How do you evaluate the definite integral #int abs(x-5) dx# from #[0,10]#?
25
you can see this in the graph
graph{abs(x-5) [-5.375, 14.625, -2.16, 7.84]}
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To evaluate the definite integral (\int_{0}^{10} |x - 5| , dx), we first need to identify the intervals where the function inside the absolute value changes its behavior. In this case, (|x - 5|) changes from (x - 5) to (-(x - 5)) at (x = 5).
So, we split the integral into two parts:
- From (x = 0) to (x = 5): (\int_{0}^{5} (5 - x) , dx)
- From (x = 5) to (x = 10): (\int_{5}^{10} (x - 5) , dx)
Integrating each part:
- (\int_{0}^{5} (5 - x) , dx = [5x - \frac{x^2}{2}]_{0}^{5} = (5(5) - \frac{5^2}{2}) - (5(0) - \frac{0^2}{2}) = (25 - \frac{25}{2}) - (0) = \frac{25}{2})
- (\int_{5}^{10} (x - 5) , dx = [\frac{x^2}{2} - 5x]_{5}^{10} = (\frac{10^2}{2} - 5(10)) - (\frac{5^2}{2} - 5(5)) = (50 - 50) - (\frac{25}{2} - 25) = 0 - (\frac{25}{2} - 25) = -(\frac{25}{2} - 25))
Adding both parts:
(\frac{25}{2} - (\frac{25}{2} - 25) = \frac{25}{2} - \frac{25}{2} + 25 = 25)
Therefore, the value of the definite integral (\int_{0}^{10} |x - 5| , dx) is (25).
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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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