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

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) (Show Source): You can put this solution on YOUR website!

The integrals are

and

Both integrals are the same, but that should be no surprise,
because is an odd function,
meaning that ,
and that makes its graph symmetrical with respect to the origin,
meaning that its graph rotated falls on to of the original graph.
The same can be said of ,
and of course, the same can be said of .

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