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

Answer 1

Paisley Johnson

I found: #-3/4#

Given: #int_0^3(x/2-1)dx=# #[x^2/4-x]_0^3=# substituting the extrema: #=(9/4-3)-0=(9-12)/4=-3/4#

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Answer 2

Christopher Agresta

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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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.