A. 0 Evaluate the integral by reversing the order of integration. OB. 0(e-6-1) O C. -2(e-³-1) D. 12(e-³-1) IT' 126-1² y dx dy

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**Exercise: Reverse the Order of Integration**

Evaluate the integral by reversing the order of integration.

\[ \int_{0}^{1} \int_{y}^{1} 12e^{-3x^2} \, dx \, dy \]

Options:

- **A.** \( 0 \)

- **B.** \( 0(e^{-6} - 1) \)

- **C.** \( -2(e^{-3} - 1) \)

- **D.** \( 12(e^{-3} - 1) \)

- **E.** \( (e^{-6} - 1) \) 

---

**Solution Explanation:**

The initial integral is set up with the limits of integration such that for each fixed \( y \), \( x \) ranges from \( y \) to 1. To reverse the order of integration, visualize or sketch the region of integration in the xy-plane:

1. **Region of Integration:**
   - The original limits describe a region where \( 0 \leq y \leq 1 \) and \( y \leq x \leq 1 \).

   - When reversing the order, the region should cover \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq x \).

2. **Reversed Integral:**
   - Rewriting the integral with these new limits becomes:
   \[ \int_{0}^{1} \int_{0}^{x} 12e^{-3x^2} \, dy \, dx \]

3. **Integration Steps:**
   - First, integrate the inner integral with respect to \( y \), leading to:
   \[ \int_{0}^{x} 12e^{-3x^2} \, dy = [12e^{-3x^2} \cdot y]_{0}^{x} = 12xe^{-3x^2} \]

   - Then, perform the outer integration with respect to \( x \):
   \[ \int_{0}^{1} 12xe^{-3x^2} \, dx \]

   - Use substitution to integrate:
     - Let \( u = -3x^2 \), then \( du = -6x \, dx \), so \( -\frac{1}{6} \, du = x \, dx
Transcribed Image Text:**Exercise: Reverse the Order of Integration** Evaluate the integral by reversing the order of integration. \[ \int_{0}^{1} \int_{y}^{1} 12e^{-3x^2} \, dx \, dy \] Options: - **A.** \( 0 \) - **B.** \( 0(e^{-6} - 1) \) - **C.** \( -2(e^{-3} - 1) \) - **D.** \( 12(e^{-3} - 1) \) - **E.** \( (e^{-6} - 1) \) --- **Solution Explanation:** The initial integral is set up with the limits of integration such that for each fixed \( y \), \( x \) ranges from \( y \) to 1. To reverse the order of integration, visualize or sketch the region of integration in the xy-plane: 1. **Region of Integration:** - The original limits describe a region where \( 0 \leq y \leq 1 \) and \( y \leq x \leq 1 \). - When reversing the order, the region should cover \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq x \). 2. **Reversed Integral:** - Rewriting the integral with these new limits becomes: \[ \int_{0}^{1} \int_{0}^{x} 12e^{-3x^2} \, dy \, dx \] 3. **Integration Steps:** - First, integrate the inner integral with respect to \( y \), leading to: \[ \int_{0}^{x} 12e^{-3x^2} \, dy = [12e^{-3x^2} \cdot y]_{0}^{x} = 12xe^{-3x^2} \] - Then, perform the outer integration with respect to \( x \): \[ \int_{0}^{1} 12xe^{-3x^2} \, dx \] - Use substitution to integrate: - Let \( u = -3x^2 \), then \( du = -6x \, dx \), so \( -\frac{1}{6} \, du = x \, dx
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