Question

If $f\left( x \right) = \int\limits_0^x {{e^{\dfrac{{ - {x^2}}}{2}}}\left( {1 - {t^2}} \right)dt} $ minimum at \[x = \]
A) \[1\]
B) \[ - 1\]
C) \[2\]
D) \[ - 2\]

Answer 1

Hint: In order to solve for the minimum value at x, we need to know about integration and taking differentiation of the value under integration. For integration we know that, it is the reverse process of differentiation which is adding or summing of the parts. And for differentiation under integration, we would use the formula for a function \[f\left( x \right) = \int\limits_{a\left( x \right)}^{b\left( x \right)} {g\left( x \right)dx} \]which gives \[f'\left( x \right) = g\left( b \right)\dfrac{{d\left( b \right)}}{{dx}} - g\left( a \right)\dfrac{{d\left( a \right)}}{{dx}}\].

Formula used:
\[\dfrac{{dx}}{{dx}} = 1\]
\[\dfrac{{d0}}{{dx}} = 0\]
\[\dfrac{{d\left( {{e^{{x^n}}}} \right)}}{{dx}} = n{x^{n - 1}}{e^{{x^n}}}\]
 \[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\]
$\dfrac{{d\left( {u.v} \right)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$

Complete answer:
We have $f\left( x \right) = \int\limits_0^x {{e^{\dfrac{{ - {x^2}}}{2}}}\left( {1 - {t^2}} \right)dt} $
Taking out the value ${e^{\dfrac{{ - {x^2}}}{2}}}$from the integration, and we get:
$f\left( x \right) = {e^{\dfrac{{ - {x^2}}}{2}}}\int\limits_0^x {\left( {1 - {t^2}} \right)dt} $
Since, we know that for a function:
\[f\left( x \right) = \int\limits_{a\left( x \right)}^{b\left( x \right)} {g\left( x \right)dx} \]
On differentiating with integral we know that the value would be:
\[f'\left( x \right) = g\left( b \right)\dfrac{{d\left( b \right)}}{{dx}} - g\left( a \right)\dfrac{{d\left( a \right)}}{{dx}}\]
Comparing \[f\left( x \right) = \int\limits_{a\left( x \right)}^{b\left( x \right)} {g\left( x \right)dx} \] with $f\left( x \right) = {e^{\dfrac{{ - {x^2}}}{2}}}\int\limits_0^x {\left( {1 - {t^2}} \right)dt} $ we get:
\[
  a = 0 \\
  b = x \\
  g\left( x \right) = 1 - {t^2} \\
 \]
Substituting the values, we get:
\[f'\left( x \right) = {e^{\dfrac{{ - {x^2}}}{2}}}\left[ {g\left( x \right)\dfrac{{d\left( x \right)}}{{dx}} - g\left( 0 \right)\dfrac{{d\left( 0 \right)}}{{dx}}} \right]\]
\[ \Rightarrow f'\left( x \right) = {e^{\dfrac{{ - {x^2}}}{2}}}\left[ {\left( {1 - {x^2}} \right)\dfrac{{d\left( x \right)}}{{dx}} - \left( {1 - {0^2}} \right)\dfrac{{d\left( 0 \right)}}{{dx}}} \right]\]
As, we know that \[\dfrac{{dx}}{{dx}} = 1\] and \[\dfrac{{d0}}{{dx}} = 0\]
\[ \Rightarrow f'\left( x \right) = {e^{\dfrac{{ - {x^2}}}{2}}}\left[ {\left( {1 - {x^2}} \right)1 - \left( {1 - {0^2}} \right)0} \right]\]
Solving it, we get:
\[ \Rightarrow f'\left( x \right) = {e^{\dfrac{{ - {x^2}}}{2}}}\left[ {\left( {1 - {x^2}} \right) - 0} \right]\]
