How do you evaluate the integral of #int (2 -1/x)dx# from 1/2 to 3?

Answer 1

Christopher Agresta

#int_(1/2)^3 (2-1/x)dx =5 - ln6#

Using the linearity of the integral:

#int_(1/2)^3 (2-1/x)dx = 2int_(1/2)^3dx - int_(1/2)^3 dx/x#

#int_(1/2)^3 (2-1/x)dx = 2(3-1/2) - (ln3-ln(1/2))#

#int_(1/2)^3 (2-1/x)dx =5 - ln6#

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Answer 2

Samantha Adkison

To evaluate the integral ∫(2 - 1/x)dx from 1/2 to 3, you first find the antiderivative of the integrand, which is 2x - ln|x|. Then, you evaluate this antiderivative at the upper limit (3) and subtract its value at the lower limit (1/2).

∫(2 - 1/x)dx = [2x - ln|x|] evaluated from 1/2 to 3 = (2(3) - ln|3|) - (2(1/2) - ln|1/2|) = (6 - ln(3)) - (1 - ln(1/2)) = 6 - ln(3) - 1 + ln(2) = 5 - ln(3) + ln(2)

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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.