Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
The problem asks to evaluate the double integral \[ \int \int e^{x^2} \, dx \, dy \] where the region \( D = \{(x, y) \mid y \leq x \leq 1, \, 0 \leq y \leq 1\} \), and also evaluate \[ \int_0^1 \int_y^1 e^{x^2} \, dx \, dy \] It further asks to explain why these two expressions are the same. **Explanation:** The problem involves evaluating a double integral of the function \( e^{x^2} \) over the region \( D \) in the plane. The region \( D \) is defined by the inequalities \( y \leq x \leq 1 \) and \( 0 \leq y \leq 1 \). In the first integral, the order of integration is \( dx \, dy \), indicating that we integrate with respect to \( x \) first and then \( y \). In the second integral, the order is reversed, \( dy \, dx \). **Why they're the same:** The reason the two integrals are equivalent is due to Fubini’s Theorem, which allows for the interchange of the order of integration, given that the function and the region are well-behaved (usually meaning the function is continuous and the region is bounded and measurable). In this scenario, the change of order corresponds to integrating over the same region \( D \) but changing the path of integration. When changing the order, the limits of integration must also be adjusted to reflect this new order while still describing the same area. For the double integral \( \int \int e^{x^2} \, dx \, dy \), first, the limits are defined by \( y \leq x \leq 1 \). For the iterated integral \( \int_0^1 \int_y^1 e^{x^2} \, dx \, dy \), we integrate from \( y \) to \( 1 \) with respect to \( x \), for each fixed \( y \), over the same area. By changing the order of integration appropriately, we confirm the two expressions evaluate to the same value.
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Transcribed Image Text:The problem asks to evaluate the double integral \[ \int \int e^{x^2} \, dx \, dy \] where the region \( D = \{(x, y) \mid y \leq x \leq 1, \, 0 \leq y \leq 1\} \), and also evaluate \[ \int_0^1 \int_y^1 e^{x^2} \, dx \, dy \] It further asks to explain why these two expressions are the same. **Explanation:** The problem involves evaluating a double integral of the function \( e^{x^2} \) over the region \( D \) in the plane. The region \( D \) is defined by the inequalities \( y \leq x \leq 1 \) and \( 0 \leq y \leq 1 \). In the first integral, the order of integration is \( dx \, dy \), indicating that we integrate with respect to \( x \) first and then \( y \). In the second integral, the order is reversed, \( dy \, dx \). **Why they're the same:** The reason the two integrals are equivalent is due to Fubini’s Theorem, which allows for the interchange of the order of integration, given that the function and the region are well-behaved (usually meaning the function is continuous and the region is bounded and measurable). In this scenario, the change of order corresponds to integrating over the same region \( D \) but changing the path of integration. When changing the order, the limits of integration must also be adjusted to reflect this new order while still describing the same area. For the double integral \( \int \int e^{x^2} \, dx \, dy \), first, the limits are defined by \( y \leq x \leq 1 \). For the iterated integral \( \int_0^1 \int_y^1 e^{x^2} \, dx \, dy \), we integrate from \( y \) to \( 1 \) with respect to \( x \), for each fixed \( y \), over the same area. By changing the order of integration appropriately, we confirm the two expressions evaluate to the same value.
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