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

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

Explanation

Solution

In order to solve this definite integral we will use the formula of integration by parts i.e., ∫(uv)dx=u∫vdx−∫(dudx∫vdx) dx\int {\left( {uv} \right)dx = u\int {vdx} - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } {\text{ }}dx where uu and vv are different function in xx . 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 ∫te−tdt\int {t{e^{ - t}}dt} from [0,6]\left[ {0,6} \right]
From this statement, we understand that the lower limit of the given definite integral is 00 and the upper limit of the given definite integral is 66
Thus, the definite integral is given as
∫06te−tdt\int\limits_0^6 {t{e^{ - t}}dt}
Let us assume
I=∫06te−tdtI = \int\limits_0^6 {t{e^{ - t}}dt}
Now we know that if there are two functions in xx those are uu and vv and we have to integrate uvuv with respect to xx then we get the value of integration using the formula of integration by parts i.e., ∫(uv)dx=u∫vdx−∫(dudx∫vdx) dx\int {\left( {uv} \right)dx = u\int {vdx} - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} } {\text{ }}dx
So, in the question
u=tu = t and v=e−tv = {e^{ - t}}
Therefore, we get
I=∫06te−tdt=[t∫e−tdt−∫(dtdt∫e−tdt)dt]06I = \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
∫e−xdx=−e−x\int {{e^{ - x}}dx} = - {e^{ - x}}
Therefore, we get
I=[t(−e−t)−∫1(−e−t)dt]06I = \left[ {t\left( { - {e^{ - t}}} \right) - \int {1\left( { - {e^{ - t}}} \right)dt} } \right]_0^6
Again, integrating we get
I=[t(−e−t)−(−(−e−t))]06I = \left[ {t\left( { - {e^{ - t}}} \right) - \left( { - \left( { - {e^{ - t}}} \right)} \right)} \right]_0^6
On simplifying, we get
I=[t(−e−t)−e−t]06I = \left[ {t\left( { - {e^{ - t}}} \right) - {e^{ - t}}} \right]_0^6
Taking common −e−t - {e^{ - t}} we get
I=[−e−t(t+1)]06I = \left[ { - {e^{ - t}}\left( {t + 1} \right)} \right]_0^6
Applying upper and lower limits, we get
I=[−e−6(6+1)−(−e0(0+1))]I = \left[ { - {e^{ - 6}}\left( {6 + 1} \right) - \left( { - {e^0}\left( {0 + 1} \right)} \right)} \right]
⇒I=[−e−6(7)+1]\Rightarrow I = \left[ { - {e^{ - 6}}\left( 7 \right) + 1} \right]
⇒I=1−7e−6\Rightarrow I = 1 - 7{e^{ - 6}}
Now we know that
a−n=1an{a^{ - n}} = \dfrac{1}{{{a^n}}}
Therefore, we get
⇒I=1−7e6\Rightarrow I = 1 - \dfrac{7}{{{e^6}}}
Therefore, the definite integral ∫te−tdt\int {t{e^{ - t}}dt} from [0,6]\left[ {0,6} \right] is equal to 1−7e61 - \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 tt is considered as the first function because it is an algebraic function and e−t{e^{ - t}} is considered as the second function because it is an exponent function.