# How do you evaluate the definite integral #int (-2x^2+3x+2)dx# from [0,2]?

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To evaluate the definite integral ( \int_{0}^{2} (-2x^2 + 3x + 2) , dx ), you need to find the antiderivative of the integrand and then evaluate it at the upper limit of integration (2) and subtract the value of the antiderivative at the lower limit of integration (0).

The antiderivative of ( -2x^2 + 3x + 2 ) with respect to ( x ) is ( -\frac{2}{3}x^3 + \frac{3}{2}x^2 + 2x ).

Now, evaluate this antiderivative at the upper limit (2) and the lower limit (0) of integration:

[ \left[ -\frac{2}{3}x^3 + \frac{3}{2}x^2 + 2x \right]_{0}^{2} ]

[ = \left( -\frac{2}{3}(2)^3 + \frac{3}{2}(2)^2 + 2(2) \right) - \left( -\frac{2}{3}(0)^3 + \frac{3}{2}(0)^2 + 2(0) \right) ]

[ = \left( -\frac{16}{3} + 6 + 4 \right) - \left( 0 + 0 + 0 \right) ]

[ = \left( -\frac{16}{3} + 10 \right) ]

[ = -\frac{16}{3} + \frac{30}{3} ]

[ = \frac{14}{3} ]

So, the value of the definite integral ( \int_{0}^{2} (-2x^2 + 3x + 2) , dx ) is ( \frac{14}{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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