# How do you find the area between #f(x)=x^2-4x, g(x)=0#?

Area:

Note that

The area enclosed by

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To find the area between ( f(x) = x^2 - 4x ) and ( g(x) = 0 ) over a given interval, you need to integrate the absolute difference between ( f(x) ) and ( g(x) ) over that interval.

First, find the points of intersection by setting ( f(x) ) equal to ( g(x) ): [ x^2 - 4x = 0 ] [ x(x - 4) = 0 ] [ x = 0 ] and ( x = 4 )

Next, integrate the absolute difference between ( f(x) ) and ( g(x) ) from ( x = 0 ) to ( x = 4 ): [ \text{Area} = \int_{0}^{4} |f(x) - g(x)| , dx ] [ \text{Area} = \int_{0}^{4} |x^2 - 4x - 0| , dx ] [ \text{Area} = \int_{0}^{4} |x^2 - 4x| , dx ] [ \text{Area} = \int_{0}^{4} x(x - 4) , dx ]

Now, compute the integral: [ \text{Area} = \int_{0}^{4} (x^2 - 4x) , dx ] [ \text{Area} = \left[ \frac{1}{3}x^3 - 2x^2 \right]_{0}^{4} ]

Evaluate at the limits: [ \text{Area} = \left( \frac{1}{3}(4)^3 - 2(4)^2 \right) - \left( \frac{1}{3}(0)^3 - 2(0)^2 \right) ] [ \text{Area} = \left( \frac{64}{3} - 32 \right) - (0 - 0) ] [ \text{Area} = \frac{64}{3} - 32 ] [ \text{Area} = \frac{64 - 96}{3} ] [ \text{Area} = \frac{-32}{3} ] [ \text{Area} = -10\frac{2}{3} ] or ( - \frac{32}{3} ) square units.

The area between ( f(x) = x^2 - 4x ) and ( g(x) = 0 ) from ( x = 0 ) to ( x = 4 ) is ( \frac{32}{3} ) square units or ( 10\frac{2}{3} ) 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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