How do you evaluate the definite integral #int (5-16x^-3) dx# from #[1,2]#?

Answer 1

#int_1^2 (5-16x^-3)d x=1#

#int_1^2 (5-16x^-3)d x=?#
#int_1^2 (5-16x^-3)d x=|5x+16*1/2x^-2|_1^2#
#int_1^2 (5-16x^-3)d x=(5*2+8*2^-2)-(5*1+8*1^-2)#
#int_1^2 (5-16x^-3)d x=(10+8/4)-(5+8/1)#
#int_1^2 (5-16x^-3)d x=14-13#
#int_1^2 (5-16x^-3)d x=1#
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Answer 2

To evaluate the definite integral ( \int_{1}^{2} (5 - 16x^{-3}) , dx ):

  1. First, integrate the function (5 - 16x^{-3}) with respect to (x) to find the antiderivative.
  2. 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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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.

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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