# Determine whether the integral is divergent or convergent. integrate from 2 to 8 of (1)/((x-6)^3)dxIf it is convergent, evaluate it. If it diverges to infinity, state your answer as INF. If it...

Determine whether the integral is divergent or convergent.

integrate from 2 to 8 of (1)/((x-6)^3)dx

If it is convergent, evaluate it. If it diverges to infinity, state your answer as INF. If it diverges to negative infinity, state your answer as MINF. If it diverges without being infinity or negative infinity, state your answer as DIV.

Asked on by master451

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mathsworkmusic | (Level 2) Educator

Posted on

`int_2^8 1/((x-6)^3) dx = -1/(2(x-6)^2)|_2^8 = -1/2(1/(2^2) - 1/(-4)^2)=-1/2(1/4 - 1/16) `

`= -1/2(3/16) = -3/32`

The integral converges to a value of -3/32

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

Since the problem requests to decide if the given integral converges or diverges, then the upper limit must be `oo`  and not 8.

Hence, you should evaluate the following limit to check if the integral converges or diverges, such that:

`int_2^oo 1/((x-6)^3)dx = lim_(n->oo) int_2^n 1/((x-6)^3)dx`

You may evaluate the integral `int 1/((x-6)^3)dx`  using the following substitution such that:

`x-6 = t => dx = dt`

Changing the variable yields:

`int 1/((x-6)^3)dx= int 1/(t^3)dt`

Using the property of negative power yields:

`int 1/(t^3)dt = int (t^(-3))dt = t^(-3+1)/(-3+1) + c`

`int 1/(t^3)dt = int (t^(-3))dt = -1/(2t^2) + c`

Substituting back `x-6`  for t yields:

`int 1/((x-6)^3)dx=-1/(2(x-6)^2) + c `

You may evaluate the definite integral using the fundamental theorem of calculus such that:

`int_2^n 1/((x-6)^3)dx = (-1/(2(x-6)^2))|_2^n`

`int_2^n 1/((x-6)^3)dx =-1/(2(n-6)^2) +1/(2(2-6)^2)`

`int_2^n 1/((x-6)^3)dx = 1/32 - 1/(2(n-6)^2)`

Evaluating the limit yields:

`lim_(n->oo) int_2^n 1/((x-6)^3)dx = lim_(n->oo)(1/32 - 1/(2(n-6)^2))`

`lim_(n->oo) int_2^n 1/((x-6)^3)dx = lim_(n->oo)(1/32) - lim_(n->oo) 1/(2(n-6)^2))`

`lim_(n->oo) int_2^n 1/((x-6)^3)dx =1/32 - (1/2)lim_(n->oo) 1/(n^2(1 - 6/n)^2)`

`lim_(n->oo) int_2^n 1/((x-6)^3)dx = 1/32 - (1/2)*(1/oo)`

`lim_(n->oo) int_2^n 1/((x-6)^3)dx = 1/32 - (1/2)*(0)`

`lim_(n->oo) int_2^n 1/((x-6)^3)dx = 1/32`

Hence, evaluating the improper integral yields `int_2^oo 1/((x-6)^3)dx = lim_(n->oo) int_2^n 1/((x-6)^3)dx = 1/32` , thus, the intgeral converges.

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