Find the area of the region bounded by the curves y = (9/10)x^2 and y = x with vertical likes at x=-7 and x=7

To find the area of the region bounded by the curves \( y = \frac{9}{10}x^2 \) and \( y = x \), between the vertical lines \( x = -7 \) and \( x = 7 \), we need to compute the integral of the difference between these two functions over the interval \([-7, 7]\).

### Step 1: Determine the points of intersection
We first need to find the points where the curves intersect. Set the two equations equal:

\[
\frac{9}{10}x^2 = x
\]

This simplifies to:

\[
\frac{9}{10}x^2 - x = 0
\]

Factor the equation:

\[
x\left(\frac{9}{10}x - 1\right) = 0
\]

This gives us:

\[
x = 0 \quad \text{or} \quad \frac{9}{10}x = 1 \quad \Rightarrow \quad x = \frac{10}{9}
\]

So, the curves intersect at \( x = 0 \) and \( x = \frac{10}{9} \).

### Step 2: Set up the integral
The area \( A \) is given by the integral of the difference between the two functions over the interval \([-7, 7]\). However, since the curves intersect, we have to consider the intervals separately:
- From \( x = -7 \) to \( x = 0 \), the line \( y = x \) lies above the parabola \( y = \frac{9}{10}x^2 \).
- From \( x = 0 \) to \( x = 7 \), the line \( y = x \) still lies above the parabola \( y = \frac{9}{10}x^2 \).

Thus, the area is:

\[
A = \int_{-7}^{7} \left( x - \frac{9}{10}x^2 \right) \, dx
\]

### Step 3: Compute the integral
Let's compute the integral:

\[
A = \int_{-7}^{7} \left( x - \frac{9}{10}x^2 \right) \, dx
\]

We can separate the integral:

\[
A = \int_{-7}^{7} x \, dx - \frac{9}{10} \int_{-7}^{7} x^2 \, dx
\]

#### Integral of \( x \):

\[
\int x \, dx = \frac{x^2}{2} \quad \text{evaluated from} \quad -7 \text{ to } 7
\]

\[
\left[ \frac{x^2}{2} \right]_{-7}^{7} = \frac{7^2}{2} - \frac{(-7)^2}{2} = \frac{49}{2} - \frac{49}{2} = 0
\]

So, the integral of \( x \) over this interval is 0.

#### Integral of \( x^2 \):

\[
\int x^2 \, dx = \frac{x^3}{3} \quad \text{evaluated from} \quad -7 \text{ to } 7
\]

\[
\left[ \frac{x^3}{3} \right]_{-7}^{7} = \frac{7^3}{3} - \frac{(-7)^3}{3} = \frac{343}{3} - \frac{-343}{3} = \frac{686}{3}
\]

So:

\[
\frac{9}{10} \int_{-7}^{7} x^2 \, dx = \frac{9}{10} \times \frac{686}{3} = \frac{6174}{30} = 205.8
\]

### Final Area:
Since the integral of \( x \) is 0, the area is:

\[
A = 205.8 \text{ square units}
\]

This is the area of the region bounded by the curves and the vertical lines.

Would you like further details or have any questions?

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Here are five related questions to expand your understanding:
1. How would the area change if the interval was different, say from \([-5, 5]\)?
2. What if the curves were \( y = \frac{9}{10}x^2 \) and \( y = 2x \)? How would the area calculation change?
3. Can you find the area using a geometric approach instead of integration?
4. How does the presence of symmetry in the problem simplify the calculations?
5. What are some other methods to solve such area problems?

**Tip:** Always check for symmetry in the curves and intervals, as it can simplify the integration process significantly.

Learn how to calculate the area of the region bounded by the curves y = (9/10)x^2 and y = x, between vertical lines at x = -7 and x = 7 using integral calculus. This detailed solution covers finding points of intersection, setting up and computing the integral, and understanding the geometric interpretation of the problem.

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