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

We start by finding the intersection points of the two functions.

#{(y = -x^2 + 4x + 2), (y = x+ 2):}#

#x+ 2 = -x^2 + 4x + 2#

#x^2 - 3x = 0#

#x(x - 3) = 0#

#x = 0 and 3#

#y = 0 + 2 and y = 3 + 2#

#y = 2 and y= 5#

Hence, the intersection points are

We now do a rudimentary sketch of the two functions.

We always proceed in the following way: AREA BETWEEN CURVES = AREA OF CURVE ABOVE - AREA OF CURVE BELOW. We find this area using integration.

We will subtract the area under

Hence, the area between the curves is

Hopefully this helps!

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To find the area between the curves ( f(x) = -x^2 + 4x + 2 ) and ( g(x) = x + 2 ), you need to find the points of intersection first. Then, you integrate the absolute difference between the functions over the interval where they intersect.

- Set the functions equal to each other and solve for ( x ) to find the points of intersection:

[ -x^2 + 4x + 2 = x + 2 ]

- Rearrange to solve for ( x ):

[ -x^2 + 3x = 0 ] [ x(-x + 3) = 0 ]

This gives ( x = 0 ) and ( x = 3 ).

- Next, integrate the absolute difference between the functions over the interval ([0, 3]):

[ \text{Area} = \int_{0}^{3} |f(x) - g(x)| , dx ]

- Determine which function is greater over the interval ( [0, 3] ) to set up the integral correctly.

[ |f(x) - g(x)| = |(-x^2 + 4x + 2) - (x + 2)| = | -x^2 + 3x | ]

Since ( f(x) = -x^2 + 4x + 2 ) is greater than ( g(x) = x + 2 ) in the interval ( [0, 3] ), the absolute value can be dropped.

[ \text{Area} = \int_{0}^{3} (-x^2 + 3x) , dx ]

- Integrate ( -x^2 + 3x ) with respect to ( x ) over the interval ([0, 3]):

[ \text{Area} = \left[ -\frac{1}{3}x^3 + \frac{3}{2}x^2 \right]_{0}^{3} ]

[ = \left[ -\frac{1}{3}(3)^3 + \frac{3}{2}(3)^2 \right] - \left[ -\frac{1}{3}(0)^3 + \frac{3}{2}(0)^2 \right] ]

[ = \left[ -9 + \frac{27}{2} \right] - \left[ 0 \right] ]

[ = -9 + \frac{27}{2} ]

[ = -\frac{18}{2} + \frac{27}{2} ]

[ = \frac{9}{2} ]

So, the area between the curves ( f(x) = -x^2 + 4x + 2 ) and ( g(x) = x + 2 ) over the interval ( [0, 3] ) is ( \frac{9}{2} ) 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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