What is the area of the region enclosed between the curves y=x^2-2x+2 and -x^2+6 ?
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We have to find the area enclosed between y=x^2 - 2x + 2 and y = -x^2 + 6.
First lets find the points of intersection
x^2 - 2x + 2 = -x^2 + 6
=> 2x^2 - 2x - 4 = 0
=> x^2 - x - 2 = 0
=> x^2 - 2x + x - 2 = 0
=> x(x - 2) +1(x - 2) = 0
=> (x + 1)(x - 2) = 0
x = -1 and x = 2
Integrate the two expressions x^2 - 2x + 2 and -x^2 + 6 between the limits x = -1 and x = 2.
Int[ x^2 - 2x + 2 ] = x^3/3 - x^2 + 2x + C
Within the limits: 2^3/3 - 2^2 + 2*2 + C - (-1)^3/3 + (-1)^2 - 2*(-1) + C
=> 8/3 - 4 + 4 + C + 1/3 + 1 + 2 + C
=> 6
Int [ -x^2 + 6 ] = -x^3/3 + 6x + C
Within the limits x = -1 and x = 2
=> -2^3/3 + 6*2 + C + (-1)^3/3 + 6*(-1) + C
=> -8/3 + 12 - 1/3 - 6
=> 3
But in this interval the two curves are curved in different directions as one curve has the term x^2 and the other has -x^2. So we have to add the areas.
The enclosed area is 9.
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To determine the area of the region between the given curves, we'll have to determine the definite integral of the difference between the expressions of the given curves.
First, we need to find out the intercepting points of the curves. For this reason, we'll equate:
x^2-2x+2 = -x^2+6
We'll move all terms to one side:
2x^2 - 2x - 4 = 0
We'll divide by 2:
x^2 - x - 2 = 0
We'll apply quadratic formula:
x1 = [1 + sqrt(1 + 8)]/2
x1 = (1+3)/2
x1 = 2
x2 = (1-3)/2
x2 = -1
We'll choose a value for x, between -1 and 2, to verify what curve is above and what curve is below.
We'll choose x = 0.
f(x) = x^2-2x+2
f(0) = 2
g(x) = -x^2+6
g(0) = 6
We notice that g(x) > f(x), between -1 and 2.
Now, we'll calculate the definite integral of g(x) - f(x), having as limits of integration x = -1 and x = 2.
g(x) - f(x) = -x^2+6-x^2+2x-2
g(x) - f(x) = -2x^2 + 2x + 4
We'll calculate the definite integral:
Int (g(x) - f(x))dx = Int (-2x^2 + 2x + 4)dx
Int (-2x^2 + 2x + 4)dx = -2x^3/3 + 2x^2/2 + 4x
F(2) = -16/3 + 4 + 8
F(2) = (36-16)/3
F(2) = 20/3
F(-1) = 2/3 + 1 - 4
F(-1) = (-9+2)/3
F(-1) = -7/3
F(2) - F(1) = 20/3 + 7/3
F(2) - F(1) = 27/3
F(2) - F(1) = 9
The area enclosed by the given curves is A = 9 square units.
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