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Question: How do you integrate \[\int {t{e^{ - t}}}dt \] using integration by parts?...

How do you integrate ∫te−tdt\int {t{e^{ - t}}}dt using integration by parts?

Explanation

Solution

In the given question, we have been given a function that needs to be integrated. The function has two functions as a product – the standard algebraic variable and the Euler’s number raised to the power of the given algebraic function. We have to integrate the whole function. To do that, we are going to solve it using Integration by Parts, or I.B.P. We are going to have to make substitutions into the formula and then solve the question.

Formula Used:
We are going to use the formula of I.B.P. which is,
∫udv=uv−∫vdu\int {udv = uv - \int {vdu} }

Complete step by step answer:
The given function is te−tdtt{e^{ - t}}dt.
Let u=t⇒dudt=1u = t \Rightarrow \dfrac{{du}}{{dt}} = 1
Then, if dvdt=e−t\dfrac{{dv}}{{dt}} = {e^{ - t}} then v=−e−tv = - {e^{ - t}}
Now, putting in the values in the formula of I.B.P.,
∫(t)(e−t)dx=(t)(−e−t)−∫(e−t)(1)dt\int {\left( t \right)\left( {{e^{ - t}}} \right)dx = \left( t \right)\left( { - {e^{ - t}}} \right)} - \int {\left( {{e^{ - t}}} \right)\left( 1 \right)dt}
=−te−t−(−e−t)+c= - t{e^{ - t}} - \left( { - {e^{ - t}}} \right) + c
Simplifying and solving,
=e−t−te−t+c= {e^{ - t}} - t{e^{ - t}} + c
=e−t(1−t)+c= {e^{ - t}}\left( {1 - t} \right) + c

Note:
In the given question, we had to solve a function by integrating it. We applied Integration by Parts on it because the function was a product of two functions. We then made the necessary substitutions, simplified the values, and got our answer. So, it is really important that we know the formulae and where, when, and how to use them so that we can get the correct result.