Question

How do you find the integration of $\int{x{{e}^{-{{x}^{2}}}}dx}$ from [0, 1]

Answer 1

Hint: Now to solve the given integral we will use the method of substitution. First we will substitute $-{{x}^{2}}=t$ and then differentiate the equation to find the substitution for xdx. Now we will check the new limits and substitute the values in the given integral. Now we can easily integrate the function as we know that $\int{{{e}^{x}}}={{e}^{x}}+C$ . Substituting the limits we will find the solution to the given problem.

Complete step by step solution:
Now we want to find the integral $\int_{0}^{1}{x{{e}^{-{{x}^{2}}}}dx}$ .
Now we can use a method of substitution to find the integration of the given function.
Let us substitute $-{{x}^{2}}=t$ .
Differentiating on both sides we get, $-2xdx=dt$
Now rearranging the terms we get $xdx=-\dfrac{dt}{2}$ .
Now again consider the equation $-{{x}^{2}}=t$ .
Now for x = 1 we have t= 1 and as x = 0 we have t = 0.
Hence the limits of the integral after substitution will be the same.
Now let us substitute the above values in the integral hence we get the integral as
$\Rightarrow \int_{0}^{1}{\dfrac{-{{e}^{t}}dt}{2}}$ .
Now we know the property of integral $\int_{a}^{b}{cf\left( x \right)}=c\int_{a}^{b}{f\left( x \right)}$ , Hence using this we get,
$\Rightarrow \int_{0}^{1}{x{{e}^{-{{x}^{2}}}}dx}=\dfrac{-1}{2}\int_{0}^{1}{{{e}^{t}}dt}$
Now we know that $\int{{{e}^{x}}}={{e}^{x}}+C$ Hence integrating we get,
$\begin{align}
  & \Rightarrow \int_{0}^{1}{x{{e}^{-{{x}^{2}}}}dx}=\dfrac{-1}{2}\left[ {{e}^{t}} \right]_{0}^{1} \\
 & \Rightarrow \int_{0}^{1}{x{{e}^{-{{x}^{2}}}}dx}=\dfrac{-1}{2}\left[ {{e}^{1}}-{{e}^{0}} \right] \\
 & \Rightarrow \int_{0}^{1}{x{{e}^{-{{x}^{2}}}}dx}=\dfrac{-1}{2}\left[ e-0 \right] \\
 & \Rightarrow \int_{0}^{1}{x{{e}^{-{{x}^{2}}}}dx}=\dfrac{-e}{2} \\
\end{align}$
Hence we have the integration of the given function is $\dfrac{-e}{2}$ .

Note: Note that whenever a method of integration by substitution is used the variable is changed. Hence we need to also change and check the limits accordingly. In the above example limits stayed the same but this will not always be the case. Hence always remember to change the limits as well as the differential element dx accordingly. Also we can substitute the value of t and then use limits of x if needed.