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...

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 info

lfryerda eNotes educator | Certified Educator

calendarEducator since 2012

write738 answers

starTop subjects are Math and Science

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).

check Approved by eNotes Editorial

Unlock This Answer Now