3. The normal curve (or the Gaussian), defined as g(x)plays an important role in probability andV27+oostatistics. It is useful to know that| g(x) dx = 1, which we will now show.-00(a) Suppose that Da is a disk, centered at the origin, with radius a. Explain why+ooe-(z²+y³)/2dx dylime-(a²+y²)/2dA.(b) Change to polar coordinates to evaluatee-(a2+y²)/2 dA, and take the limit as a → +o.Da(c) Explain whytoo+oo+ootoo"|e-=²+s²)/2 dA =x² /2 dx-y²/2dy-00-00(d) Conclude that|e-a*/2 dx = V27, so thatdx = 1, aswe hoped.

A coordinate system is a convention that is used to represent a point uniquely using a set of…

. The normal curve (or the Gaussian), defined as g(x) = 27 plays an important role in probability and statistics. It is useful to know that | g(x) dx = 1, which we will now show. (a) Suppose that Da is a disk, centered at the origin, with radius a. Explain why *e-*+v*)/2 dr dy = lim // e-(z²+u*)/2 dA. (b) Change to polar coordinates to evaluate || e-(x²+y²)/2 dA, and take the limit as a –→+∞. (c) Explain why +oo e-(2²+y³)/2 dA = 1² /2 dx e-v³/2 dy e -00 (d) Conclude that e-2* /2 dx = V27, so that g(x) dx = 1, as we hoped.

. The normal curve (or the Gaussian), defined as g(x) = 27 plays an important role in probability and statistics. It is useful to know that | g(x) dx = 1, which we will now show. (a) Suppose that Da is a disk, centered at the origin, with radius a. Explain why e-+v)/2 dr dy = lim // e-(z²+u)/2 dA. (b) Change to polar coordinates to evaluate || e-(x²+y²)/2 dA, and take the limit as a –→+∞. (c) Explain why +oo e-(2²+y³)/2 dA = 1² /2 dx e-v³/2 dy e -00 (d) Conclude that e-2* /2 dx = V27, so that g(x) dx = 1, as we hoped.

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

3. The normal curve (or the Gaussian), defined as g(x)
plays an important role in probability and
V27
+oo
statistics. It is useful to know that
| g(x) dx = 1, which we will now show.
-00
(a) Suppose that Da is a disk, centered at the origin, with radius a. Explain why
+oo
e-(z²+y³)/2
dx dy
lim
e-(a²+y²)/2
dA.
(b) Change to polar coordinates to evaluate
e-(a2+y²)/2 dA, and take the limit as a → +o.
Da
(c) Explain why
too
+oo
+oo
too
"|e-=²+s²)/2 dA =
x² /2 dx
-y²/2
dy
-00
-00
(d) Conclude that
|e-a*/2 dx = V27, so that
dx = 1, as
we hoped.

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