Question

Find the area enclosed by the curves: y=x^3-4x, y=0, x=0,x=2... I know how to solve it but im not sure if my answer is correct

Answer 1

are u in connexus?

Answer 2

what?

Answer 3

What did you start with to get an answer?

Answer 4

the integral of x^3-4x-0 [0,2]

Answer 5

Right. So we are looking for the absolute area under the curve.
But if I am not mistaken, x^3 - 4x factors like this: x (x^2 - 4) = x(x + 2)(x - 2), meaning its sign changes a few times over the graph.

specifically, between x=0 and x=2, x^3 - 4x is negative. (ex. x=1, 1^3 - 4(1) = -3)
Integrals give us a signed area, but when we want to find an absolute area, we need to split the integral so that the larger function is subtracted by the lesser, in this case y=0 is actually larger than x^3 - 4x.

Answer 6

So, I assume you would end up with an answer that is negative given that setup. In this case, it would just be an absolute value. Sometimes, though, the area can be tricky as in [0, 4] for the same function.

Answer 7

so does that mean ill have to do two integrals?

Answer 8

In this case, y = x^3 - 4x is always less than y = 0. So you only need this one, but we subtract it from 0 because it is less than 0:
integral [0, 2] 0 - (x^3 - 4x) dx

If we did have an interval like [0, 4] where the functions intersected, you would have to split it up so you always have the bigger function subtracting the smaller. This one was just nicer. :)

Answer 9

lol okay, so how do the sign changes affect it

Answer 10

Here is the example of where you have [0, 4].
This is a situation where you have y=0 > y=x^3-4x over [0,2] and y=x^3-4x > y=0 over [2,4].
Graph: http://www.wolframalpha.com/input/?i=y+%3D+x%5E3+-+4x%2C+y+%3D+0+from+x%3D0+to+x%3D4
You want to find the area in those regions between y=0 and y=x^3-4x.
So you just do this:
integral from 0 to 4 of (x^3 - 4x) - 0 dx
So for every value between x=0 and x=2, you have this:
small positive width (dx) * negative function (x^3 - 4x)
and these are all added together for a joint negative value.

Then you start adding from x=2 to x=4. Suddenly you are adding:
small positive width (dx) * positive function
And now you're adding these positive values on top of the negative values. They're cancelling each other out!

Answer 11

So the moral of the story: y = 0 > y = x^3 - 4x, over [0, 2]
So you take the difference that makes these values always positive:
integral [0,2] 0 - (x^3 - 4x) dx
Now THAT is positive!

And if it were from [2, 4], you then have a positive value already for x^3 - 4x.
integral [0,2] 0 - (x^3 - 4x) dx + integral [2,4] (x^3 - 4x) - 0 dx

Am I making any sense? I might just be rambling to myself. >.>

Answer 12

yeah i understand , im just having trouble figuring out where you got the [2,4]

Answer 13

that part is just extending to the case we had [0,4]. [0,4] splits into [0, 2] and [2, 4].
or how to know that x=2 was the breaking point?

Answer 14

the choice of breaking it at x=2 is because y=0 and y=x^3 - 4x intersect there. its the point where they are equal and then y=x^3 - 4x stops being negative becomes positive.

Answer 15

basically in any problem of the sort where we want "an area enclosed by a region", it is important just to know that integration gives you an area with a sign + or -. you would never want to measure the area with negative values, so you would check and make sure.
(a) do the functions intersect in the region?
(b) which one is larger? the larger value minus the smaller is a positive value overall!

Answer 16

"an area enclosed by curves", not region*

Answer 17

hmm okay i think i have a better understanding now , thank you :)

Answer 18

glad to help, and yea i realize after the fact that i'm throwing a lot of information that doesn't quite pertain to the problem. hope it at least made sense. :p