# How do you integrate #(1/2)x-1dx# for the interval (0,3)?

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To integrate (1/2)x - 1 dx over the interval (0,3), you can use the definite integral formula. First, integrate (1/2)x - 1 with respect to x:

∫((1/2)x - 1) dx = (1/4)x^2 - x + C

Next, evaluate the antiderivative at the upper and lower bounds of the interval (0,3):

F(3) - F(0) = [(1/4)(3)^2 - 3] - [(1/4)(0)^2 - 0] = [(1/4)(9) - 3] - [0 - 0] = (9/4 - 3) - 0 = (9/4 - 12/4) = -3/4

So, the definite integral of (1/2)x - 1 dx over the interval (0,3) is -3/4.

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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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