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Let f(x) be a continuous function defined on the interval [2, infinity)such that...

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hockeyfan54 | Student | Salutatorian

Posted November 2, 2012 at 3:35 AM via web

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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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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted November 13, 2012 at 6:55 AM (Answer #1)

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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` .

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