13.2 Double Integrals over General Regions Problem 25 **17–26. Evaluating Integrals**Evaluate the following integrals as they are written. **25. Evaluate the double integral:**\[\int_0^1 \int_0^x 2e^{x^2} \, dy \, dx\]**Explanation:**This is a double integral where the function $2e^{x^2}$ is integrated first with respect to $y$ and then with respect to $x$. - The limits of the inner integral are from 0 to $x$, indicating that for each fixed $x$, $y$ varies between 0 and $x$.- The outer integral varies $x$ from 0 to 1.In the context of double integrals, this setup suggests calculating the area under a surface $z = 2e^{x^2}$ over the triangular region defined by $0 \leq y \leq x \leq 1$. The evaluation involves first integrating the function over $y$, and then using that result to integrate over $x$.

Double Integral: ∫01∫0x2ex2dydx While integrating with respect to y, consider x as constant and similarly while integrating with respect to x, consider y as constant. Given integral is rearranged as: ∫012ex2∫0xdydx Now evaluate the integral inside, with respect to y: ∫012ex2∫0xdydx=∫012ex2y0xdx=∫012xex2dxNow, integrate obtained term with respect to x: ∫012xex2dx Replace x2=t Then, 2xdx=dt Change the limits according to the substitution: For x=0, value of t is t=0 For x=1, value of t is t=1 Now, evaluate the integral in t: ∫01etdt=et01=e1-e0=e-1 Therefore, the value of the given double integral is: ∫01∫0x2ex2dydx=e-1

Answered: 25. fď S7 2e²dy dz

Math

Advanced Math

25. fď S7 2e²dy dz

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

13.2 Double Integrals over General Regions Problem 25

**17–26. Evaluating Integrals**

Evaluate the following integrals as they are written.

**25. Evaluate the double integral:**

\[
\int_0^1 \int_0^x 2e^{x^2} \, dy \, dx
\]

**Explanation:**

This is a double integral where the function $2e^{x^2}$ is integrated first with respect to $y$ and then with respect to $x$.

- The limits of the inner integral are from 0 to $x$, indicating that for each fixed $x$, $y$ varies between 0 and $x$.
- The outer integral varies $x$ from 0 to 1.

In the context of double integrals, this setup suggests calculating the area under a surface $z = 2e^{x^2}$ over the triangular region defined by $0 \leq y \leq x \leq 1$. The evaluation involves first integrating the function over $y$, and then using that result to integrate over $x$.

With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.

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