# 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

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)+...

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