# Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, give the answer -1.   integrate from 5 to infinity of xe^(-3x)dx

You should evaluate the following limit to check if the integral diverges or converges such that:

int_5^oo x*e^(-3x)dx

You should use the following substitution such that:

-3x = t => -3dx = dt => dx = -(dt)/3

Changing the variable yields:

int x*e^(-3x)dx = (1/9)int te^t dt

You need to use integration by parts such that:

int udv = uv - int vdu

u = t => du = dt

dv = e^t dt => v = e^t

int te^t dt = te^t - int e^t dt

int te^t dt = te^t - e^t + c

int te^t dt = e^t(t - 1) + c

Substituting back -3x for t yields:

int x*e^(-3x)dx = (1/9)e^(-3x)(-3x-1) + c

You may evaluate the definite integral yields:

int_5^oo x*e^(-3x)dx = lim_(n->oo) int_5^n x*e^(-3x)dx

int_5^oo x*e^(-3x)dx = lim_(n->oo) ((1/9)(e^(-3n)(-3n-1) - e^(-15)*(-16)))

lim_(n->oo) ((1/9)(e^(-3n)(-3n-1) - e^(-15)*(-16)))

You should evaluate the limit lim_(n->oo) (-3n-1)/(e^(3n))  such that:

lim_(n->oo) (-3n-1)/(e^(3n)) = oo/oo

Using l'Hospital's theorem yields:

lim_(n->oo) ((-3n-1)')/((e^(3n))') = lim_(n->oo) -3/(3e^(3n)) = -3/oo = 0

lim_(n->oo) ((1/9)(e^(-3n)(-3n-1) - e^(-15)*(-16))) = 16e^(-15)/9

Since the result of limit is finite, lim_(n->oo) int_5^n x*e^(-3x)dx =16e^(-15)/9 , hence, the given improper integral converges.

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