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How to Find Simple Areas With Root Finding and Integration

How to Find Simple Areas With Root Finding and Integration
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  • 0:10 Simple Areas
  • 4:48 Solving the Integral
  • 7:36 Lesson Summary
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Lesson Transcript
Instructor: Heather Higinbotham
Combine your calculus tools in this lesson! Find the area enclosed by multiple arbitrary functions by first finding the root of an equation and then integrating over the resulting range.

Simple Areas

5.6 is the point at which the river meets the road
Simple Areas River Meets Road Point

We know that the area underneath a curve is equal to the integral for that region, and we know that we can use it to find areas such as the amount of property that you own. So let's take another look at this in a little more detail. Let's say that you have a piece of property that you found that's right next to the mall, and it's bordered in the back and on the east side by a river. On the south side, it's bordered by a road and, of course, on the west side is the fantastic mall and superplex.

The river follows some function, f(x), where x is a variable along the road and y, or f(x), is the distance from the road. That function is equal to f(x)=4 for x less than 4 and 4 - (x - 4)^3 for values of x that are greater than or equal to 4. Given this function, you can calculate the area of your property, right? Just by finding the integral from zero to - wait a second. Where does your property end on the east side? It ends where the river meets the road, but where is that point? We can't find the area if we don't have an upper bound to this integral.

So before we can do anything - before we can calculate this area at all - we need to determine where that point is. What value of x is right here? Let's take a look at the river and, in particular, the function f(x) at that eastern edge of the property. The function of the river over on that side is equal to 4 - (x - 4)^3. You know this because when f(x)=4, you're actually running parallel to the road, so you're never going to come down and hit the road. You hit the road for a value of x that's greater than 4. So what is this value of x? Well, if the road is at y=0 and the river is at y=f(x), then you're looking for the point where 0 = 4 - (x - 4)^3. That is, where f(x)=0, where these two points meet, the river and the road. So what is that point? Well, if you set 0 = 4 - (x - 4)^3 you can solve for x.

F(x) has two equations, so the integral should be split into two parts
Simple Areas Divide Regions

First, I'm going to add (x - 4)^3 to both sides of the equation, and then I'm going to take the cube root of both sides of the equation. I end up with x - 4 = cube root of 4. At this point, I'm going to add 4 to both sides, so I get x = 4 + the cube root of 4, and this is about 5.6. So let's take our integral from 0 to x=5.6. This point here, where the river meets the road - how do we find this integral exactly? Well, this integral follows f(x), but f(x) has two different equations: one for values of x less than 4 and one for values of x greater than 4. So what I need to do is split my integral up into two different parts: one for x less than 4 and one for x greater than 4.

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