How do you evaluate the definite integral #int (5-16x^-3) dx# from #[1,2]#?
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To evaluate the definite integral ( \int_{1}^{2} (5 - 16x^{-3}) , dx ):
- First, integrate the function (5 - 16x^{-3}) with respect to (x) to find the antiderivative.
- Evaluate the antiderivative at the upper limit of integration (2) and subtract the value of the antiderivative at the lower limit of integration (1).
The antiderivative of (5 - 16x^{-3}) with respect to (x) is (5x + 16x^{-2}/2 + C).
Evaluating the antiderivative at the upper limit (2): [ (5 \cdot 2) + \frac{16}{2(2)^2} = 10 + 4 = 14 ]
Evaluating the antiderivative at the lower limit (1): [ (5 \cdot 1) + \frac{16}{2(1)^2} = 5 + 8 = 13 ]
Subtract the lower limit value from the upper limit value: [ 14 - 13 = 1 ]
Therefore, the value of the definite integral ( \int_{1}^{2} (5 - 16x^{-3}) , dx ) is 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.
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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