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NARA [144]
2 years ago
13

8ntergrate by parts |x^3 e^-x dx​

Mathematics
1 answer:
Lana71 [14]2 years ago
7 0

I assume the integral is

\displaystyle \int x^3 e^{-x} \, dx

Integrate by parts with

u = x^3 \implies du = 3x^2 \, dx

dv = e^{-x} \, dx \implies v = -e^{-x}

so that

\displaystyle \int u \, dv = uv - \int v \, du

i.e.

\displaystyle \int x^3 e^{-x} \, dx = -x^3 e^{-x} + 3 \int x^2 e^{-x} \, dx

Integrate by parts again, this time with

u = x^2 \implies du = 2x \, dx

dv = e^{-x} \, dx \implies v = -e^{-x}

and so

\displaystyle \int x^2 e^{-x} \, dx = -x^2 e^{-x} + 2 \int x e^{-x} \, dx

\implies \displaystyle \int x^3 e^{-x} \, dx = -x^3 e^{-x} + 3 \left(-x^2 e^{-x} + 2 \int x e^{-x} \, dx\right)

\implies \displaystyle \int x^3 e^{-x} \, dx = -x^3 e^{-x} - 3x^2 e^{-x} + 6 \int x e^{-x} \, dx

Once more, with

u = x \implies du = dx

dv = e^{-x} \, dx \implies v = -e^{-x}

\displaystyle \int x e^{-x} \, dx = -xe^{-x} + \int e^{-x} \, dx

\implies \displaystyle \int x^3 e^{-x} \, dx = -x^3 e^{-x} - 3x^2 e^{-x} + 6 \left(-xe^{-x} + \int e^{-x} \, dx\right)

\implies \displaystyle \int x^3 e^{-x} \, dx = -x^3 e^{-x} - 3x^2 e^{-x} -6xe^{-x} + 6 \int e^{-x} \, dx

Finally,

\displaystyle \int e^{-x} \, dx = -e^{-x} + C

and so

\displaystyle \int x^3 e^{-x} \, dx = -x^3 e^{-x} - 3x^2 e^{-x} -6xe^{-x} - 6 e^{-x} + C

or equivalently,

\displaystyle \int x^3 e^{-x} \, dx = \boxed{-(x^3+3x^2+6x+6) e^{-x} + C}

We can also approach the integral more generally by considering

I_n = \displaystyle \int x^n e^{-x} \, dx

Integrate by parts with

u = x^n \implies du = nx^{n-1} \, dx

dv = e^{-x} \, dx \implies v = -e^{-x}

Then

\displaystyle I_n = -x^n e^{-x} + n \int x^{n-1} e^{-x} \, dx = -x^n e^{-x} + n I_{n-1}

We can solve the recurrence using the initial "value" I_0 = -e^{-x}+C to find a general formula for I_n. Then we get the integral we want I_3 "for free".

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