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How do you evaluate the integral #int 1/(2x-1)# from 1 to 2?

Answer 1

Ezekiel Alexander

I found: #ln(3)/2#

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Answer 2

Charles Arocha

To evaluate the integral (\int_{1}^{2} \frac{1}{2x-1} , dx):

Perform a substitution: Let ( u = 2x - 1 ). Then, ( du = 2 , dx ) or ( \frac{1}{2} du = dx ).
Substitute the limits of integration: When ( x = 1 ), ( u = 2(1) - 1 = 1 ). When ( x = 2 ), ( u = 2(2) - 1 = 3 ).
Rewrite the integral in terms of ( u ): [ \int \frac{1}{u} \cdot \frac{1}{2} , du ]
Integrate: [ \frac{1}{2} \int \frac{1}{u} , du = \frac{1}{2} \ln|u| + C ] Where ( C ) is the constant of integration.
Substitute back for ( u ): [ \frac{1}{2} \ln|2x-1| ]
Evaluate from 1 to 2: [ \frac{1}{2} \ln|2(2)-1| - \frac{1}{2} \ln|2(1)-1| ] [ = \frac{1}{2} \ln(3) - \frac{1}{2} \ln(1) ] [ = \frac{1}{2} \ln(3) ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.