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Question

A minimum value of integral from 0 to xte-t2dt is


A

1

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B

2

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C

3

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D

0

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Solution

The correct option is D

0


Explanation for the correct option:

Step 1. Finding the minimum value:

Given, f(x)=0xte-t2dt
Substitute t2=u and differentiate it with respect to t

2t=dudt

2tdt=du

tdt=12du

Now, f(x)=120x2e-udu

Integrate it with respect to u

f(x)=12e-u-10x2=-12e-x2-e-0=-12e-x2-1 e-x=-e-x

Step 2. Find the minimum value, differentiate f(x) and put f'(x)=0

ddx121-e-x2=0

0+e-x2ddx(x2)=0 d(e-x)dx=-e-x

2xe-x2=0

if x=0, then 2xe-x2=0

f'(x)change sign from -ve to +ve ,so at x=0,f(x) has local minima.

Hence, it is a point of minima

The minimum value is f(0)=00te-t2dt=0

Hence, The correct option is option (D).


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