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The following definite integral evaluates to

$$\int_{-\infty}^{0} e^ {-\left(\frac{x^2}{20} \right )}dx$$

1. $\frac{1}{2}$
2. $\pi \sqrt{10}$
3. $\sqrt{10}$
4. $\pi$
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+1
+1
in question its not 20 but 2

$\dpi{120} \\Let: x = \sqrt{20t} \qquad\text{Then, } \mathrm{d}x = \frac{\sqrt 5}{\sqrt t}\,\mathrm{d}t$

Now, the given integral can be written as:

$\dpi{120} \int\limits_{0}^{\infty}{ e^{-t} \cdot \frac{\sqrt 5}{\sqrt t}\, \mathrm{d}t}\qquad =\qquad \sqrt{5} \int\limits_{0}^{\infty}{\frac1{e^t\sqrt{t}}\,\mathrm{d}t}$

$\dpi{120} = \sqrt{5} \cdot \sqrt \pi \qquad \qquad \Bigg\{\int\limits_{0}^{\infty}{\frac1{e^t\sqrt{t}}\,\mathrm{d}t} = \sqrt{\pi} \text{ is a famous } \gamma \text{-reduction}$

Since none of the options are equal to $\dpi{120} \sqrt{5\pi}$, the given options are incorrect. You can verify the result here: wolframalpha(Click to view)

by Active (3.8k points)
edited
+3

The original answer, in case needed, can be obtained from what follows:

let put x = (20t)1/2
Equation become     dx = ((51/2)/t1/2)dt
51/2e-tt-1/2dt================== very famous gamma reduction
51/2pie1/2

0
How to do gamma reduction. Please provide full solution.
0
It is not needed for GATE
0
Is gamma reduction in syllabus?
+1 vote
You can simply view $\int_{-\infty}^{0} e^ {-\left(\frac{x^2}{20} \right )}dx$ as normal distribution  with mean = 0 and $\sigma =\sqrt{10}$. Multiply and divide by $\sqrt{2\pi}\sigma$.
by Active (2.6k points)