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### Course: AP®︎/College Calculus AB>Unit 8

Lesson 4: Finding the area between curves expressed as functions of x

# Composite area between curves

Sometimes we need to break the area into multiple different areas, because the enclosing functions change. Created by Sal Khan.

## Want to join the conversation?

• Hey, so why is the second part minus x_2/4-1 at around ?
• It's below the x axis, so its area is negative, you subtract it in order to cancel out the negative, effectively adding it.
If you didn't have the minus sign, you would end up with a much smaller, incorrect value, telling you difference of area ABOVE the x axis. Possibly useful under certain circumstances, but not what we want here.
• If I were only given the functions, is there a mathematical way to determine where an intersection is like at ?
• You set one function equal to another and then solve for x. That will necessarily be a point or points of intersection.
• In a previous video, when we determined the area under a curve of cos x, the areas above and below x-axis cancelled each other out. In this example we are adding the areas above and below the x-axis. Why?
• because your lower boundary is curve y=x^2/4 - 1, not the x axis.
• Can I do it with three integrals? One for the sqrt(x) from 0 to 1 plus integral of (2 - x) from 1 to 2 minus (because it's under the x axis, so the result will be negative) integral of ((x^2/4) - 1) from 0 to 2? The result is the same and the technique seems correct to me, but that doesn't mean it's right.
• Yes, you can do what you just described , but it will require more work. There is nothing wrong with your reasoning.
• At why is he subtracting ? u have to add the 2 function to get the height of the thin blue rectangle right ?
• The height of the rectangle is the difference of the two functions.

Try it by graphing two very simple functions say:
y = 5
y = 2

The height between them will always be the difference: 3.
• Shouldn't the answer be 4 units squared? The definite integral of the second part from 0 to 1 of (2 - x) - ( (x^2)/4 + 1) is 29/12 and the definite integral of the first section is 19/12. So, the combined area would be (29 + 19)/12 = 48/12 = 4 units squared.
• The second part is from 1 to 2. The first part just encompasses the region of x between 0 and 1, but the second part is the region of x between 1 and 2 not 0 and 1 again.
• What do I do when I want to know the area of the function when x<0, hm, an example would be: definite integral of x² between x=0 and x= -2 ? What do I do? Because the final result is negative if I do it in the normal way, is there a different way that the result will be positive?
(1 vote)
• Just do it the normal way and, if you do it correctly, the result will be positive. Are you remembering that the lower bound will be -2 and the upper bound 0? And that two negatives cancel to make a positive?
• why aren't we taking net area here as in previous area problems?
(1 vote)
• If you look carefully at how the area between curves is found, you'll see that we're always subtracting the ''smaller'' function from the ''greater'' one, say, `f(x)-g(x)`.

By finding the net area, you are taking into account the signs separately but not the behavior of both/all functions. It's not the same `f(x)-g(x)` when `f(x)>g(x)` as when `f(x)<g(x)` (in the latter case `f(x)` covers less area under its curve).

In the video, the orange function is the smallest function (on the interval, it's the only one shifted toward the negative side of the y axis). Since it'll be substracting (because is the smallest), when it has a negative area we'll be adding area (subtract a negative value is the same as adding it). Just what we want.

If you find the net area separately, you'd have to figure out the correct sign by hand, instead of avoiding that if you had set which is the smaller and which the greater from the get-go.