# How do you find the definite integral for: #(2-x^2)dx# for the intervals [2,-1]?

Definite integral would be

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To find the definite integral of ( (2 - x^2) , dx ) for the interval ([2, -1]), we first need to find the antiderivative of ( (2 - x^2) ).

The antiderivative of ( 2 ) with respect to ( x ) is ( 2x ), and the antiderivative of ( x^2 ) with respect to ( x ) is ( \frac{x^3}{3} ).

Therefore, the antiderivative of ( (2 - x^2) ) is:

[ \int (2 - x^2) , dx = 2x - \frac{x^3}{3} + C ]

Now, to find the definite integral for the interval ([2, -1]), we evaluate the antiderivative at the upper and lower limits and subtract:

[ \left[2x - \frac{x^3}{3}\right]_{-1}^2 ]

Now, plug in the upper limit:

[ = 2(2) - \frac{2^3}{3} ]

[ = 4 - \frac{8}{3} ]

And plug in the lower limit:

[ -2(-1) - \frac{(-1)^3}{3} ]

[ = -2 + \frac{1}{3} ]

Now, subtract the lower limit from the upper limit:

[ = \left(4 - \frac{8}{3}\right) - \left(-2 + \frac{1}{3}\right) ]

[ = \left(4 - \frac{8}{3}\right) + \left(2 - \frac{1}{3}\right) ]

[ = \frac{12}{3} - \frac{8}{3} + \frac{6}{3} - \frac{1}{3} ]

[ = \frac{18}{3} - \frac{9}{3} ]

[ = \frac{9}{3} ]

[ = 3 ]

Therefore, the definite integral of ( (2 - x^2) ) for the interval ([2, -1]) is ( 3 ).

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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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