SOLUTION: The two functions are f(x) = x^3 -x and g(x) = 3x I must find the area bounded by both of them. First, I set them equal to each other. I get x = 0, 2, -2 Therefore, by the

Algebra ->  Surface-area -> SOLUTION: The two functions are f(x) = x^3 -x and g(x) = 3x I must find the area bounded by both of them. First, I set them equal to each other. I get x = 0, 2, -2 Therefore, by the       Log On


   



Question 1006409: The two functions are f(x) = x^3 -x and g(x) = 3x I must find the area bounded by both of them.
First, I set them equal to each other. I get x = 0, 2, -2
Therefore, by the addition property of integrals I have [-2,0] and [0,2].
Now, I must test points in those intervals to see where the graph is bigger.
for [-2,0]
f(-1) = (-1)^3 - (-1) = 0
g(-1) = (-1)(3) = - 3
so, f is bigger.
for [0,2]
f(1) = (1)^3 - 1 = 0
g(1) = 3(1) = 3
so, g is bigger
Now, to set up the integrals
int[from -2 to 0] (f(x)-g(x))dx + int[from 0 to 2] (g(x)-f(x))dx
integrating the first I get:
(x^4)/4 -(4x^2)/2
integrating the second I get:
(3x^2)/2 - (x^4)/4 + (x^2)/2
evaluating the first I get the value -4
evaluating the second I get the value 0
Area = |-4| = 4?
Something seems off.
Please help
Thank you

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29-g%28x%29=x%5E3-x-3x=x%5E3-4x
g%28x%29-f%28x%29=-%28f%28x%29-g%28x%29%29=-x%5E3%2B4x
The integrals are

and
%28-2%5E4%2F4%2B2%2A2%5E2%29-%28-0%5E4%2F4%2B2%2A0%5E2%29=%28-16%2F4%2B2%2A4%29-0=-4%2B8=4
Both integrals are the same, but that should be no surprise,
because f%28x%29 is an odd function,
meaning that f%28-x%29=-f%28x%29+for+all+values+of+%7B%7B%7Bx ,
and that makes its graph symmetrical with respect to the origin,
meaning that its graph rotated 180%5Eo falls on to of the original graph.
The same can be said of g%28x%29 ,
and of course, the same can be said of f%28x%29-g%28x%29 .
graph%28300%2C300%2C-3%2C3%2C-8%2C8%2Cx%5E3-x%2C3x%29