# What is the net area between #f(x) = (x-2)^3 # and the x-axis over #x in [1, 5 ]#?

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To find the net area between ( f(x) = (x-2)^3 ) and the x-axis over ( x ) in ([1, 5]), you need to calculate the definite integral of ( f(x) ) from 1 to 5. The integral represents the area bounded by the function and the x-axis within the specified interval.

[ \text{Net area} = \int_{1}^{5} (x-2)^3 , dx ]

To find this integral, you can use the power rule for integration:

[ \int x^n , dx = \frac{x^{n+1}}{n+1} + C ]

Applying this rule to ( (x-2)^3 ), you'll get:

[ \int (x-2)^3 , dx = \frac{(x-2)^4}{4} + C ]

Now, evaluate this expression at the upper and lower bounds of the interval and subtract the lower bound value from the upper bound value:

[ \text{Net area} = \left[\frac{(5-2)^4}{4} - \frac{(1-2)^4}{4}\right] ]

[ = \left[\frac{3^4}{4} - \frac{(-1)^4}{4}\right] ]

[ = \left[\frac{81}{4} - \frac{1}{4}\right] ]

[ = \frac{80}{4} ]

[ = 20 ]

Therefore, the net area between ( f(x) = (x-2)^3 ) and the x-axis over ( x ) in ([1, 5]) is ( 20 ) square units.

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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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