# Math Help - Find the area between the curves: y=x^3-12x^2+20x and y=-x^3+12x^2-20x

1. ## Find the area between the curves: y=x^3-12x^2+20x and y=-x^3+12x^2-20x

Ok I am pretty sure what I have done is correct various things are telling me wrong so can someone tell me where I went wrong and the correct way to do it. Thanks.

dA = (top-bottom) dx
= (x^3-12x^2+20x)-(-x^3+12x^2-20x) dx
= (2x^3-24x^2+40x) dx

therefore..

∫ (2x^3-24x^2+40x) dx from x=0 to x=2

{(1/2)x^4-8x^3+20x^2} from x=0 to x=2

therefore..

((1/2)(2)^4-8(2)^3+20(2)^2) - 0

= 24

this seems to be incorrect so can someone explain to me why and how to correctly do it? Thanks

2. Hi Kitizhi,

Your limits of integration look a bit off.
to find the area between $: y=x^3-12x^2+20x \;and \; y=-x^3+12x^2-20x$ first find the intersection of the two graphs by setting the two equations equal to each other

$x^3-12x^2+20x = -x^3+12x^2-20x$
$2x^3-24x^2+40x=0$
$2x(x-2)(x-12)=0$
$x=0 \; x=2\; x=12$
$between \ { } x=0 \ { } x=2 \ { } x^3-12x^2+20x > -x^3+12x^2-20x$ and
$between \ { } x=2 \ { } x=12 \ { } x^3-12x^2+20x < -x^3+12x^2-20x$

$\int_{0}^{2} [x^3-12x^2+20x -(-x^3+12x^2-20x)]) + \int_{2}^{12} [(-x^3+12x^2-20x)-(x^3-12x^2+20x)]$

Hope this helps

3. Ahh I think I get what you mean but you mis-typed the equation so the x values I got are really different..

(x^3-12x^2+20x)=(-x^3+12x^2-20x)
2x^3-24x^2+40x=0
2x(x-10)(x-2)=0

therefore x=0,2,10

so my limits should be..

$

\int_{0}^{2} 2x^3-24x^2+40x + \int_{2}^{10} 2x^3-24x^2+40x
$

I believe...correct me if I am wrong.

which should be
$
{(1/2)x^4-8x^3+20x^2}=
$

$
({(1/2)(2)^4-8(2)^3+20(2)^2}-{(1/2)(0)^4-8(0)^3+20(0)^2}) +$
$({(1/2)(10)^4-8(10)^3+20(10)^2}-{(1/2)(2)^4-8(2)^3+20(2)^2})
$

4. Yes I mistyped the equations, but was able to correct them.
Your limit values are correct but be aware that we must subtract the lesser of the two functions in the particular interval note

$\int_{0}^{2} [x^3-12x^2+20x -(-x^3+12x^2-20x)] + \int_{2}^{12} [(-x^3+12x^2-20x)-(x^3-12x^2+20x)]$
$
\int_{0}^{2} [-24x^2+40x] dx+ \int_{2}^{12} [-2x^3+24x^2-40x]dx
$

5. thanks!
what did you get for a final answer just so I can compare incase i made a mistake

6. I got $[(1/2)x((x-16)x^2+80]$ for the integral so I did..

$
([(1/2)(2)((2-16)(2)^2+80]-0)+$
$([(1/2)(10)((10-16)(10)^2+80]-[(1/2)(2)((2-16)(2)^2+80])
$

which turns out to be -2600 but its wrong...so I am stumped now..