The function e is an example of a Gaussian (bell curve)'. A very useful result in probability/statistics is that dr = VT. In this problem, we'll use a double integral in polar coordinates to compute this integral. Let f(x, y) = e-¤²-y´ and let DR be the disk of radius R centered at the origin, x² + y² < R². Compute /I f(x, y) dA. Let R → o in your answer to part (a). The result is dA. 8. Use the result of part (b), together with the result of #2(a), to show that -72 dx VT. (Hint: the name of the variable in an integral does not matter.)
The function e is an example of a Gaussian (bell curve)'. A very useful result in probability/statistics is that dr = VT. In this problem, we'll use a double integral in polar coordinates to compute this integral. Let f(x, y) = e-¤²-y´ and let DR be the disk of radius R centered at the origin, x² + y² < R². Compute /I f(x, y) dA. Let R → o in your answer to part (a). The result is dA. 8. Use the result of part (b), together with the result of #2(a), to show that -72 dx VT. (Hint: the name of the variable in an integral does not matter.)
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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![The function \( e^{-x^2} \) is an example of a Gaussian (bell curve)\(^1\). A very useful result in probability/statistics is that
\[
\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}.
\]
In this problem, we'll use a double integral in polar coordinates to compute this integral.
Let \( f(x, y) = e^{-x^2-y^2} \) and let \( D_R \) be the disk of radius \( R \) centered at the origin, \( x^2 + y^2 \leq R^2 \). Compute \(\int\int_{D_R} f(x, y) \, dA\).
Let \( R \to \infty \) in your answer to part (a). The result is
\[
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \, dA.
\]
Use the result of part (b), together with the result of \(\#2(a)\), to show that
\[
\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}.
\]
(Hint: the name of the variable in an integral does not matter.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F66e8a441-1535-4b77-8170-ec0b022e4104%2Fdeb28eb7-82fe-4847-96f8-adcc9b7ff3bb%2Fz3xo1k_processed.png&w=3840&q=75)
Transcribed Image Text:The function \( e^{-x^2} \) is an example of a Gaussian (bell curve)\(^1\). A very useful result in probability/statistics is that
\[
\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}.
\]
In this problem, we'll use a double integral in polar coordinates to compute this integral.
Let \( f(x, y) = e^{-x^2-y^2} \) and let \( D_R \) be the disk of radius \( R \) centered at the origin, \( x^2 + y^2 \leq R^2 \). Compute \(\int\int_{D_R} f(x, y) \, dA\).
Let \( R \to \infty \) in your answer to part (a). The result is
\[
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \, dA.
\]
Use the result of part (b), together with the result of \(\#2(a)\), to show that
\[
\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}.
\]
(Hint: the name of the variable in an integral does not matter.)
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