How do you evaluate the definite integral \[\int {t{e^{ - t}}dt} \] from \[\left[ {0,6} \right]?\]

Hint: In order to solve this definite integral we will use the formula of integration by parts i.e., \[\int {\left( {uv} \right)dx = u\int {vdx} - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } {\text{ }}dx\] where \[u\] and \[v\] are different function in \[x\] . After finding the integral we will apply the upper limit and lower limit. Further we will simplify and hence we will get the result.

Complete answer:
Given that we have to evaluate the definite integral of \[\int {t{e^{ - t}}dt} \] from \[\left[ {0,6} \right]\]
From this statement, we understand that the lower limit of the given definite integral is \[0\] and the upper limit of the given definite integral is \[6\]
Thus, the definite integral is given as
\[\int\limits_0^6 {t{e^{ - t}}dt} \]
Let us assume
\[I = \int\limits_0^6 {t{e^{ - t}}dt} \]
Now we know that if there are two functions in \[x\] those are \[u\] and \[v\] and we have to integrate \[uv\] with respect to \[x\] then we get the value of integration using the formula of integration by parts i.e., \[\int {\left( {uv} \right)dx = u\int {vdx} - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } {\text{ }}dx\]
So, in the question
\[u = t\] and \[v = {e^{ - t}}\]
Therefore, we get
\[I = \int\limits_0^6 {t{e^{ - t}}dt} = \left[ {t\int {{e^{ - t}}dt - \int {\left( {\dfrac{{dt}}{{dt}}\int {{e^{ - t}}dt} } \right)dt} } } \right]_0^6\]
We know that
\[\int {{e^{ - x}}dx} = - {e^{ - x}}\]
Therefore, we get
\[I = \left[ {t\left( { - {e^{ - t}}} \right) - \int {1\left( { - {e^{ - t}}} \right)dt} } \right]_0^6\]
Again, integrating we get
\[I = \left[ {t\left( { - {e^{ - t}}} \right) - \left( { - \left( { - {e^{ - t}}} \right)} \right)} \right]_0^6\]
On simplifying, we get
\[I = \left[ {t\left( { - {e^{ - t}}} \right) - {e^{ - t}}} \right]_0^6\]
Taking common \[ - {e^{ - t}}\] we get
\[I = \left[ { - {e^{ - t}}\left( {t + 1} \right)} \right]_0^6\]
Applying upper and lower limits, we get
\[I = \left[ { - {e^{ - 6}}\left( {6 + 1} \right) - \left( { - {e^0}\left( {0 + 1} \right)} \right)} \right]\]
\[ \Rightarrow I = \left[ { - {e^{ - 6}}\left( 7 \right) + 1} \right]\]
\[ \Rightarrow I = 1 - 7{e^{ - 6}}\]
Now we know that
\[{a^{ - n}} = \dfrac{1}{{{a^n}}}\]
Therefore, we get
\[ \Rightarrow I = 1 - \dfrac{7}{{{e^6}}}\]
Therefore, the definite integral \[\int {t{e^{ - t}}dt} \] from \[\left[ {0,6} \right]\] is equal to \[1 - \dfrac{7}{{{e^6}}}\]

Note: In such types of problems of integrals we need to recall the formula of integration by parts. In accordance with this concept, then we have to decide the first and second term. One of the most important rule to decide the first and second term is the ILATE rule which means: Inverse, Logarithmic, Algebraic, Trigonometric and Exponent. Like in this problem \[t\] is considered as the first function because it is an algebraic function and \[{e^{ - t}}\] is considered as the second function because it is an exponent function.