Question

Let f(x) be a continuous function defined on the interval [2, infinity)such that
f(3)=5
abs(f(x)) is less than (x^6)+1
 and
integrate from 3 to infinity of f(x)e^(-x/6)dx=-8
Determine the value of
integrate from 3 to infinity of f'(x)e^(-x/6)dx

Accepted Answer

You should solve the improper integral int_3^oo f'(x)e^(-x/6)dx , using integration by parts such that:
int udv = uv - int vdu
You should consider u = f(x) => du = f'(x)dx  and dv= e^(-x/6)dx => v = int e^(-x/6)dx Using the substitution -x/6 = t => -dx/6 = dt => dx = -6dt  yields:
int e^t*(-6dt) = -6e^t + c => v = -6e^(-x/6)
Hence, evaluating the integral yields:
int_3^n f(x)e^(-x/6)dx = -6f(x)e^(-x/6)+ 6int_3^n f'(x)*e^(-x/6)dx
Since the problem provides the information that lim_(n->oo)int_3^n f(x)e^(-x/6)dx = -8 , hence, you may evaluate int_3^n f'(x)*e^(-x/6)dx  such that:
-8 = lim_(n->oo)(-6f(n)e^(-n/6)+ 6f(3)e^(-3/6))+ 6int_3^oo f'(x)*e^(-x/6)dx
6int_3^oo f'(x)*e^(-x/6)dx = -8- lim_(n->oo)(-6f(n)e^(-n/6) + 6f(3)e^(-3/6))
Since the problem provides the information that f(3) = 5  yields:
6int_3^oo f'(x)*e^(-x/6)dx = -8 - 6*5*e^(-1/2) - lim_(n->oo)(-6f(n)e^(-n/6))
Since the problem provides the information that |f(x)|<x^6+1  yields:
6int_3^oo f'(x)*e^(-x/6)dx = -8 - 6*5*e^(-1/2) - lim_(n->oo)(-6(n^6+1)e^(-n/6))
6int_3^oo f'(x)*e^(-x/6)dx = -8 - 30/sqrt e+ 6lim_(n->oo)((n^6+1)e^(-n/6))
You need to evaluate the limit lim_(n->oo)((n^6+1)e^(-n/6))  such that:
lim_(n->oo)((n^6+1)e^(-n/6)) = lim_(n->oo)(n^6+1)/(e^(n/6)) = oo/oo
Since evaluating the limit yields the indetermination oo/oo, then you may use l'Hospital's theorem such that:
lim_(n->oo)(n^6+1)/(e^(n/6)) = lim_(n->oo)((n^6+1)')/((e^(n/6)) ')
lim_(n->oo)(n^6+1)/(e^(n/6)) = lim_(n->oo)(6n^5)/((1/6)e^(n/6)) = oo/oo
lim_(n->oo)(30n^4)/((1/36)e^(n/6)) = oo/oo
lim_(n->oo)(30*4*3*2*1)/((1/36)*(1/36)*(1/36)*e^(n/6)) = 720/oo = 0
6int_3^oo f'(x)*e^(-x/6)dx = -8 - 30/sqrt e + 0
You need to divide by 6 such that:
int_3^oo f'(x)*e^(-x/6)dx = -4/3 - 5/sqrt e
Hence, evaluating the improper integral  int_3^oo f'(x)e^(-x/6)dx,  using the information provided by the problem, yields int_3^oo f'(x)*e^(-x/6)dx = -4/3 - 5/sqrt e .

Math

Let f(x) be a continuous function defined on the interval [2, infinity)such that f(3)=5 abs(f(x)) is less than (x^6)+1 and integrate from 3 to infinity of f(x)e^(-x/6)dx=-8 Determine the value of integrate from 3 to infinity of f'(x)e^(-x/6)dx

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