Find the area between two curves which are the graphs of functions in either the x or y direction.
Sample problems:
• f(x) = 6/x^2+1 , g(x) = 3x^2

To find the area between two curves, \( f(x) \) and \( g(x) \), you typically need to follow these steps:

1. **Identify the region of intersection**: You need to find the points where the two curves intersect. These points give the boundaries of integration.
  
2. **Set up the integral**: Depending on the orientation of the functions, you subtract one function from the other (whichever is on top in the region) and integrate over the interval.

### Sample Problem:
Given:
- \( f(x) = \frac{6}{x^2+1} \)
- \( g(x) = 3x^2 \)

#### Step 1: Find the points of intersection
Set \( f(x) = g(x) \) to find where the two curves intersect:
\[
\frac{6}{x^2+1} = 3x^2
\]

Multiply both sides by \( x^2 + 1 \) to eliminate the denominator:
\[
6 = 3x^2(x^2 + 1)
\]
\[
6 = 3x^4 + 3x^2
\]
Divide by 3:
\[
2 = x^4 + x^2
\]
Let \( y = x^2 \), then:
\[
y^2 + y - 2 = 0
\]
Factor the quadratic equation:
\[
(y - 1)(y + 2) = 0
\]
So, \( y = 1 \) or \( y = -2 \).

Since \( y = x^2 \), we discard the negative value and have:
\[
x^2 = 1 \implies x = \pm 1
\]

Thus, the points of intersection are at \( x = -1 \) and \( x = 1 \).

#### Step 2: Set up the integral
The area between the curves from \( x = -1 \) to \( x = 1 \) is:
\[
A = \int_{-1}^{1} \left( \frac{6}{x^2 + 1} - 3x^2 \right) dx
\]

#### Step 3: Compute the integral
Now, let's evaluate the integral:

\[
A = \int_{-1}^{1} \frac{6}{x^2 + 1} \, dx - \int_{-1}^{1} 3x^2 \, dx
\]

##### First integral:
\[
\int_{-1}^{1} \frac{6}{x^2 + 1} \, dx = 6 \cdot \left[ \tan^{-1}(x) \right]_{-1}^{1} = 6 \cdot \left( \frac{\pi}{4} - \left( -\frac{\pi}{4} \right) \right) = 6 \cdot \frac{\pi}{2} = 3\pi
\]

##### Second integral:
\[
\int_{-1}^{1} 3x^2 \, dx = 3 \cdot \left[ \frac{x^3}{3} \right]_{-1}^{1} = 3 \cdot (0 - 0) = 0
\]

#### Step 4: Final result
Thus, the area is:
\[
A = 3\pi - 0 = 3\pi
\]

### Final Answer:
The area between the curves is \( 3\pi \).

---

Would you like more details on any step?

### Related Questions:
1. How do you handle cases where the curves intersect multiple times?
2. How do you find the area between curves when functions are given in terms of \( y \)?
3. How does the symmetry of a region affect the computation of areas?
4. How do you compute the area when the curves are defined by parametric equations?
5. How would the approach change if the curves are piecewise-defined?

### Tip:
When the curves have symmetry, you can often simplify the integral by calculating the area for one part and doubling it!

Find the area between two curves which are the graphs of functions in either the x or y direction.
Sample problem: f(x) = 6/(x^2 + 1), g(x) = 3x^2

Learn how to find the area between the curves f(x) = 6/(x^2 + 1) and g(x) = 3x^2 with a step-by-step solution. This problem demonstrates how to calculate the intersection points and evaluate the definite integral, using key calculus concepts like integration and arctangent functions. Suitable for Grades 11-12 and college-level students.

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