I am trying to do this integral sign going from 0 to 4 (0 on the bottom, 4 on the top) 1/(4-x) dx i got stuck. Thanks a lot!!!
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$\displaystyle \int_0^4\frac1{4-x}~dx=\lim_{a\to4}\int_0^a\frac1{4-x}~dx$
Originally Posted by Krizalid $\displaystyle \int_0^4\frac1{4-x}~dx=\lim_{a\to4}\int_0^a\frac1{4-x}~dx$ Yeah i was able to get that far, but not so sure how to solve it.
Originally Posted by davecs77 Yeah i was able to get that far, but not so sure how to solve it. If you got that far, what's the problem? $\displaystyle \lim_{a \to 4} \int_0^a \frac {1}{4 - x}~dx$ $\displaystyle = \lim_{a \to 4} \left[ - \ln (4 - x) \right]_{0}^{a}$ $\displaystyle = \lim_{a \to 4} \left( - \ln (4 - a) + \ln 4 \right)$ Now take the limit
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