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Question: How do you integrate \({t^5}{e^{ - t}}dt\) ?...

How do you integrate t5e−tdt{t^5}{e^{ - t}}dt ?

Explanation

Solution

Here we have to integrate the function. The function has two functions- the standard algebraic variable and the Euler’s number raised to the power of the given algebraic function. We will integrate the function using Integration By Parts or I.B.P. which states that if uu and vv are the two function then integration of the product of these function can be calculated by the formula;
∫uv dx=u×∫vdx−∫(u′×∫vdx)dx\int {uv\,dx} = u \times \int {vdx - \int {\left( {u' \times \int {vdx} } \right)} } dx

Complete step by step answer:
Integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions in smaller parts. Differentiation is the process of finding the derivative and integration is the process of finding the antiderivative of a function. So, these processes are inverse of each other.

In this question we will use integration by parts to integrate the function t5e−tdt{t^5}{e^{ - t}}dt.
Integration By Parts rule states that if the integrand function can be represented as a multiple of two or more functions say u(x)v(x)u(x)v(x) then the integration of this function can be calculated by the formula;
∫u(x)v (x)dx=u(x)×∫v(x)dx−∫(u′(x)×∫v(x)dx)dx\int {u(x)v\,(x)dx} = u(x) \times \int {v(x)dx - \int {\left( {u'(x) \times \int {v(x)dx} } \right)} } dx
where uu is the function of xx, vv is the function of xx and u′u' is the derivative of the function u(x)u(x).

The given function is t5e−tdt{t^5}{e^{ - t}}dt.
Let u=t5u = {t^5} and v=e−tv = {e^{ - t}}
Substituting these values in the above formula. We get,
⇒∫t5e−tdt=t5×∫e−tdt−∫(d(t5)dt×∫e−tdt)dt\Rightarrow \int {{t^5}{e^{ - t}}} dt = {t^5} \times \int {{e^{ - t}}dt - \int {\left( {\dfrac{{d({t^5})}}{{dt}} \times \int {{e^{ - t}}dt} } \right)} } dt
Firstly, we will the derivative of the function u(x)u(x) i.e., t5{t^5}
We know that d(xn)dx=nxn−1\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}
So, ⇒d(t5)dt=5t4 \Rightarrow \dfrac{{d({t^5})}}{{dt}} = 5{t^4}

Putting this value in the above equation. We get,
⇒∫t5e−tdt=t5×∫e−tdt−∫(5t4×∫e−tdt)dt\Rightarrow \int {{t^5}{e^{ - t}}} dt = {t^5} \times \int {{e^{ - t}}dt - \int {\left( {5{t^4} \times \int {{e^{ - t}}dt} } \right)} } dt
We know that ∫et=et\int {{e^t}} = {e^t}
so, ∫e−t=−e−t\int {{e^{ - t}}} = - {e^{ - t}}
putting this value in the equation. We get,
⇒∫t5e−tdt=t5×(−e−t)−∫(5t4×(−e−t))dt\Rightarrow \int {{t^5}{e^{ - t}}} dt = {t^5} \times ( - {e^{ - t}}) - \int {\left( {5{t^4} \times ( - {e^{ - t}})} \right)} dt
Let I1=t5×(−e−t)−∫(5t4×(−e−t))dt{I_1} = {t^5} \times ( - {e^{ - t}}) - \int {\left( {5{t^4} \times ( - {e^{ - t}})} \right)} dt

