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How do you evaluate the definite integral #int (1/2x-1) dx# from #[0, 3]#?

Answer 1

Paisley Johnson

The answer is #=-3/4#

Use #intx^ndx=x^(n+1)/(n+1)+C# #int_0^3(x/2-1)dx=int_0^3x/2dx-int_0^3 1*dx=(x^2/4-x)_0^3# #=(9/4-3)-0=-3/4#

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Answer 2

Ezekiel Alexander

To evaluate the definite integral (\int_{0}^{3} \frac{1}{2x-1} , dx), you first need to find the antiderivative of (\frac{1}{2x-1}), which is (\frac{1}{2} \ln|2x-1|). Then, you evaluate this antiderivative at the upper and lower limits of integration and subtract the lower value from the upper value.

So, evaluate (\frac{1}{2} \ln|2x-1|) at (x = 3) and (x = 0), then subtract the result of the evaluation at (x = 0) from the result of the evaluation at (x = 3).

[ \int_{0}^{3} \frac{1}{2x-1} , dx = \left[\frac{1}{2} \ln|2x-1|\right]_{0}^{3} = \frac{1}{2} \ln|2(3)-1| - \frac{1}{2} \ln|2(0)-1| ]

[ = \frac{1}{2} \ln|6-1| - \frac{1}{2} \ln|-1| = \frac{1}{2} \ln(5) - \frac{1}{2} \ln(1) = \frac{1}{2} \ln(5) - \frac{1}{2} \cdot 0 = \frac{1}{2} \ln(5) ]

So, (\int_{0}^{3} \frac{1}{2x-1} , dx = \frac{1}{2} \ln(5)).

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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.