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
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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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