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

Answer 1

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 #(0, 2)# and #(3, 5)#.

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 #y = x + 2# from #y = -x^2 + 4x + 2#.

#=>int_0^3(-x^2 + 4x + 2 - (x + 2))dx#

#=>int_0^3(-x^2 + 3x)#

#=>-1/3x^3 + 3/2x^2|_0^3#

#=>-1/3(3)^3 + 3/2(3)^2 - (-1/3(0)^3 + 3/2(0)^2)#

#=> -1/3(27) + 3/2(9)#

#=> -9 + 27/2#

#=> 9/2#

Hence, the area between the curves is #9/2" u"^2#.

Hopefully this helps!

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Answer 2

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.

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

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

  1. Rearrange to solve for ( x ):

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

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

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

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

  1. 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 ]

  1. 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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Answer from HIX Tutor

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.

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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