Question

Which of the following is the minimum value of $\int\limits_0^x {t \times {e^{ - {t^2}}}} dt$
( 1) $1$
( 2) $2$
( 3) $3$
( 4) $0$

Answer 1

Hint : For any function $f\left( x \right)$ if we want to know the minimum or maximum or we can say extremum value first we have to take derivative of the function and equate that with zero .We will get the value of $x$ for which first derivative is zero those are the points where the function is extremum . Now we have to take the second derivative of the function and put the points on the second derivative and have to check whether the value is positive or negative . If the value is positive then the point is the minimum point . At that point we will get the minimum value of the function.
Formula used :
Leibniz Integral Rule
$\dfrac{d}{{dx}}\int\limits_{u\left( x \right)}^{v\left( x \right)} {f\left( t \right)} \,dt\, = f\left( {v\left( x \right)} \right) \times v'\left( x \right) - f\left( {u\left( x \right)} \right) \times u'\left( x \right)$
$u\left( x \right)$ and $v\left( x \right)$ are the lower and upper limits of integration and also function of $x$ .

Complete step-by-step answer :
Here the function $f\left( x \right)$= $\int\limits_0^x {t{e^{ - {t^2}}}} dt$ is integration of a function $f\left( t \right)$ so to take derivative of that we need to follow Leibniz rule
Here $f\left( x \right) = \int\limits_0^x {t{e^{ - {t^2}}}} dt$
$f\left( t \right) = t{e^{ - {t^2}}}$
$u\left( x \right)$ and $v\left( x \right)$ are the lower and upper limits of integration and also function of $x$.
Here $u\left( x \right) = 0$
$v\left( x \right) = x$
So $v'\left( x \right) = 1$
And $u'\left( x \right) = 0$
Leibniz Rule
$\dfrac{d}{{dx}}\int\limits_{u\left( x \right)}^{v\left( x \right)} {f\left( t \right)} \,dt\, = f\left( {v\left( x \right)} \right) \times v'\left( x \right) - f\left( {u\left( x \right)} \right) \times u'\left( x \right)$
$f\left( t \right) = t{e^{ - {t^2}}}$
Putting $u\left( x \right)$ and $v\left( x \right)$ in the function $f\left( t \right)$ we get
$f\left( {v\left( x \right)} \right) = x{e^{ - {x^2}}}$
$f\left( {u\left( x \right)} \right) = 0$
Putting all this value and using Leibniz rule we get
$f'\left( x \right) = x{e^{ - {x^2}}} \times 1 - 0$
$= x{e^{ - {x^2}}}$
Now we have to equate $f'\left( x \right)$ with zero .
$x{e^{ - {x^2}}} = 0$
By doing that we get $x = 0$ is an extremum point as at x = 0$$$$f'\left( x \right) is equal to zero.
Now we have to find the second derivative .
Number of dashes represents the number of times we took a derivative .
Following the rule for derivation of multiplication of two function we get
So $f''\left( x \right) = {e^{ - {x^2}}} \times 1 + x{e^{ - {x^2}}} \times \left( { - 2x} \right)$
$= {e^{ - {x^2}}} - 2{x^2}{e^{ - {x^2}}}$
At x = 0$$$$f''\left( x \right)is
$f''\left( 0 \right) = 1$
We get positive value so $x = 0$is maximum point for function $f\left( x \right)$= $\int\limits_0^x {t{e^{ - {t^2}}}} dt$
And Minimum value of the function is $f\left( 0 \right) = 0$
So answer is option 4
So, the correct answer is “Option 4”.

Note : We have to remember how to obtain minimum and maximum points. We have to keep in mind the basic rule of derivation. We must apply the Leibniz rule carefully . Take care of the calculations in order to be sure of the final answer.