Question

How do you integrate $\int_{0}^{1}{{{x}^{2}}{{e}^{x}}dx}$ by integration by parts?

Answer 1

Hint: Now we know that according to integration by parts we have $\int{u.vdx}=u\int{vdx}-\int{u'\left( \int{vdx} \right)dx}$ . Hence considering ${{x}^{2}}$ as u and ${{e}^{x}}$ as v we can easily solve the integration. Now we will again solve the integral obtained by the same method and hence will find the solution of the equation.

Complete step by step solution:
Now we are given with the integral $\int_{0}^{1}{{{x}^{2}}{{e}^{x}}dx}$ .
To solve the integral we will use Integration by parts.
Now integration by parts tells us that the integration of $\int{u.vdx}=u\int{vdx}-\int{u'\left( \int{vdx} \right)dx}$
Now first let us understand the ILATE rule.
ILATE is an acronym for Inverse, Logarithmic, Algebraic, Trigonometric and Exponential. The rule gives us order in which the first function must be chosen.
Now in the given integral we have two functions one is ${{x}^{2}}$ and ${{e}^{x}}$ . Now according to ILATE rule we must take the first function as ${{x}^{2}}$ and second function as ${{e}^{x}}$ as algebraic functions are preferred before exponential.
Now substituting the functions in the formula for Integration by parts we get,
$\Rightarrow \int{{{x}^{2}}{{e}^{x}}dx}={{x}^{2}}\int{{{e}^{x}}dx}-\int{\left( {{x}^{2}} \right)'\left( \int{{{e}^{x}}dx} \right)dx}$
Now we know that the integration of the function ${{e}^{x}}$ is ${{e}^{x}}$ and the differentiation of the function ${{x}^{2}}$ is 2x. Hence we get the integration as,
$\Rightarrow \int{{{x}^{2}}{{e}^{x}}dx}={{x}^{2}}{{e}^{x}}-\int{2x{{e}^{x}}dx}$
Now taking the constant out from the integration we get,
$\Rightarrow \int{{{x}^{2}}{{e}^{x}}dx}={{x}^{2}}{{e}^{x}}-2\int{x{{e}^{x}}dx}......................\left( 1 \right)$
Now we will again use Integration by parts to solve $\int{x{{e}^{x}}dx}$
Hence we get,
$\begin{align}
  & \Rightarrow \int{x{{e}^{x}}dx}=x{{e}^{x}}-\int{{{e}^{x}}dx} \\
 & \Rightarrow \int{x{{e}^{x}}dx}=x{{e}^{x}}-{{e}^{x}} \\
\end{align}$
Now substituting the value of $\int{x{{e}^{x}}dx}$ in equation (1) we get,
\[\Rightarrow \int{{{x}^{2}}{{e}^{x}}dx}={{x}^{2}}{{e}^{x}}-2\left( x{{e}^{x}}-{{e}^{x}} \right)+C\]
Hence putting the limits we get,
$\begin{align}
  & \Rightarrow \int_{0}^{1}{{{x}^{2}}{{e}^{x}}}=\left[ {{x}^{2}}{{e}^{x}}-2\left( x{{e}^{x}}-{{e}^{x}} \right) \right]_{0}^{1} \\
 & \Rightarrow \int_{0}^{1}{{{x}^{2}}{{e}^{x}}}=\left[ {{e}^{1}}-2\left( {{e}^{1}}-{{e}^{1}} \right) \right]-\left[ 0-2\left( 0-{{e}^{0}} \right) \right] \\
 & \Rightarrow \int_{0}^{1}{{{x}^{2}}{{e}^{x}}}={{e}^{1}}-\left[ 2 \right] \\
 & \Rightarrow \int_{0}^{1}{{{x}^{2}}{{e}^{x}}}=e-2 \\
\end{align}$

Hence the solution of the given integral is e – 2.

Note: Now note that the integral by parts can also be written as $\int{udv}=uv-\int{vdu}$ . Also note that whenever we have functions multiplied to each other we can always use integration by parts.