Simplifying the above equation. We get,
I1=−e−tt5+5∫(t4e−t)dt ……(1){I_1} = - {e^{ - t}}{t^5} + 5\int {\left( {{t^4}{e^{ - t}}} \right)} dt\, \ldots \ldots \left( 1 \right)
Solving ∫(t4e−t)dt\int {\left( {{t^4}{e^{ - t}}} \right)} dt using Integration By Parts. We get,
⇒∫(t4e−t)dt=t4∫e−tdt−∫(d(t4)dt×∫e−tdt)dt\Rightarrow \int {\left( {{t^4}{e^{ - t}}} \right)} dt = {t^4}\int {{e^{ - t}}} dt - \int {\left( {\dfrac{{d({t^4})}}{{dt}} \times \int {{e^{ - t}}} dt} \right)} dt
Putting d(t4)dt=4t3\dfrac{{d({t^4})}}{{dt}} = 4{t^3} and ∫e−t=−e−t\int {{e^{ - t}}} = - {e^{ - t}} in the above equation. We get,
⇒∫(t4e−t)dt=−e−tt4−∫(4t3×(−e−t))dt\Rightarrow \int {\left( {{t^4}{e^{ - t}}} \right)} dt = - {e^{ - t}}{t^4} - \int {\left( {4{t^3} \times ( - {e^{ - t}})} \right)} dt
Substituting this value in equation (1)\left( 1 \right). We get,
⇒I1=−t5e−t+5(−e−tt4−∫(4t3×(−e−t)))dt\Rightarrow {I_1} = - {t^5}{e^{ - t}} + 5\left( { - {e^{ - t}}{t^4} - \int {\left( {4{t^3} \times ( - {e^{ - t}})} \right)} } \right)dt

Simplifying the above equation. We get,
⇒I1=−t5e−t−5e−tt4+20∫(t3e−t)dt ……(2)\Rightarrow {I_1} = - {t^5}{e^{ - t}} - 5{e^{ - t}}{t^4} + 20\int {\left( {{t^3}{e^{ - t}}} \right)} dt\, \ldots \ldots \left( 2 \right)
Solving ∫(t3e−t)dt\int {\left( {{t^3}{e^{ - t}}} \right)} dt using Integration By Parts. We get,
⇒∫(t3e−t)dt=t3∫e−tdt−∫(d(t3)dt×∫e−tdt)dt\Rightarrow \int {\left( {{t^3}{e^{ - t}}} \right)} dt = {t^3}\int {{e^{ - t}}} dt - \int {\left( {\dfrac{{d({t^3})}}{{dt}} \times \int {{e^{ - t}}} dt} \right)} dt
Putting d(t3)dt=3t2\dfrac{{d({t^3})}}{{dt}} = 3{t^2} and ∫e−t=−e−t\int {{e^{ - t}}} = - {e^{ - t}} in the above equation. We get,
⇒∫(t3e−t)dt=−e−tt3−∫(3t2×(−e−t))dt\Rightarrow \int {\left( {{t^3}{e^{ - t}}} \right)} dt = - {e^{ - t}}{t^3} - \int {\left( {3{t^2} \times ( - {e^{ - t}})} \right)} dt

Substituting this value in equation (2)\left( 2 \right). We get,
⇒I1=−t5e−t−5e−tt4+20(−e−tt3−∫(3t2×(−e−t))dt)\Rightarrow {I_1} = - {t^5}{e^{ - t}} - 5{e^{ - t}}{t^4} + 20\left( { - {e^{ - t}}{t^3} - \int {(3{t^2} \times ( - {e^{ - t}}))dt} } \right)
Simplifying the above equation. We get,
⇒I1=−t5e−t−5e−tt4−20e−tt3+60∫(t2e−t)dt ……(3)\Rightarrow {I_1} = - {t^5}{e^{ - t}} - 5{e^{ - t}}{t^4} - 20{e^{ - t}}{t^3} + 60\int {({t^2}{e^{ - t}})dt} \, \ldots \ldots \left( 3 \right)
Solving ∫(t2e−t)dt\int {\left( {{t^2}{e^{ - t}}} \right)} dt using Integration By Parts. We get,
⇒∫(t2e−t)dt=t2∫e−tdt−∫(d(t2)dt×∫e−tdt)dt\Rightarrow \int {\left( {{t^2}{e^{ - t}}} \right)} dt = {t^2}\int {{e^{ - t}}} dt - \int {\left( {\dfrac{{d({t^2})}}{{dt}} \times \int {{e^{ - t}}} dt} \right)} dt

