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

2. Originally Posted by waite3
Find the area between the curves:
y=x^3-12x^2+20x and
y=-x^3+12x^2-20x
first, find where the curves intersect. by setting them equal to each other and solving for x. let's say they intersect for $\displaystyle x = a$ and $\displaystyle x = b$, where $\displaystyle a<b$. then these are your limits of integration.

now we need to check which graph is above which. you can simply plug in a value between $\displaystyle a$ and $\displaystyle b$ and see which function gives the higher value. (it is an extremely good idea to graph the curves. just in case they enclose more than one area. you can use a graphing utility to do this)

Now, let the area between the curves be $\displaystyle A$.

we have $\displaystyle A = \int_a^b (\mbox{Top function } - \mbox{ Bottom function})~dx$

can you continue?

3. i keep getting either 0 or 940 but the answer is not correct. any help would be appreciated.

4. Originally Posted by waite3
i keep getting either 0 or 940 but the answer is not correct. any help would be appreciated.
Let $\displaystyle f(x) = x^3 - 12x^2 + 20x$

Let $\displaystyle g(x) = -x^3 + 12x^2 - 20x$

set $\displaystyle f(x) = g(x) \implies f(x) - g(x) = 0$

the solutions are $\displaystyle x = 0, ~x = 2, \mbox{ and }x = 10$

with the aid of a graph, we see that the curves enclose areas between $\displaystyle [0,2]$ and $\displaystyle [2,10]$

by plugging in $\displaystyle x = 1$ and $\displaystyle x = 3$, respectively. we see that:

for $\displaystyle [0,2]$, $\displaystyle f(x) > g(x)$, and,
for $\displaystyle [2,10]$, $\displaystyle g(x) > f(x)$.

Thus, the area is given by:

$\displaystyle A = \int_0^2 [f(x) - g(x)]~dx + \int_2^{10}[g(x) - f(x)]~dx$

now continue