Use a computer algebra system to find the exact area enclosed by the curves y=x^5-6 x^3+4 x and

Use a computer algebra system to find the exact area...

Key Concepts

Intersection Points of Curves

This concept involves finding the points where two curves meet, which typically serve as the limits of integration for calculating the area between the curves. It requires solving an equation where the two functions are set equal to each other, and it is crucial for determining the exact boundaries of the region of interest.

Definite Integration

Definite integration is the process of computing the integral of a function within specified limits. In the context of finding areas between curves, it is used to sum the infinitesimal differences between the curves over the interval defined by their intersection points, yielding the exact area enclosed.

Area Between Curves

The area between curves is found by integrating the difference of the functions that define the upper and lower boundaries of the region. This concept involves setting up the integral correctly by identifying which function is on top and subtracting the lower from the upper function over the interval where they enclose the area.

Computer Algebra Systems (CAS)

A Computer Algebra System is a software tool that automates symbolic mathematics, including the solving of equations to find intersection points, performing integrations, and simplifying algebraic expressions. CAS tools are particularly useful for obtaining exact solutions in problems involving intricate algebraic manipulations, such as computing the area enclosed by complex curves.

Transcript

00:01 We want to find the area of the region enclosed by these two functions, the first one in blue.

00:06 Y equals x to the 5 minus 6x to the 3 plus 4x, and the second function in green, y equals x.

00:14 So using a graphing calculator, we can get a quick idea of what the region looks like.

00:21 And we find actually that there are four enclosed regions with five points of intersection.

00:26 So the points of intersection are here, here, here, the origin.

00:31 One here and one here.

00:34 So we actually want to find the area of all four enclosed regions.

00:37 However, what we notice is that our two functions here are odd functions, which means that this red region here has the same area as this green region here, and this region here has the same area as this large region over here.

01:03 So actually we only need to determine the area of two of the regions.

01:07 We're going to be looking at the red regions.

01:10 And then once we determine that area, all we have to do is double our answer.

01:15 Okay, so we're going to do area equals area 1 plus area.

01:26 Okay, we do want to figure out the exact area.

01:29 So we're not going to use the approximations that the graphing calculator give us for the points of intersection.

01:36 So we're going to have to determine those by ourselves.

01:39 So let's do that.

01:41 We want to find at what points the two functions cross.

01:44 So the first function was x to the 5 minus 6x to the 3 plus 4x.

01:51 And the second function was just x.

01:54 So what we're going to do is solve this equation by bringing everything to one side.

01:59 0 equals x to the 5 minus 6x to the 3 plus 3x.

02:04 We can factor out an x.

02:07 We're left with x to the 4 minus 6x squared plus 3.

02:13 Okay, so right away, we see that one solution coming from here is that x equals 0.

02:19 So that's what we expected.

02:20 We saw that the graphs crossed at the origin.

02:23 But now we want to solve this quadratic.

02:25 We want to determine at what x values is this quartic function equal to 0.

02:32 But notice that if we write a equals x squared, this shows that we actually have a squared minus 6a plus 3 equals 0.

02:45 So this is actually just a quadratic in disguise.

02:48 So we can solve this.

02:51 So to solve this, we can use the quadratic formula, which after using the quadratic formula will give us that a is equal to 3 plus or minus the square root of 6.

03:05 But remember, a is x squared.

03:08 So this tells us that x is going to be the square root of 3, plus or minus root 6.

03:17 But since we took the square root, we have to put a plus or minus here.

03:22 So we find actually that we have all five solutions explicitly.

03:26 The first one was x equals 0...

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