Question

How do you integrate $\int {{x^2}{e^{ - x}}dx}$ using integration by parts?

Answer 1

Integration by parts means that one needs to break the function into two parts and then solve it by the formula of integration by parts which is as follows:
$\int {\left( {u \cdot v} \right)} dx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}} \right)} } \cdot \left( {\int {vdx} } \right)dx$ where ‘u’ and ‘v’ are both functions of ‘x’.
Also, while selecting ‘u’ and ‘v’ we follow the ILATE convention.
Formula Used:
$\int {\left( {u \cdot v} \right)} dx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}} \right)} } \cdot \left( {\int {vdx} } \right)dx$ where ‘u’ and ‘v’ are both functions of ‘x’.

Complete step by step solution:
When doing integration by parts, we need to select two functions such that when combined, they complete the equation, as such, becoming parts of the equation. Of them, one becomes the first function, represented by ‘u’ and the other becomes the second function represented by ‘v’
For selecting the first and second function, we follow the ILATE convention wherein I stand for inverse, L stands for logarithmic function, A stands for arithmetic function, T stands for trigonometric function and E stands for an exponential function. We check the original equation and assign the first and second functions on the basis of the order in which they appear in ILATE.
The original equation is $\int {{x^2}{e^{ - x}}dx}$ which can be broken into two parts which are ${x^2}$ and ${e^{ - x}}$. Here ${x^2}$ being an arithmetic function comes before ${e^{ - x}}$ which is an exponential function, making it the first function and ${e^{ - x}}$ becomes the second function.
Putting it into the equation, $\int {\left( {u \cdot v} \right)} dx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}} \right)} } \cdot \left( {\int {vdx} } \right)dx$ it becomes
$\ \int {{x^2}{e^{ - x}}} dx = {x^2}\int {{e^{ - x}}dx - \int {\left( {\dfrac{{d\left( {{x^2}} \right)}}{{dx}}} \right)} } \cdot \left( {\int {{e^{ - x}}dx} } \right)dx \\\ = - {x^2}{e^{ - x}} - \int {2x\left( { - {e^{ - x}}} \right)} dx \\\ = - {x^2}{e^{ - x}} + \int {2x \cdot {e^{ - x}}} dx \\\ \$
Applying integration by parts on the second half of the equation again, it becomes

\int {{x^2}{e^{ - x}}} dx = - {x^2}{e^{ - x}} + \int {2x \cdot {e^{ - x}}} dx \\\ = - {x^2}{e^{ - x}} + 2x\int {{e^{ - x}}} dx - \int {\left( {\dfrac{{d\left( {2x} \right)}}{{dx}}} \right)} \cdot \left( {\int {{e^{ - x}}} dx} \right)dx \\\ = - {x^2}{e^{ - x}} - 2x{e^{ - x}} - \int {2\left( { - {e^{ - x}}} \right)dx} \\\ = - {x^2}{e^{ - x}} - 2x{e^{ - x}} + \int {2{e^{ - x}}dx} \\\ = - {x^2}{e^{ - x}} - 2x{e^{ - x}} - 2{e^{ - x}} \\\ = - {e^{ - x}}\left( {{x^2} + 2x + 2} \right)+C \\\ \ $$ Where C is integral constant. **Note:** Here in integration by parts, sometimes we face questions with multiple integrations. In this case, we had to do integration by parts only two times so it was fairly easy but if there are more integrations involved, following this simple algorithm will give the solution $\int {\left( {u \cdot v} \right)dx} = u{v_1} - u'{v_2} + u''{v_3} - u'''{v_4} + ........$ where dash above ‘u’ represents the number of times it is differentiated whereas the subscript of ‘v’ denoted the number of times it has been integrated.