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Chapter 6: Problem 95

(a) Use a graphing utility to graph the function \(y=e^{-x^{2}}\). (b) Show that \(\int_{0}^{\infty} e^{-x^{2}} d x=\int_{0}^{1} \sqrt{-\ln y} d y\).

Short Answer

Expert verified
The verification of the equality of the two integrals \(\int_{0}^{鈭瀩 e^{-x^{2}} dx\) and \(\int_{0}^{1} \sqrt{-\ln y} dy\) can be completed using a change of variable in the integral.

Step by step solution

01

Graph the function \(y=e^{-x^{2}}\)

You can use a graphing utility, like a graphing calculator or an online tool, to draw the function \(y=e^{-x^{2}}\). The basic shape of the graph should be a bell curve. The peak of the graph is at x=0, and as x moves away from 0 in either direction, the value of y approaches 0.
02

Recognize the consecutive integrals and make a variable substitution

The task here is to show that the integrals \(\int_{0}^{鈭瀩 e^{-x^{2}} dx\) and \(\int_{0}^{1} \sqrt{-\ln y} dy\) are equivalent. Make the following substitution: let \(y=e^{-x^{2}}\), which implies \(x=\sqrt{-\ln y}\). The differential dx can be determined by differentiating the latter equation: \(dx=d\sqrt{-\ln y}={$\frac{-1}{2y\sqrt{-\ln y}}} dy\.
03

Rewrite the initial integral using the new variable

The integral \(\int_{0}^{鈭瀩 e^{-x^{2}} dx\) becomes \(\int_{1}^{0} -\sqrt{-\ln y}(1/2y\sqrt{-\ln y}) dy\) after the substitution. This simplifies to \(\int_{0}^{1} \sqrt{-\ln y} dy\), thus verifying the initial claim.

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