{3} \frac{15x^2 - 8x^3 + x^4}{2} \, dx$

This can be split into three separate integrals:

$\frac{1}{2} \left( \int_{0}^{3} 15x^2 \, dx - \int_{0}^{3} 8x^3 \, dx + \int_{0}^{3} x^4 \, dx \right)$

Now evaluate each term:

1. $\int_{0}^{3} 15x^2 \, dx = 15 \cdot \frac{x^3}{3} \bigg|_0^3 = 15 \cdot \frac{27}{3} = 135$
2. $\int_{0}^{3} 8x^3 \, dx = 8 \cdot \frac{x^4}{4} \bigg|_0^3 = 8 \cdot \frac{81}{4} = 162$
3. $\int_{0}^{3} x^4 \, dx = \frac{x^5}{5} \bigg|_0^3 = \frac{243}{5}$

Thus, the integral becomes:

$\frac{1}{2} \left( 135 - 162 + \frac{243}{5} \right) = \frac{1}{2} \left( -27 + \frac{243}{5} \right) = \frac{1}{2} \left( \frac{-135 + 243}{5} \right) = \frac{1}{2} \cdot \frac{108}{5} = \frac{54}{5}$

Final Answer:

The value of the double integral is $\frac{54}{5}$.

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Would you like further details on any of the steps or have additional questions? Here are some related questions you might find interesting:

1. How would the answer change if the bounds were reversed?
2. What would happen if the region was bounded by two parabolas instead of a line and a parabola?
3. How would you handle a triple integral over a similar region?
4. Can you change the order of integration in this problem? If so, how?
5. What is the geometric interpretation of this double integral?

Tip: Always check if the region of integration is bounded and visualize the curves before setting up your integrals. This ensures correct limits and makes the process smoother!