# How do you evaluate the definite integral #int s^3-3s^2+2# from #[-2,2]#?

Evaluate for s = 2 and subtract the evaluation for s = - 2

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To evaluate the definite integral (\int_{-2}^{2} (s^3 - 3s^2 + 2) , ds), you first integrate the function (s^3 - 3s^2 + 2) with respect to (s) and then evaluate it at the upper and lower limits of integration (in this case, -2 and 2).

(\int_{-2}^{2} (s^3 - 3s^2 + 2) , ds = \left[\frac{1}{4}s^4 - s^3 + 2s\right]_{-2}^{2})

Now, substitute the upper limit of integration (2) into the integrated function:

(\left[\frac{1}{4}(2)^4 - (2)^3 + 2(2)\right])

Then, substitute the lower limit of integration (-2) into the integrated function:

(-\left[\frac{1}{4}(-2)^4 - (-2)^3 + 2(-2)\right])

Now, calculate the values:

(= \left[\frac{1}{4}(16) - 8 + 4\right] - \left[\frac{1}{4}(16) + 8 - 4\right])

(= \left[4 - 8 + 4\right] - \left[4 + 8 + 4\right])

(= 0 - 16)

(= -16)

So, the value of the definite integral is (-16).

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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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