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Solving integrals/ x^n*e^(-x)

Integral of x^n*e^(-x) dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  1          
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 |   n  -x   
 |  x *E   dx
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∫01e−xxn dx\int\limits_{0}^{1} e^{- x} x^{n}\, dx 
Integral(x^n*E^(-x), (x, 0, 1))
Detail solution

    UpperGammaRule(a=-1, e=n, context=E**(-x)*x**n, symbol=x)

  1. Add the constant of integration:

    −Γ(n+1,x)+constant- \Gamma\left(n + 1, x\right)+ \mathrm{constant}

The answer is:

−Γ(n+1,x)+constant- \Gamma\left(n + 1, x\right)+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                               
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 |  n  -x                         
 | x *E   dx = C - Gamma(1 + n, x)
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∫e−xxn dx=C−Γ(n+1,x)\int e^{- x} x^{n}\, dx = C - \Gamma\left(n + 1, x\right)
Simplify
The answer [src] 
Gamma(1 + n)*lowergamma(1 + n, 1)   n*Gamma(1 + n)*lowergamma(1 + n, 1)
--------------------------------- + -----------------------------------
           Gamma(2 + n)                         Gamma(2 + n)
nΓ(n+1)γ(n+1,1)Γ(n+2)+Γ(n+1)γ(n+1,1)Γ(n+2)\frac{n \Gamma\left(n + 1\right) \gamma\left(n + 1, 1\right)}{\Gamma\left(n + 2\right)} + \frac{\Gamma\left(n + 1\right) \gamma\left(n + 1, 1\right)}{\Gamma\left(n + 2\right)} 
=
= 
Gamma(1 + n)*lowergamma(1 + n, 1)   n*Gamma(1 + n)*lowergamma(1 + n, 1)
--------------------------------- + -----------------------------------
           Gamma(2 + n)                         Gamma(2 + n)
nΓ(n+1)γ(n+1,1)Γ(n+2)+Γ(n+1)γ(n+1,1)Γ(n+2)\frac{n \Gamma\left(n + 1\right) \gamma\left(n + 1, 1\right)}{\Gamma\left(n + 2\right)} + \frac{\Gamma\left(n + 1\right) \gamma\left(n + 1, 1\right)}{\Gamma\left(n + 2\right)} 
gamma(1 + n)*lowergamma(1 + n, 1)/gamma(2 + n) + n*gamma(1 + n)*lowergamma(1 + n, 1)/gamma(2 + n)

    Use the examples entering the upper and lower limits of integration.

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