# How do you find the integral of #x^2-6x+5# from the interval [0,3]?

-3

We can use the reverse power rule:

Therefore,

This is equivalent to:

Therefore, the answer is -3. Note that the definite integral gives the net area under the graph. The negative sign implies that there is more area below the x-axis (than above) for the interval from 0 to 3 (inclusive).

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To find the integral of (x^2 - 6x + 5) over the interval ([0, 3]), you would first integrate the function with respect to (x), then evaluate the result at the upper and lower bounds of the interval and subtract the lower bound result from the upper bound result. The integral of (x^2 - 6x + 5) with respect to (x) is (\frac{1}{3}x^3 - 3x^2 + 5x). Evaluating this expression at the upper bound (3) yields ( \frac{1}{3}(3)^3 - 3(3)^2 + 5(3) = 9 - 27 + 15 = -3), and at the lower bound (0) yields ( \frac{1}{3}(0)^3 - 3(0)^2 + 5(0) = 0). Therefore, the definite integral of (x^2 - 6x + 5) from (0) to (3) is (-3 - 0 = -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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