• Feb 26th 2008, 11:47 AM
waite3
Find the area between the curves:
y=x^3-12x^2+20x and
y=-x^3+12x^2-20x
• Feb 26th 2008, 12:14 PM
Jhevon
Quote:

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 $x = a$ and $x = b$, where $a. 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 $a$ and $b$ 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 $A$.

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

can you continue?
• Feb 26th 2008, 12:21 PM
waite3
i keep getting either 0 or 940 but the answer is not correct. any help would be appreciated.
• Feb 26th 2008, 12:33 PM
Jhevon
Quote:

Originally Posted by waite3
i keep getting either 0 or 940 but the answer is not correct. any help would be appreciated.

Let $f(x) = x^3 - 12x^2 + 20x$

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

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

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

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

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

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

Thus, the area is given by:

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

now continue