25. fď S7 2e²dy dz

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13.2 Double Integrals over General Regions Problem 25

**17–26. Evaluating Integrals**

Evaluate the following integrals as they are written.
Transcribed Image Text:**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 \(x\).
Transcribed Image Text:**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 \(x\).
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