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.  

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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).

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