# What is the average value of a function #y=x^2-2x+4 # on the interval #[0,8]#?

By definition

so here

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To find the average value of the function ( y = x^2 - 2x + 4 ) on the interval ([0, 8]), you need to calculate the definite integral of the function over the interval ([0, 8]), and then divide that result by the length of the interval, which is (8 - 0 = 8).

The definite integral of the function ( y = x^2 - 2x + 4 ) from (0) to (8) is:

[ \int_{0}^{8} (x^2 - 2x + 4) , dx ]

[ = \left[\frac{x^3}{3} - x^2 + 4x \right]_0^8 ]

[ = \left(\frac{8^3}{3} - 8^2 + 4(8)\right) - \left(\frac{0^3}{3} - 0^2 + 4(0)\right) ]

[ = \left(\frac{512}{3} - 64 + 32\right) - (0 - 0 + 0) ]

[ = \frac{512}{3} - 32 ]

[ = \frac{512}{3} - \frac{96}{3} ]

[ = \frac{512 - 96}{3} ]

[ = \frac{416}{3} ]

Now, to find the average value, divide this integral by the length of the interval:

[ \text{Average value} = \frac{\frac{416}{3}}{8} ]

[ = \frac{416}{24} ]

[ = \frac{52}{3} ]

So, the average value of the function ( y = x^2 - 2x + 4 ) on the interval ([0, 8]) is ( \frac{52}{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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