\[ \Rightarrow f'\left( x \right) = {e^{\dfrac{{ - {x^2}}}{2}}}\left( {1 - {x^2}} \right)\]
To find the critical points:
Putting \[f'(x) = 0\],we get:
\[1 - {x^2} = 0\]
Adding both sides by \[{x^2}\]:
\[
  1 - {x^2} + {x^2} = 0 + {x^2} \\
   \Rightarrow 1 = {x^2} \\
   \Rightarrow {x^2} = 1 \\
 \]
Taking square root both the sides:
\[ \Rightarrow \sqrt {{x^2}} = \sqrt 1 \]
We get \[x = \pm 1\]
Since, we have two functions inside the given function, so using the product rule, we are differentiating \[f'(x)\] with respect to \[x\] we get:
As, we know that:
$\dfrac{{d\left( {u.v} \right)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
Considering \[u = {e^{\dfrac{{ - {x^2}}}{2}}}{\text{ and v = }}\left( {1 - {x^2}} \right)\], substituting the values:
$f''\left( x \right) = {e^{\dfrac{{ - {x^2}}}{2}}}\dfrac{{d\left( {1 - {x^2}} \right)}}{{dx}} + \left( {1 - {x^2}} \right)\dfrac{{d\left( {{e^{\dfrac{{ - {x^2}}}{2}}}} \right)}}{{dx}}$
Since, we know that \[\dfrac{{d\left( {{e^{{x^n}}}} \right)}}{{dx}} = n{x^{n - 1}}{e^{{x^n}}}\] and \[\dfrac{{d\left( {1 - {x^2}} \right)}}{{dx}} = \dfrac{{d1}}{{dx}} - \dfrac{{d{x^2}}}{{dx}} = 0 - 2x = - 2x\].
Substituting the values, we get:
\[f''(x) = {e^{\dfrac{{^{ - {x^2}}}}{2}}}( - 2x) + (1 - {x^2}){e^{\dfrac{{^{ - {x^2}}}}{2}}}\left( {\dfrac{{ - 2x}}{2}} \right)\]
Finding the value for critical points by substituting the value at $x = 1$:
\[
  f''(1) = {e^{\dfrac{{^{ - {1^2}}}}{2}}}( - 2 \times 1) + (1 - {1^2}){e^{\dfrac{{^{ - {1^2}}}}{2}}}\left( {\dfrac{{ - 2 \times 1}}{2}} \right) \\
   \Rightarrow f''(1) = - 2{e^{\dfrac{{^{ - {1^2}}}}{2}}} + 0 \\
   \Rightarrow f''(1) = - 2{e^{\dfrac{{^{ - {1^2}}}}{2}}} < 0 \\
 \]
Therefore \[x = 1\]is a point of maxima.
The critical value at \[x = - 1\]
\[
  f''( - 1) = {e^{\dfrac{{^{{1^2}}}}{2}}}( - 2 \times - 1) + (1 - {\left( { - 1} \right)^2}){e^{\dfrac{{^{{1^2}}}}{2}}}\left( {\dfrac{{ - 2 \times - 1}}{2}} \right) \\
   \Rightarrow f''( - 1) = 2{e^{\dfrac{{^{ - {1^2}}}}{2}}} + 0 \\
   \Rightarrow f''( - 1) = 2{e^{\dfrac{{^{ - {1^2}}}}{2}}} > 0 \\
 \]
\[f''(x) = f''( - 1) = {e^{\dfrac{{^{ - 1}}}{2}}}(2) > 0\]
Therefore, \[x = - 1\] is a point of minima.
Therefore, \[f(x) = \int\limits_0^x {{e^{\dfrac{{^{ - {x^2}}}}{2}}}} (1 - {t^2})dt\]is minimum at\[x = - 1\].
Hence, Option (2) is the correct answer.

Note:
As we know that to find the critical point, we need to find the first order derivative of a function and to find the maximum and minimum value, find the second order derivative of the function and substitute the critical points. If the value is less than zero, the condition will be maxima and if the value is more than zero, the condition would be minima.