Find the area of the region between the curves y=x+1 y=9−x^2 x=−1 and x=2
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For teaching purposes, I am going to call the first curve f(x) and the second curve g(x).
f(x) = x + 1, g(x) = 9 - x^2, x = -1, x = 2
Your first step is to determine whether the curves f(x) and g(x) intersect between the points -1 and 2.
To find the points of intersection, simply equate the functions to each other.
f(x) = g(x)
x + 1 = 9 - x^2
x^2 + x - 8 = 0
x = [-1 +/- sqrt(1 - 4(1)(-8))]/2
x = [-1 +/- sqrt(33)]/2
Now, [-1 + sqrt(33)]/2 is roughly equal to 2.372, which is clearly outside of the bounds -1 and 2.
Additionally, [-1 - sqrt(33)]/2 is roughly equal to -3.37, which is also outside of our bounds.
That means there are no points of intersection, which makes the problem a whole lot simpler.
Our next step is to determine which of f(x) or g(x) is greater, on the interval -1 and 2. After all, the area formula is going to be
A = Integral(a to b, [higher curve] - [lower curve] dx)
So we ask ourselves: which of f(x) or g(x) is greater on the interval -1 and 2? All you have to do is test ANY value between -1 and 2, and the ideal value to test is 0. Plug 0 into both f(x) and g(x); whichever has the greater result is the greater function.
f(x) = x + 1, so if x = 0,
f(0) = 0 + 1 = 1.
g(x) = 9 - x^2, so if x = 0,
g(x) = 9 - 0^2 = 9 - 0 = 9.
Clearly, g(x) is greater on that interval, making the formula
A = Integral (-1 to 2, [g(x) - f(x)] dx )
Now, plug in your functions,
A = Integral (-1 to 2, [ (9 - x^2) - (x + 1) ] dx )
And simplify.
A = Integral (-1 to 2, [9 - x^2 - x - 1] dx)
A = Integral (-1 to 2, [8 - x^2 - x] dx )
Which is an easy and straightforward integral; just use the reverse power rule.
A = [8x - (1/3)x^3 - (1/2)x^2] {evaluated at -1 to 2}
A = [8(2) - (1/3)(2)^3 - (1/2)(2)^2] - [8(-1) - (1/3)(-1)^3 - (1/2)(-1)^2]
A = [16 - (1/3)(8) - (1/2)(4)] - [-8 - (1/3)(-1) - (1/2)]
A = [16 - (8/3) - 2] - [-8 + (1/3) - (1/2)]
A = 16 - (8/3) - 2 + 8 - (1/3) + (1/2)
A = 22 - (9/3) + (1/2)
A = 22 - 3 + (1/2)
A = 19 + (1/2)
A = (38/2) + (1/2)
A = 39/2
f(x) = x + 1, g(x) = 9 - x^2, x = -1, x = 2
Your first step is to determine whether the curves f(x) and g(x) intersect between the points -1 and 2.
To find the points of intersection, simply equate the functions to each other.
f(x) = g(x)
x + 1 = 9 - x^2
x^2 + x - 8 = 0
x = [-1 +/- sqrt(1 - 4(1)(-8))]/2
x = [-1 +/- sqrt(33)]/2
Now, [-1 + sqrt(33)]/2 is roughly equal to 2.372, which is clearly outside of the bounds -1 and 2.
Additionally, [-1 - sqrt(33)]/2 is roughly equal to -3.37, which is also outside of our bounds.
That means there are no points of intersection, which makes the problem a whole lot simpler.
Our next step is to determine which of f(x) or g(x) is greater, on the interval -1 and 2. After all, the area formula is going to be
A = Integral(a to b, [higher curve] - [lower curve] dx)
So we ask ourselves: which of f(x) or g(x) is greater on the interval -1 and 2? All you have to do is test ANY value between -1 and 2, and the ideal value to test is 0. Plug 0 into both f(x) and g(x); whichever has the greater result is the greater function.
f(x) = x + 1, so if x = 0,
f(0) = 0 + 1 = 1.
g(x) = 9 - x^2, so if x = 0,
g(x) = 9 - 0^2 = 9 - 0 = 9.
Clearly, g(x) is greater on that interval, making the formula
A = Integral (-1 to 2, [g(x) - f(x)] dx )
Now, plug in your functions,
A = Integral (-1 to 2, [ (9 - x^2) - (x + 1) ] dx )
And simplify.
A = Integral (-1 to 2, [9 - x^2 - x - 1] dx)
A = Integral (-1 to 2, [8 - x^2 - x] dx )
Which is an easy and straightforward integral; just use the reverse power rule.
A = [8x - (1/3)x^3 - (1/2)x^2] {evaluated at -1 to 2}
A = [8(2) - (1/3)(2)^3 - (1/2)(2)^2] - [8(-1) - (1/3)(-1)^3 - (1/2)(-1)^2]
A = [16 - (1/3)(8) - (1/2)(4)] - [-8 - (1/3)(-1) - (1/2)]
A = [16 - (8/3) - 2] - [-8 + (1/3) - (1/2)]
A = 16 - (8/3) - 2 + 8 - (1/3) + (1/2)
A = 22 - (9/3) + (1/2)
A = 22 - 3 + (1/2)
A = 19 + (1/2)
A = (38/2) + (1/2)
A = 39/2