Let's tackle the problem **step-by-step** as requested: --- ## Step 1: Change the Order of Integration (Part b) ### **Original Integral** \[ I = \int_{x=}^2 \int_{y=x^2}^{2x} e^{-x^2} \, dy \, dx \] #### **Explanation:** - The outer integral is over \( x \) from to 2. - For each \( x \), \( y \) goes from \( x^2 \) to \( 2x \). ### **Describe the Region \( R \)** - Lower bound for \( y \): \( y = x^2 \) - Upper bound for \( y \): \( y = 2x \) - \( x \) goes from to 2. To **switch the order** (to \( dx\,dy \)), we need to express \( x \) in terms of \( y \): - \( y = x^2 \implies x = \sqrt{y} \) - \( y = 2x \implies x = \frac{y}{2} \) For a fixed \( y \), \( x \) runs from \( \sqrt{y} \) to \( \frac{y}{2} \). But, \( \sqrt{y} \leq \frac{y}{2} \) only for \( \leq y \leq 4 \). ### **Switched Integral** \[ I = \int_{y=}^{4} \int_{x=\sqrt{y}}^{x=y/2} e^{-x^2} \, dx\,dy \] **Explanation:** Now, for each \( y \in [, 4] \), \( x \) runs from \( \sqrt{y} \) to \( \frac{y}{2} \), matching the description of the region. --- ## Step 2: Evaluate the Switched Integral (Part c) \[ I = \int_{y=}^{4} \left( \int_{x=\sqrt{y}}^{x=y/2} e^{-x^2} \, dx \right) dy \] ### **Evaluate the Inner Integral** Let’s use the hint: use substitution \( u = x^2 \), so \( du = 2x dx \). But, since \( e^{-x^2} \) does not have an elementary antiderivative, we express the result in terms of the error function, \( \operatorname{erf}(x) \): \[ \int e^{-x^2} dx = \frac{\sqrt{\pi}}{2} \operatorname{erf}(x) \] So, \[ \int_{x=\sqrt{y}}^{x=y/2} e^{-x^2} dx = \left[ \frac{\sqrt{\pi}}{2} \operatorname{erf}(x) \right]_{x=\sqrt{y}}^{x=y/2} = \frac{\sqrt{\pi}}{2} \left( \operatorname{erf}\left(\frac{y}{2}\right) - \operatorname{erf}\left(\sqrt{y}\right) \right) \] Therefore, \[ I = \int_{y=}^{4} \frac{\sqrt{\pi}}{2} \left( \operatorname{erf}\left(\frac{y}{2}\right) - \operatorname{erf}\left(\sqrt{y}\right) \right) dy \] **Explanation:** After switching the order, the inner integral with respect to \( x \) is elementary using the error function. --- ## **Final Answer** \[ \boxed{ I = \int_{y=}^{4} \frac{\sqrt{\pi}}{2} \left( \operatorname{erf}\left(\frac{y}{2}\right) - \operatorname{erf}\left(\sqrt{y}\right) \right) dy } \] This is the switched and simplified integral as required.

# Solution to the Problem ## Step 1: Analyze the Region \( R \) and Change the Order of Integration ### (a) Sketch Region \( R \) The region \( R \) is defined as: \[ R = \{(x,y) : 0 < x < 2, \, x^2 < y < 2x \} \] ### **Identify the Intersection Points** To find the intersection points of \( y = x^2 \) and \( y = 2x \): 1. Set \( x^2 = 2x \). 2. Rearranging gives \( x^2 - 2x = 0 \). 3. Factor to get \( x(x - 2) = 0 \). Thus, the solutions are \( x = 0 \) and \( x = 2 \). **Intersection Points:** - At \( x = 0 \): \( y = 0^2 = 0 \) - At \( x = 2 \): \( y = 2^2 = 4 \) The intersection points within the range \( 0 < x < 2 \) are \( (0, 0) \) and \( (2, 4) \). ### **Sketch** The sketch should show \( y = x^2 \) as a parabola opening upwards and \( y = 2x \) as a straight line, intersecting at the points discussed. --- ### (b) Change the Order of Integration The original integral is given by: \[ I = \int_{0}^{2} \int_{x^2}^{2x} e^{-x^2} \, dy \, dx \] ### **Switching the Order** To switch to \( dy \, dx \): 1. Identify bounds for \( y \): - Lower bound: \( y = x^2 \) - Upper bound: \( y = 2x \) 2. For \( y \), \( y \) runs from \( 0 \) to \( 4 \) (since \( y \) goes up to \( 2 \times 2 \)). 3. For a fixed \( y \), \( x \) varies: - From \( x = \sqrt{y} \) (from \( y = x^2 \)) - To \( x = \frac{y}{2} \) (from \( y = 2x \)) ### **Switched Integral** Thus, the switched integral is: \[ I = \int_{0}^{4} \int_{\sqrt{y}}^{\frac{y}{2}} e^{-x^2} \, dx \, dy \] **Explanation:** The region \( R \) is now described in terms of \( y \) first, allowing for the inner integral to be evaluated with respect to \( x \). --- ## Step 2: Evaluate the Switched Integral ### (c) Evaluate the Integral Now we have: \[ I = \int_{0}^{4} \left( \int_{\sqrt{y}}^{\frac{y}{2}} e^{-x^2} \, dx \right) dy \] ### **Inner Integral Evaluation** Using the substitution \( u = x^2 \) gives \( du = 2x \, dx \), or \( dx = \frac{du}{2\sqrt{u}} \). The limits change as follows: - When \( x = \sqrt{y} \), \( u = y \) - When \( x = \frac{y}{2} \), \( u = \left(\frac{y}{2}\right)^2 = \frac{y^2}{4} \) The inner integral transforms to: \[ \int_{\sqrt{y}}^{\frac{y}{2}} e^{-x^2} \, dx = \int_{y}^{\frac{y^2}{4}} e^{-u} \frac{du}{2\sqrt{u}} \] This integral does not have a simple elementary form, so we express it using the error function: \[ \int e^{-u} \, du = -e^{-u} \] ### **Final Expression for \( I \)** The result of the inner integral: \[ I = \int_{0}^{4} \left[ -\frac{1}{2} e^{-u} \right]_{y}^{\frac{y^2}{4}} dy \] ### **Final Answer** Putting it all together, the value of \( I \) is: \[ \boxed{I = \int_{0}^{4} \left( -\frac{1}{2} e^{-\frac{y^2}{4}} + \frac{1}{2} e^{-y} \right) dy} \] This is the final form of the integral as required.