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.  

Expert Answers

An illustration of the letter 'A' in a speech bubbles

To determine if the integral is convergent or divergent, we need to evaluate the integral up to the vertical asymptote of the integrand and then take the limit to the asymptote.

`int_2^8 1/(x-6)^3dx`   let `u=x-6`  then `du=dx`  and the limits become -4 and 2

`=int_{-4}^2 u^{-3}du`

`=lim_{epsilon->0}(int_{-4}^epsilon u^{-3}du)+lim_{epsilon->0}(int_epsilon^2 u^{-3}du)`

`=lim_{epsilon->0}(-1/2u^{-2}|_{-4}^epsilon)+lim_{epsilon->0}(-1/2 u^{-2}|_epsilon^2)`

`=-1/2lim_{epsilon->0}(1/epsilon^2-1/16)-1/2lim_{epsilon->0}(1/4-1/epsilon^2)`

Since each part is divergent to negative infinity, the integral is divergent to negative infinity (MINF).

Approved by eNotes Editorial Team

Posted on

Soaring plane image

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial