
Evaluate the integral $\int {{x^2}{e^x}dx} $.
Answer
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Hint: In order to solve this integral we need to integrate it by parts to get the right answer. We should also need to know that integration of ${e^x}$ is ${e^x}$ and differentiation of ${x^2}$ is $2x$ and that of $2x$ is $2$. If we need to integrate uv with respect to x where u and v are different function in x then we integrate it by part as $\int {\left( {uv} \right)dx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\left( {\int {vdx} } \right)} \right)} } } dx$. Doing this will solve your problems and will give you the right answer.
Complete step-by-step solution:
We have to integrate $\int {{x^2}{e^x}dx} $.
We know that if there are two functions in x those are f(x) and g(x) and we have to integrate f(x)g(x) with respect to x then we get the value of integration as,
$ \Rightarrow \int {\left( {f(x)g(x)} \right)dx = f(x)\int {g(x)dx - \int {\left( {f'(x)\left( {\int {g(x)dx} } \right)} \right)} } } dx$.
On putting $f(x) = {x^2}$ and $g(x) = {e^x}$ we get the value of above equation as,
\[
\Rightarrow \int {\left( {{x^2}{e^x}} \right)dx = {x^2}\int {{e^x}dx - \int {\left( {2x\left( {\int {{e^x}dx} } \right)} \right)} } } dx \\
\Rightarrow \int {\left( {{x^2}{e^x}} \right)dx = {x^2}{e^x} - \int {2x{e^x}dx} } \\
\]
We again got the term \[\int {2x{e^x}dx} \] to integrate it by parts. So, on integrating it by parts we get the value of the equation as,
$ \Rightarrow \int {2x{e^x}dx} = 2x{e^x} - 2{e^x}$
So, the answer is \[{x^2}{e^x} - 2x{e^x} - 2{e^x} + C\] where $C$ is the constant of integration.
Note: In such types of problems of integrals we need to recall integration by parts formula. Then we have to decide which is to be integrated and which term should be differentiated so that we can get the answer easily. In accordance with this concept, we decide the 1st and 2nd term first term is differentiated and the second is integrated. One of the most famous rules to decide first and second term integrate it by part is the ILATE rule which means: Inverse, Logarithmic, Algebraic, Trigonometric, Exponent which states that the inverse function should be assumed as the first function while performing the integration. A useful rule of integral by parts is ‘ILATE’. Knowing this will solve your problems and will give you the right answers.
Complete step-by-step solution:
We have to integrate $\int {{x^2}{e^x}dx} $.
We know that if there are two functions in x those are f(x) and g(x) and we have to integrate f(x)g(x) with respect to x then we get the value of integration as,
$ \Rightarrow \int {\left( {f(x)g(x)} \right)dx = f(x)\int {g(x)dx - \int {\left( {f'(x)\left( {\int {g(x)dx} } \right)} \right)} } } dx$.
On putting $f(x) = {x^2}$ and $g(x) = {e^x}$ we get the value of above equation as,
\[
\Rightarrow \int {\left( {{x^2}{e^x}} \right)dx = {x^2}\int {{e^x}dx - \int {\left( {2x\left( {\int {{e^x}dx} } \right)} \right)} } } dx \\
\Rightarrow \int {\left( {{x^2}{e^x}} \right)dx = {x^2}{e^x} - \int {2x{e^x}dx} } \\
\]
We again got the term \[\int {2x{e^x}dx} \] to integrate it by parts. So, on integrating it by parts we get the value of the equation as,
$ \Rightarrow \int {2x{e^x}dx} = 2x{e^x} - 2{e^x}$
So, the answer is \[{x^2}{e^x} - 2x{e^x} - 2{e^x} + C\] where $C$ is the constant of integration.
Note: In such types of problems of integrals we need to recall integration by parts formula. Then we have to decide which is to be integrated and which term should be differentiated so that we can get the answer easily. In accordance with this concept, we decide the 1st and 2nd term first term is differentiated and the second is integrated. One of the most famous rules to decide first and second term integrate it by part is the ILATE rule which means: Inverse, Logarithmic, Algebraic, Trigonometric, Exponent which states that the inverse function should be assumed as the first function while performing the integration. A useful rule of integral by parts is ‘ILATE’. Knowing this will solve your problems and will give you the right answers.
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