# How do you find the average value of #f(x)=-2x^3+10x^2-7x+5# as x varies between #[1/2,3]#?

The average value of f(x) is found using.

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To find the average value of ( f(x) = -2x^3 + 10x^2 - 7x + 5 ) over the interval ( [1/2, 3] ), we use the formula:

[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) , dx ]

where ( a ) and ( b ) are the endpoints of the interval.

In this case, ( a = 1/2 ) and ( b = 3 ), and the function is ( f(x) = -2x^3 + 10x^2 - 7x + 5 ).

So, the average value is:

[ \text{Average value} = \frac{1}{3 - 1/2} \int_{1/2}^{3} (-2x^3 + 10x^2 - 7x + 5) , dx ]

[ = \frac{1}{5/2} \int_{1/2}^{3} (-2x^3 + 10x^2 - 7x + 5) , dx ]

[ = \frac{2}{5} \left[ -\frac{1}{2}x^4 + \frac{10}{3}x^3 - \frac{7}{2}x^2 + 5x \right]_{1/2}^{3} ]

Now, we substitute the endpoints:

[ = \frac{2}{5} \left[ \left( -\frac{1}{2}(3)^4 + \frac{10}{3}(3)^3 - \frac{7}{2}(3)^2 + 5(3) \right) - \left( -\frac{1}{2}(1/2)^4 + \frac{10}{3}(1/2)^3 - \frac{7}{2}(1/2)^2 + 5(1/2) \right) \right] ]

[ = \frac{2}{5} \left[ \left( -\frac{1}{2}(81) + \frac{10}{3}(27) - \frac{7}{2}(9) + 15 \right) - \left( -\frac{1}{2}(1/16) + \frac{10}{3}(1/8) - \frac{7}{2}(1/4) + \frac{5}{2} \right) \right] ]

[ = \frac{2}{5} \left[ \left( -\frac{81}{2} + \frac{270}{3} - \frac{63}{2} + 15 \right) - \left( -\frac{1}{32} + \frac{10}{24} - \frac{7}{8} + \frac{5}{2} \right) \right] ]

[ = \frac{2}{5} \left[ \left( -\frac{81}{2} + 90 - \frac{63}{2} + 15 \right) - \left( -\frac{1}{32} + \frac{5}{6} - \frac{7}{8} + \frac{5}{2} \right) \right] ]

[ = \frac{2}{5} \left[ -\frac{81}{2} + 90 - \frac{63}{2} + 15 + \frac{3}{96} + \frac{40}{96} - \frac{42}{96} + \frac{120}{96} \right] ]

[ = \frac{2}{5} \left[ -\frac{81}{2} + 90 - \frac{63}{2} + 15 + \frac{-3 + 40 - 42 + 120}{96} \right] ]

[ = \frac{2}{5} \left[ -\frac{81}{2} + 90 - \frac{63}{2} + 15 + \frac{115}{96} \right] ]

[ = \frac{2}{5} \left[ \frac{-162 + 180 - 63 + 30 + 115}{96} \right] ]

[ = \frac{2}{5} \left[ \frac{100}{96} \right] ]

[ = \frac{5}{12} ]

So, the average value of ( f(x) = -2x^3 + 10x^2 - 7x + 5 ) over the interval ( [1/2, 3] ) is ( \frac{5}{12} ).

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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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