Graph the equations y = x2,y = 4and shade the region in the first quadrant bounded by the two equations. Label all important points on the graph. а. b. Set up the integral(s) that represent the area shaded in part a, with respect to x. С. Set up the integral(s) that represent the area shaded in part a, with respect to y.

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### Calculus Problem: Area Between Curves

1. **Task Overview:**

   a. Graph the equations \( y = x^2 \) and \( y = 4 \). Shade the region in the first quadrant bounded by these two equations. Label all important points on the graph.

   b. Set up the integral(s) that represent the area shaded in part a, with respect to \( x \).

   c. Set up the integral(s) that represent the area shaded in part a, with respect to \( y \).

   d. Integrate your integrals from parts b and c and show that the two integrals give the same area.

### Graph Explanation:

- **Equation 1**: \( y = x^2 \)
  - This is a parabolic curve opening upwards, intersecting the x-axis at the origin \((0,0)\).
  
- **Equation 2**: \( y = 4 \)
  - This is a horizontal line intersecting the y-axis at \( y = 4 \).
  
- **Region of Interest**: 
  - The shaded region is the area in the first quadrant bounded above by the line \( y = 4 \) and below by the parabola \( y = x^2 \).
  - Important points: The intersection of \( y = x^2 \) and \( y = 4 \) occurs where \( x^2 = 4 \), giving \( x = 2 \) in the first quadrant \((2,4)\).

### Setting Up Integrals:

b. **Integrals with Respect to \( x \):**

- Lower Limit: \( x = 0 \)
- Upper Limit: \( x = 2 \)
- Integral: \( \int_0^2 (4 - x^2) \, dx \)

c. **Integrals with Respect to \( y \):**

- Lower Limit: \( y = 0 \)
- Upper Limit: \( y = 4 \)
- The inverse of \( y = x^2 \) is \( x = \sqrt{y} \).
- Integral: \( \int_0^4 (2 - \sqrt{y}) \, dy \)

### Solving the Integrals:

d. **Compute both integrals and confirm they yield the same area**:

1. \[
   \int_0^2 (4 - x^2
Transcribed Image Text:### Calculus Problem: Area Between Curves 1. **Task Overview:** a. Graph the equations \( y = x^2 \) and \( y = 4 \). Shade the region in the first quadrant bounded by these two equations. Label all important points on the graph. b. Set up the integral(s) that represent the area shaded in part a, with respect to \( x \). c. Set up the integral(s) that represent the area shaded in part a, with respect to \( y \). d. Integrate your integrals from parts b and c and show that the two integrals give the same area. ### Graph Explanation: - **Equation 1**: \( y = x^2 \) - This is a parabolic curve opening upwards, intersecting the x-axis at the origin \((0,0)\). - **Equation 2**: \( y = 4 \) - This is a horizontal line intersecting the y-axis at \( y = 4 \). - **Region of Interest**: - The shaded region is the area in the first quadrant bounded above by the line \( y = 4 \) and below by the parabola \( y = x^2 \). - Important points: The intersection of \( y = x^2 \) and \( y = 4 \) occurs where \( x^2 = 4 \), giving \( x = 2 \) in the first quadrant \((2,4)\). ### Setting Up Integrals: b. **Integrals with Respect to \( x \):** - Lower Limit: \( x = 0 \) - Upper Limit: \( x = 2 \) - Integral: \( \int_0^2 (4 - x^2) \, dx \) c. **Integrals with Respect to \( y \):** - Lower Limit: \( y = 0 \) - Upper Limit: \( y = 4 \) - The inverse of \( y = x^2 \) is \( x = \sqrt{y} \). - Integral: \( \int_0^4 (2 - \sqrt{y}) \, dy \) ### Solving the Integrals: d. **Compute both integrals and confirm they yield the same area**: 1. \[ \int_0^2 (4 - x^2
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