# Using integration by parts, we find that int x^(n)e^(-x) dx=

The integral int x^n e^-x dx has to be determined.

Integration by parts gives us the rule: int u dv = u*v - int v du

Let u = x^n and dv = e^-x dx

du = n*x^(n-1) dx

v = -e^(-x)

int x^n e^-x dx

= x^n*-1*e^-x -...

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The integral int x^n e^-x dx has to be determined.

Integration by parts gives us the rule: int u dv = u*v - int v du

Let u = x^n and dv = e^-x dx

du = n*x^(n-1) dx

v = -e^(-x)

int x^n e^-x dx

= x^n*-1*e^-x - int -1*e^-x*n*x^(n-1) dx

= x^n*-1*e^-x + n*int e^-x*x^(n-1) dx

= -x^n*e^-x + n*int e^-x*x^(n-1) dx

int e^-x*x^(n-1) dx

= -x^(n-1)*e^-x + (n-1)*int e^-x*x^(n-2) dx

Substituting this in the original integral

int x^n*e^-x dx

= -x^n*e^-x + n*(-x^(n-1)*e^-x + (n-1)*int e^-x*x^(n-2) dx)

= -x^n*e^-x- n*x^(n-1)*e^-x + n*(n-1)*int e^-x*x^(n-2) dx

= -e^-x*(x^n+n*x^(n-1))+n*(n-1)*int e^-x*x^(n-2) dx

This can be continued n times to yield the final result.

= -e^-x*(x^n+n*x^(n-1)+ n*(n-1)x^(n-2)+...n!)

The integral int x^n*e^(-x)dx = -e^-x*(x^n+n*x^(n-1)+ n*(n-1)x^(n-2)+...n!) + C

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The formula for the integration by parts is as follows:

int udv = uv - int vdu

To take the given integral by parts, let

u = x^n

and dv = e^(-x) dx

Then, du = nx^(n- 1) and v = -e^(-x)

Plugging these into the formula, we get

int x^n e^(-x) dx = x^n(-e^(-x)) - int (-e^(-x))nx^(n-1)dx

= -x^n e^(-x) + n int x^(n-1)e^(-x) dx

The resultant integral can be taken by parts again in a similar way, resulting in the following:

int x^(n-1)e^(-x) dx = -x^(n-1)e^(-x) + (n-1)int x^(n-2)e^(-x) dx

For a given value of n, this process can be repeated times until the power of x in the integral is 0. Then, the final integral can be taken as follows:

int e^(-x) dx = -e^(-x)

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We need to find intx^n e^(-x)dx.

Let's solve the problem for n=1.

int xe^(-x)dx=|(u=x, dv=e^(-x)),(du=dx, v=-e^(-x))|=-xe^(-x)+e^(-x)=-e^(-x)(x+1) 

Let's now solve the problem for n=2.

int x^2 e^(-x)dx=|(u=x^2, dv=e^(-x)),(du=2xdx, v=-e^(-x))|=-x^2e^(-x)+2int xe^(-x)dx=

-x^2e^(-x)-2e^(-x)(x+1)=-e^(-x)(x^2+2x+2)

Similarly for n=3 we would have

int x^3e^(-x)dx=-e^(-x)(x^3+3x^2+6x+6)

We can now see the pattern

int x^n e^(-x)dx=-e^(-x) sum_(k=0)^n (x^n)^((k))   <-- Solution

where (x^n)^((k)) is the kth derivation of x^n. If we were to expand the above formula it would look like this

int x^n e^(-x)dx=-e^(-x)(x^n+nx^(n-1)+n(n-1)x^(n-2)+cdots+n!x+n!)`

If you wish you can prove the formula formally by using mathematical induction.

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