# What is the net area between #f(x) = x^2-x+2# and the x-axis over #x in [1, 2 ]#?

Area =

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To find the net area between the curve ( f(x) = x^2 - x + 2 ) and the x-axis over the interval ( x ) in ([1, 2]), you need to calculate the definite integral of ( f(x) ) with respect to ( x ) over the given interval and take the absolute value of the result.

[ \text{Net area} = \left| \int_{1}^{2} (x^2 - x + 2) , dx \right| ]

Now, integrate ( f(x) = x^2 - x + 2 ) with respect to ( x ) over the interval ([1, 2]):

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

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

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

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

[ = \left( \frac{14}{3} \right) - \left( \frac{1}{3} - \frac{1}{2} \right) ]

[ = \frac{14}{3} + \frac{1}{2} - \frac{1}{3} ]

[ = \frac{14}{3} + \frac{2}{6} - \frac{1}{6} ]

[ = \frac{14}{3} + \frac{1}{6} ]

[ = \frac{28}{6} + \frac{1}{6} ]

[ = \frac{29}{6} ]

Thus, the net area between ( f(x) = x^2 - x + 2 ) and the x-axis over the interval ( x ) in ([1, 2]) is ( \frac{29}{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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