Find the area of the region that is enclosed between y= 4x^2 - x^3 + x and y = x^3 +3x

To find the area of the region enclosed between the curves \( y = 4x^2 - x^3 + x \) and \( y = x^3 + 3x \), we can follow these steps:

### 1. Find the Points of Intersection
First, we need to determine the points where the two curves intersect. This requires setting the two equations equal to each other:

\[
4x^2 - x^3 + x = x^3 + 3x
\]

Simplify this equation by moving all terms to one side:

\[
4x^2 - x^3 + x - x^3 - 3x = 0
\]

\[
-2x^3 + 4x^2 - 2x = 0
\]

Factor out the common factor:

\[
-2x(x^2 - 2x + 1) = 0
\]

Further factor the quadratic expression:

\[
-2x(x - 1)^2 = 0
\]

So, the solutions are \( x = 0 \) and \( x = 1 \). These are the points of intersection between the two curves.

### 2. Set Up the Integral
To find the area of the region enclosed between the curves, we integrate the difference between the functions from \( x = 0 \) to \( x = 1 \):

\[
A = \int_{0}^{1} \left[(4x^2 - x^3 + x) - (x^3 + 3x)\right] dx
\]

Simplify the integrand:

\[
A = \int_{0}^{1} \left[4x^2 - x^3 + x - x^3 - 3x\right] dx
\]

\[
A = \int_{0}^{1} \left[4x^2 - 2x^3 - 2x\right] dx
\]

### 3. Evaluate the Integral
Now, integrate each term separately:

\[
A = \int_{0}^{1} \left(4x^2\right) dx - \int_{0}^{1} \left(2x^3\right) dx - \int_{0}^{1} \left(2x\right) dx
\]

\[
A = \left[\frac{4x^3}{3}\right]_{0}^{1} - \left[\frac{2x^4}{4}\right]_{0}^{1} - \left[x^2\right]_{0}^{1}
\]

Substituting the limits:

\[
A = \left[\frac{4(1)^3}{3} - \frac{4(0)^3}{3}\right] - \left[\frac{2(1)^4}{4} - \frac{2(0)^4}{4}\right] - \left[1^2 - 0^2\right]
\]

\[
A = \left(\frac{4}{3}\right) - \left(\frac{1}{2}\right) - 1
\]

\[
A = \frac{4}{3} - \frac{1}{2} - \frac{3}{3}
\]

To combine these fractions, convert them to have a common denominator:

\[
A = \frac{8}{6} - \frac{3}{6} - \frac{6}{6} = \frac{8 - 3 - 6}{6} = \frac{-1}{6}
\]

Since area cannot be negative, we take the absolute value:

\[
A = \frac{1}{6}
\]

### Final Answer:
The area of the region enclosed by the curves is \( \boxed{\frac{1}{6}} \).

---

Would you like more details on any part of this process, or do you have any questions?

### Related Questions:
1. How do you find the points of intersection for more complex functions?
2. What is the geometric interpretation of the integral when finding areas between curves?
3. Can you use symmetry to simplify the integration process in some cases?
4. What happens if the curves cross each other more than twice?
5. How do you handle finding areas between curves when the functions are piecewise or discontinuous?

### Tip:
Always check the points of intersection carefully. Misidentifying these points can lead to incorrect bounds for your integrals and therefore incorrect areas.

Learn how to calculate the area of the region enclosed between the curves y = 4x^2 - x^3 + x and y = x^3 + 3x using calculus. Detailed step-by-step solution provided. Suitable for college-level students and calculus enthusiasts.

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