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

By the Fundamental Theorem of Calculus, we know that,

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To evaluate the definite integral (\int_{-1}^{2} (-x^2+x+2) , dx), you need to find the antiderivative of the function (-x^2+x+2), and then evaluate it at the upper and lower limits of integration, and finally subtract the value at the lower limit from the value at the upper limit.

First, find the antiderivative: [\int (-x^2+x+2) , dx = -\frac{1}{3}x^3 + \frac{1}{2}x^2 + 2x + C]

Now evaluate this antiderivative at the upper and lower limits of integration: [F(2) - F(-1) = \left(-\frac{1}{3}(2)^3 + \frac{1}{2}(2)^2 + 2(2)\right) - \left(-\frac{1}{3}(-1)^3 + \frac{1}{2}(-1)^2 + 2(-1)\right)]

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

[= \left(-\frac{8}{3} + 6\right) - \left(\frac{1}{6} - 2\right)]

[= \left(\frac{10}{3}\right) - \left(-\frac{11}{6}\right)]

[= \frac{10}{3} + \frac{11}{6}]

[= \frac{20}{6} + \frac{11}{6}]

[= \frac{31}{6}]

So, (\int_{-1}^{2} (-x^2+x+2) , dx = \frac{31}{6}).

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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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