Area between two functions math

[regnar]

y= 1/2 x2 and y=-x2+6. I'm trying to find the area enclosed between the two functions and the answer is eight but I keep getting 16.

 

[berkeman]

y= 1/2 x2 and y=-x2+6. I'm trying to find the area enclosed between the two functions and the answer is eight but I keep getting 16.

Show us your work, and we'll figure out where you're going wrong.

 

[regnar]

I found the bounds to be 2 and -2.

int[(-x^2+6)-(1/2 x^2)]dx = int[(-3/2 x^2 + 6)]dx
I integrated it and got:
(-3/2)(x^3/3) + 6x + C = (-1/2)x^3 + 6
I used FTC part 2 and got:
[-1/2(8) + 12] - [-1/2(-8) - 12]
8 + 8 = 16

 

[berkeman]

I found the bounds to be 2 and -2.

int[(-x^2+6)-(1/2 x^2)]dx = int[(-3/2 x^2 + 6)]dx
I integrated it and got:
(-3/2)(x^3/3) + 6x + C = (-1/2)x^3 + 6
I used FTC part 2 and got:
[-1/2(8) + 12] - [-1/2(-8) - 12]
8 + 8 = 16

There is a small typo at the end of the 4th from the last line

(-1/2)x^3 + 6 --> (-1/2)x^3 + 6x

But in the next line, you correctly use 6x in the substitutions, so that is not the mistake.

Your math looks okay to me. When you draw a sketch of the lines and estimate the area, what do you get? The difference between your answer and the book's answer is big enough that you should be able to figure out graphically which is correct.

Another issue that the sketch may help with is if the lines cross someplace in that x=[-2,2] range. If they do, you will need to calculate where they cross, and set up separate integrals for the sub-ranges of x.

 

[berkeman]

BTW, you can use the Latex editer to make your equations easier to read. There is a \Sigma button in the Advanced Reply (and original start of thread) dialog, to the right above the edit box. Click on that, and you will have access to the Latex graphics.

So int[(-x^2+6)-(1/2 x^2)]dx can become

\int^{2}_{-2}{(6-x^2)}-{(\frac{x^2}{2})} dx

You can use the QUOTE button on this post of mine to see the tex tags that were used to do this formatting.