Putting d(t2)dt=2t\dfrac{{d({t^2})}}{{dt}} = 2t and ∫e−t=−e−t\int {{e^{ - t}}} = - {e^{ - t}} in the above equation. We get,
⇒∫(t2e−t)dt=−e−tt2−∫(2t×(−e−t))dt\Rightarrow \int {\left( {{t^2}{e^{ - t}}} \right)} dt = - {e^{ - t}}{t^2} - \int {\left( {2t \times ( - {e^{ - t}})} \right)} dt
Substituting this value in equation (3)\left( 3 \right). We get,
⇒I1=−t5e−t−5e−tt4−20e−tt3+60(−e−tt2−∫(2t×(−e−t))dt)\Rightarrow {I_1} = - {t^5}{e^{ - t}} - 5{e^{ - t}}{t^4} - 20{e^{ - t}}{t^3} + 60\left( { - {e^{ - t}}{t^2} - \int {(2t \times ( - {e^{ - t}}} ))dt} \right)
Simplifying the above equation. We get,
⇒I1=−t5e−t−5e−tt4−20e−tt3−60e−tt2+120∫(te−t)dt ……(4)\Rightarrow {I_1} = - {t^5}{e^{ - t}} - 5{e^{ - t}}{t^4} - 20{e^{ - t}}{t^3} - 60{e^{ - t}}{t^2} + 120\int {(t{e^{ - t}}} )dt\, \ldots \ldots \left( 4 \right)

Solving ∫(te−t)dt\int {\left( {t{e^{ - t}}} \right)} dt using Integration By Parts. We get,
⇒∫(te−t)dt=t∫e−tdt−∫(d(t)dt×∫e−tdt)dt\Rightarrow \int {\left( {t{e^{ - t}}} \right)} dt = t\int {{e^{ - t}}} dt - \int {\left( {\dfrac{{d({t^{}})}}{{dt}} \times \int {{e^{ - t}}} dt} \right)} dt
Putting d(t2)dt=1\dfrac{{d({t^2})}}{{dt}} = 1 and ∫e−t=−e−t\int {{e^{ - t}}} = - {e^{ - t}} in the above equation. We get,
⇒∫(te−t)dt=−e−tt−∫(1×(−e−t))dt\Rightarrow \int {\left( {t{e^{ - t}}} \right)} dt = - {e^{ - t}}t - \int {\left( {1 \times ( - {e^{ - t}})} \right)} dt
Substituting this value in equation (4)\left( 4 \right). We get,
⇒I1=−t5e−t−5e−tt4−20e−tt3−60e−tt2+120(−e−tt−∫(1×(−e−t))dt)\Rightarrow {I_1} = - {t^5}{e^{ - t}} - 5{e^{ - t}}{t^4} - 20{e^{ - t}}{t^3} - 60{e^{ - t}}{t^2} + 120\left( { - {e^{ - t}}t - \int {(1 \times ( - {e^{ - t}}} ))dt} \right)

Simplifying the above equation. We get,
⇒I1=−t5e−t−5e−tt4−20e−tt3−60e−tt2−120e−tt+120∫e−tdt\Rightarrow {I_1} = - {t^5}{e^{ - t}} - 5{e^{ - t}}{t^4} - 20{e^{ - t}}{t^3} - 60{e^{ - t}}{t^2} - 120{e^{ - t}}t + 120\int {{e^{ - t}}} dt
Putting ∫e−t=−e−t\int {{e^{ - t}}} = - {e^{ - t}} in the above equation. We get,
⇒I1=−t5e−t−5e−tt4−20e−tt3−60e−tt2−120e−tt−120e−t+C\Rightarrow {I_1} = - {t^5}{e^{ - t}} - 5{e^{ - t}}{t^4} - 20{e^{ - t}}{t^3} - 60{e^{ - t}}{t^2} - 120{e^{ - t}}t - 120{e^{ - t}} + C
where CC is a constant of integration.
Taking −e−t - {e^{ - t}} common from the above equation. We get,
∴I1=−e−t(t5+5t4+20t3+60t2+120t+120)+C\therefore {I_1} = - {e^{ - t}}\left( {{t^5} + 5{t^4} + 20{t^3} + 60{t^2} + 120t + 120} \right) + C

Hence, the integration of t5e−tdt{t^5}{e^{ - t}}dt is −e−t(t5+5t4+20t3+60t2+120t+120)+C - {e^{ - t}}\left( {{t^5} + 5{t^4} + 20{t^3} + 60{t^2} + 120t + 120} \right) + C.

Note: In Integration By Parts we use ILATE rule that helps us to select the first function and second function by this we can decide which term we should differentiate and which term should we integrate. ILATE stands for Inverse Logarithmic Algebraic Trigonometric Exponential. The term which is closer to I is differentiated first and the term which is closer to E is integrated first.