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The solution to question I've been given is:

$$P=\frac{1}{\sqrt{\pi}a}\int_{-a}^ae^{-x^2/a^2} dx$$

$$P=\frac{1}{\sqrt{\pi}a}\int_{-a}^ae^{-u^2} du$$

$$P=\frac{1}{\sqrt{\pi}a} \times 1.494a \approx 0.84$$ given that, using the substitution $u = \frac{x}{a}$:

$$\int_{-a}^ae^{-x^2/a^2} dx=a\int_{-1}^1e^{-u^2} du = 1.494a$$

I thought it would be:

$$\int_{-a}^ae^{-x^2/a^2} dx=a\int_{-a}^ae^{-u^2} du = a^2\int_{-1}^1e^{-u^2} du =1.494a^2$$

I would appreciate it if someone could explain why this isn't the case. Thanks.

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$$P=\frac{1}{\sqrt{\pi}a}\int_{-a}^ae^{-x^2/a^2} dx$$

$u = x / a \Rightarrow a du = dx$

$x = a \Rightarrow u = 1$

$$ \begin{aligned} P &=\frac{1}{\sqrt{\pi}a} \int_{-1}^1 e^{-u^2} \, a du \\ &= \frac{a}{\sqrt{\pi}a} \int_{-1}^1 e^{-u^2} \, du \\ &= \frac{1}{\sqrt{\pi}} \int_{-1}^1 e^{-u^2} \, du \\ &\approx \frac{1}{\sqrt{\pi}} 1.494 \end{aligned} $$

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  • $\begingroup$ Thanks for the response. I understand that the substitution of $dx$ for $du$ gives us an extra $a$. But why do the limits of the integral change to $1$ and $-1$? Sorry if I'm missing a trick. $\endgroup$ – Mike A Jun 3 '17 at 15:32
  • $\begingroup$ Wouldn't changing the limits require $a=1$ rather than $u=1$? $\endgroup$ – Mike A Jun 3 '17 at 15:37
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    $\begingroup$ The original integral is over $x$ with limits $x =\pm a$ after changing variable to get the new integral we also need to convert the limits. Because $x/a=u$ we need to use the limits $u=\pm 1$ in the converted integral. $\endgroup$ – Paul Aljabar Jun 3 '17 at 15:47
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note that $dx=adu$ thus integral turns to $\frac{1}{\sqrt{\pi} a} \int _{-1} ^ 1 e^{-u^2} .adu$ thus the result follows

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  • $\begingroup$ Thanks for the response. I understand this step, but why do we go from $\frac{1}{\sqrt{\pi} a} \int _{-a} ^ a e^{-u^2} .adu$ to $\frac{1}{\sqrt{\pi} a} \int _{-1} ^ 1 e^{-u^2} .adu$. Apologies if I am missing something obvious. $\endgroup$ – Mike A Jun 3 '17 at 15:34
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    $\begingroup$ No when you substitute integral is in u not x so limits change from -a,a to -1,1 $\endgroup$ – Archis Welankar Jun 3 '17 at 15:39
  • $\begingroup$ Would you mind explaining that particular process for me? My attempt would be $a=\frac{x}/{u}$. So the limits would become + and - $\frac{x}{u}$. Then that would have to equal + and - $\frac{x}{u}=1$ and so $x=u$? As you can probably tell I'm very confused about the limit change in particular. $\endgroup$ – Mike A Jun 3 '17 at 15:48
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    $\begingroup$ when x=a we have $u=\frac{x}{a}=\frac{a}{a}=1$ similarly for other limit $\endgroup$ – Archis Welankar Jun 3 '17 at 15:49
  • $\begingroup$ Is $x=a$ because $\int_{x_{1}}^{x_2} = \int_{-a}^a$? If so, I think we have found the source of my idiocy. $\endgroup$ – Mike A Jun 3 '17 at 15:53

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