set up the definite integral for y1 = 3(x^3-x) and y2 = 0

I set y1 = y2 and found my bounds to be 0-1 and the integral being -3x^3 + 3x.... In the book it says the answer is -6 integral 0-1 x^3-x... It would make sense if it were -3, but why -3? Did the 2 3's add together? I didn't think that would make sense since we are pulling a 3 out?

Question

set up the definite integral for y1 = 3(x^3-x) and y2 = 0

I set y1 = y2 and found my bounds to be 0-1 and the integral being -3x^3 + 3x....  In the book it says the answer is -6 integral 0-1 x^3-x...  It would make sense if it were -3, but why -3?  Did the 2 3's add together?  I didn't think that would make sense since we are pulling a 3 out?

kinggeorge · Answer

Is it possible the book is multiplying by 2? If you look at the graph, $x=-1$ is also a solution. to $y_1=y_2$

konradzuse · Answer

oh yeah we did learn about that...  If it is semetrical, then it is really 2 * the integral I believe...