Question

The value of the double integral \( \iint_R e^{x^2} dxdy \), where R is a region given by \( 2y \le x \le 2 \) and \( 0 \le y \le 1 \), is:

• A. \( (e^4-1) \)
• B. \( \frac{1}{4}(e^4-1) \)
• C. \( \frac{1}{4}(e^4+1) \)
• D. \( \frac{1}{2}(e^4-1) \)

Hint:

If you encounter a double integral where the inner integral is impossible to compute with elementary functions (like \( \int e^{x^2} dx \) or \( \int \frac{\sin x}{x} dx \)), your first thought should be to try changing the order of integration. Sketching the region of integration is an essential first step for this process.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires evaluating a double integral over a specified region R. The integrand, \( e^{x^2} \), does not have an elementary antiderivative with respect to \(x\). This suggests that we must change the order of integration.

Step 2: Key Formula or Approach:
The region of integration R is given by \( 2y \le x \le 2 \) and \( 0 \le y \le 1 \). This is a Type II region (integrating x first). To change the order of integration, we need to re-describe this region as a Type I region (integrating y first). The boundaries are the lines \( x=2y \) (or \( y=x/2 \)), \( x=2 \), \( y=0 \), and \( y=1 \). Sketching the region: It's a triangle with vertices at (0,0), (2,1), and (2,0). To describe this as a Type I region, we let \(x\) vary from its minimum to maximum value, and for each \(x\), \(y\) varies from a lower curve to an upper curve. - The minimum \(x\) value is 0. - The maximum \(x\) value is 2. - The lower bound for \(y\) is the x-axis, \(y=0\). - The upper bound for \(y\) is the line \( y=x/2 \). So the new limits are \( 0 \le x \le 2 \) and \( 0 \le y \le x/2 \).

Step 3: Detailed Explanation:
The integral with the order changed is: \[ \int_{0}^{2} \int_{0}^{x/2} e^{x^2} dy \, dx \] First, integrate with respect to \(y\): \[ \int_{0}^{x/2} e^{x^2} dy = e^{x^2} [y]_{0}^{x/2} = e^{x^2} \left(\frac{x}{2} - 0\right) = \frac{x}{2} e^{x^2} \] Now, integrate the result with respect to \(x\): \[ \int_{0}^{2} \frac{x}{2} e^{x^2} dx = \frac{1}{2} \int_{0}^{2} x e^{x^2} dx \] To solve this, use a u-substitution. Let \( u = x^2 \), then \( du = 2x dx \), or \( x dx = \frac{1}{2} du \). Change the limits of integration: - When \( x=0 \), \( u=0^2=0 \). - When \( x=2 \), \( u=2^2=4 \). The integral becomes: \[ \frac{1}{2} \int_{0}^{4} e^u \left(\frac{1}{2} du\right) = \frac{1}{4} \int_{0}^{4} e^u du \] \[ = \frac{1}{4} [e^u]_{0}^{4} = \frac{1}{4} (e^4 - e^0) = \frac{1}{4}(e^4 - 1) \]
Step 4: Final Answer:
The value of the double integral is \( \frac{1}{4}(e^4 - 1) \).