$\displaystyle \int_{1}^2 (6e^{3x}-1/x)\, dx $
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= 2*e^3x-ln(x) = 2e^6-ln(2)-2*e^3
Originally Posted by foodmmm $\displaystyle \int_{1}^2 (6e^{3x}-1/x)\, dx $ You're integrating standard forms. Read this carefully (most of the formulae should be in your class notes or textbook): Table of Integrals
what did you get as the final answer? 2e^6 - ln(2) - 2e^3 ?
Originally Posted by foodmmm what did you get as the final answer? 2e^6 - ln(2) - 2e^3 ? Yes.
there is no + c on the end?
Originally Posted by foodmmm there is no + c on the end? It's a definite integral. There's only a '+ c on the end' when it's an indefininte integral.
If you add c (F(b)+c)-(F(a)+c) =F(b) -F(a) so c is omitted from definite integrals
thank you
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