# How do you test the improper integral #int (2x-1)^-3dx# from #[0,1/2]# and evaluate if possible?

The integral is divergent.

To calculate the improper integral, proceed as follows :

First, calculate the indefinite integral

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To test the improper integral ( \int_{0}^{1/2} (2x - 1)^{-3} , dx ), first check if the function ( (2x - 1)^{-3} ) is continuous on the interval ([0, 1/2]). Since the function is continuous except at (x = 1/2), the integral is improper due to a discontinuity at (x = 1/2).

To evaluate the improper integral, split it into two integrals: ( \int_{0}^{1/2} (2x - 1)^{-3} , dx = \lim_{t \to 1/2^-} \int_{0}^{t} (2x - 1)^{-3} , dx + \lim_{t \to 1/2^+} \int_{t}^{1/2} (2x - 1)^{-3} , dx ).

Now, compute each limit separately. For the first limit, integrate ( (2x - 1)^{-3} ) from (0) to (t), then take the limit as (t) approaches (1/2) from the left. For the second limit, integrate ( (2x - 1)^{-3} ) from (t) to (1/2), then take the limit as (t) approaches (1/2) from the right.

After evaluating the integrals and limits, you'll have the value of the improper integral if it exists.

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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.

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.

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.

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