# How do you evaluate #int abs(x-1)dx# for [0, 5]?

Therefore,

graph{|x-1| [-1.926, 8.074, -0.84, 4.16]}

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To evaluate ∫|x-1| dx for [0, 5], we first identify the intervals where the absolute value function changes behavior, which is at x = 1. Then we split the integral into two parts:

∫|x-1| dx = ∫(x-1) dx from 0 to 1 + ∫(1-x) dx from 1 to 5

For the interval [0, 1]:

∫(x-1) dx = (x^2/2 - x) evaluated from 0 to 1 = (1/2 - 1) - (0 - 0) = -1/2

For the interval [1, 5]:

∫(1-x) dx = (x - x^2/2) evaluated from 1 to 5 = (5 - 25/2) - (1 - 1/2) = 3/2

Therefore, the value of ∫|x-1| dx for [0, 5] is the sum of the two intervals:

∫|x-1| dx = -1/2 + 3/2 = 1.

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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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