find the area between y=x^2, y=8-2x, y=2x/3+16/3 and the intersections are -2,4  1,6  2,4 find the area from -2 to 2 and up to 6

To find the area between the curves \( y = x^2 \), \( y = 8 - 2x \), and \( y = \frac{2x}{3} + \frac{16}{3} \) within the limits \( x = -2 \) to \( x = 2 \) and up to \( y = 6 \), we will proceed step by step. Here's the approach:

### 1. **Sketch the Graph and Analyze the Region**:
The curves intersect at the given points: \( (-2, 4) \), \( (1, 6) \), and \( (2, 4) \). This indicates the boundaries of the region enclosed by these curves.

### 2. **Determine the Region of Interest**:
We are asked to find the area from \( x = -2 \) to \( x = 2 \), with the upper limit being \( y = 6 \). Let's now break this down by determining which function bounds the area at each section of \( x \)-values.

### 3. **Find the Areas Under Each Curve**:
We'll calculate the areas between the curves using definite integrals.

#### Between \( y = x^2 \) and \( y = 8 - 2x \):
For \( x = -2 \) to \( x = 1 \), the top function is \( y = 8 - 2x \) and the bottom function is \( y = x^2 \). The area between these curves is given by:

\[
\text{Area 1} = \int_{-2}^{1} \left( (8 - 2x) - x^2 \right) dx
\]

#### Between \( y = \frac{2x}{3} + \frac{16}{3} \) and \( y = x^2 \):
For \( x = 1 \) to \( x = 2 \), the top function is \( y = \frac{2x}{3} + \frac{16}{3} \) and the bottom function is \( y = x^2 \). The area between these curves is given by:

\[
\text{Area 2} = \int_{1}^{2} \left( \left( \frac{2x}{3} + \frac{16}{3} \right) - x^2 \right) dx
\]

### 4. **Calculate the Areas**:

Let me compute these integrals.The total area between the curves \( y = x^2 \), \( y = 8 - 2x \), and \( y = \frac{2x}{3} + \frac{16}{3} \) from \( x = -2 \) to \( x = 2 \) and up to \( y = 6 \) is 28 square units.

This is the sum of:
- Area from \( x = -2 \) to \( x = 1 \) between \( y = 8 - 2x \) and \( y = x^2 \): 24 square units.
- Area from \( x = 1 \) to \( x = 2 \) between \( y = \frac{2x}{3} + \frac{16}{3} \) and \( y = x^2 \): 4 square units.

Would you like further details or have any questions?

Here are 5 related questions:
1. How do we set up integrals for regions bounded by curves in 3D?
2. What happens when the curves intersect at multiple points within the interval?
3. How can we find the volume under a surface defined by these curves?
4. How does the choice of intervals affect the integral setup?
5. How does symmetry simplify the calculation of area between curves?

**Tip:** Always identify which curve is the upper and lower boundary before setting up integrals to ensure correct subtraction.

Learn how to find the area between the curves y=x^2, y=8-2x, and y=2x/3+16/3, including intersections at key points like (-2, 4), (1, 6), and (2, 4). This step-by-step solution uses definite integrals to compute areas between the curves from x=-2 to x=2. Suitable for students in Grades 11-12.

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