Find the area of the region between y=7x^2 and y=4+x^2

To find the area of the region between the two curves \( y = 7x^2 \) and \( y = 4 + x^2 \), we need to compute the integral of the difference between the two functions over the interval where they intersect.

### Step 1: Find the points of intersection
To find where the curves intersect, we set the two functions equal to each other:

\[
7x^2 = 4 + x^2
\]

Solve for \(x\):

\[
7x^2 - x^2 = 4
\]

\[
6x^2 = 4
\]

\[
x^2 = \frac{4}{6} = \frac{2}{3}
\]

\[
x = \pm \sqrt{\frac{2}{3}}
\]

So, the curves intersect at \( x = -\sqrt{\frac{2}{3}} \) and \( x = \sqrt{\frac{2}{3}} \).

### Step 2: Set up the integral
The area between the two curves is given by the integral of the difference between the upper and lower functions over the interval from \( -\sqrt{\frac{2}{3}} \) to \( \sqrt{\frac{2}{3}} \). The upper curve is \( y = 4 + x^2 \), and the lower curve is \( y = 7x^2 \).

\[
\text{Area} = \int_{-\sqrt{\frac{2}{3}}}^{\sqrt{\frac{2}{3}}} \left( (4 + x^2) - (7x^2) \right) \, dx
\]

Simplify the integrand:

\[
\text{Area} = \int_{-\sqrt{\frac{2}{3}}}^{\sqrt{\frac{2}{3}}} \left( 4 - 6x^2 \right) \, dx
\]

### Step 3: Evaluate the integral
We can now evaluate the integral term by term:

\[
\int_{-\sqrt{\frac{2}{3}}}^{\sqrt{\frac{2}{3}}} 4 \, dx - \int_{-\sqrt{\frac{2}{3}}}^{\sqrt{\frac{2}{3}}} 6x^2 \, dx
\]

Since the integrand is symmetric about \(x = 0\), we can double the integral from 0 to \( \sqrt{\frac{2}{3}} \):

\[
\text{Area} = 2 \left( \int_0^{\sqrt{\frac{2}{3}}} 4 \, dx - \int_0^{\sqrt{\frac{2}{3}}} 6x^2 \, dx \right)
\]

Now evaluate each integral:
1. \( \int_0^{\sqrt{\frac{2}{3}}} 4 \, dx = 4x \Big|_0^{\sqrt{\frac{2}{3}}} = 4\sqrt{\frac{2}{3}} \)
2. \( \int_0^{\sqrt{\frac{2}{3}}} 6x^2 \, dx = 6 \cdot \frac{x^3}{3} \Big|_0^{\sqrt{\frac{2}{3}}} = 2 \left( \frac{2}{3} \right)^{3/2} = \frac{4}{3} \sqrt{\frac{2}{3}} \)

Thus, the area becomes:

\[
\text{Area} = 2 \left( 4\sqrt{\frac{2}{3}} - \frac{4}{3} \sqrt{\frac{2}{3}} \right)
\]

Factor out \( \sqrt{\frac{2}{3}} \):

\[
\text{Area} = 2 \cdot \sqrt{\frac{2}{3}} \cdot \left( 4 - \frac{4}{3} \right)
\]

Simplify the expression inside the parentheses:

\[
\text{Area} = 2 \cdot \sqrt{\frac{2}{3}} \cdot \frac{8}{3} = \frac{16}{3} \cdot \sqrt{\frac{2}{3}}
\]

Thus, the area of the region between the curves is:

\[
\boxed{\frac{16}{3} \cdot \sqrt{\frac{2}{3}}}
\]

Do you want further details on this? Here are five related questions to consider:

1. How can we verify the points of intersection between two curves?
2. What is the significance of symmetric functions in definite integrals?
3. How do we handle regions where one curve is above another over an interval?
4. What happens if the two curves intersect at multiple points?
5. Can this approach be extended to find the area between more complex functions?

**Tip:** Always sketch the curves to better understand the region you're working with!

Discover how to find the area of the region between the curves y=7x^2 and y=4+x^2 using integral calculus. This detailed solution walks through finding the points of intersection, setting up the integral, and evaluating the area. Suitable for Grades 11-12 or early college students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation