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## Integral Calculus

### Unit 3: Lesson 4

Area: vertical area between curves- Area between a curve and the x-axis
- Area between a curve and the x-axis: negative area
- Area between a curve and the x-axis
- Area between curves
- Worked example: area between curves
- Area between two curves given end points
- Area between two curves
- Composite area between curves

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# Area between curves

AP.CALC:

CHA‑5 (EU)

, CHA‑5.A (LO)

, CHA‑5.A.1 (EK)

By integrating the difference of two functions, you can find the area between them. Created by Sal Khan.

## Want to join the conversation?

- I've plugged this integral into my TI-84 Plus calculator and never quite got 1/3, instead I get a number very close to 1/3 (e.g. 0.3333335436) is there a reason for this? does it matter at all?(4 votes)
- The error comes from the inaccuracy of the calculator. The area is exactly 1/3.(37 votes)

- How can I integrate expressions like (ax+b)^n, for example 16-(2x+1)^4 ? Are there any videos explaining these?(3 votes)
- The exact details of the problem matter, so there cannot be a one-size-fits all solution. But, in general here are your best options:

Method 1: If it is possible to convert the problem to a`∫u^(n) du`

form, then you can simply use the power rule. This is the easier method.

Method 2: If it is not possible to convert the problem to a`∫u^(n) du`

nor to some other standard integral from, then you can expand out the polynomial and integrate each term separately. This can be a tedious process that is easy to make mistakes on, but sometimes it is what you need to do.

The important thing to remember in integration is that you have to have an EXACT match of a standard integral form in order to use it. You will often see beginners (and sometimes people who have enough experience to know better) ignore the`du`

as if it were just some trivial bit of notation: it is NOT. The`du`

of the standard integral forms is the derivative of whatever you decided to call`u`

. If it is not present, you cannot use the standard integral form at all.

So, here is how to solve the example you mentioned. I will include some extra steps to make things perfectly clear, but you don't actually have to show this much detail:`∫ {16-(2x+1)⁴} dx`

←in order to use a standard form, split the integral by its terms`=∫ 16 dx − ∫(2x+1)⁴ dx`

← the first part is easily integrated to`16x+C`

, so let us just focus on the second integral.`Let u= 2x+1, thus du= 2dx`

← notice that the integral does not have a`2dx`

, but only a`dx`

, so I must divide by`2`

in order to create an exact match to the standard integral form.`½ du = ½ (2 dx)`

So the substitution is:`−∫(2x+1)⁴ dx = −∫ u⁴ (½ du)`

Now, factor out the`½`

to get an EXACT match for the standard integral form.`= −½ ∫ u⁴du`

This is now an exact match for the standard integral form for the power rule. So, let us apply that to integrate:`= −½ [⅕u⁵] + C`

`= −(1/10)u⁵ + C`

← now back substitute`= −(1/10)(2x+1)⁵ + C`

And, finally remember that this is only one term of the integral, we need to add in the`16x + C`

for a final answer of:`∫ {16-(2x+1)⁴} dx = 16x − (1/10)(2x+1)⁵ + C`

Note: notice that I did not use two +C terms. The +C represents ANY constant, not a particular constant, so it will absorb all of the constants generated by the integration.

As mentioned above, once you become proficient at it, there is no need to show this much detail.(11 votes)

- if we cannot sketch the curve how do we know which curve is on the top and which one is below??(6 votes)
- If we have two functions f(x) and g(x), we can find solutions to the equation f(x)=g(x) to find their intersections, and to find which function is on the top or on the bottom we can either plug in values or compare the slopes of the functions to see which is larger at an intersection.(7 votes)

- Can I still find the area if I used horizontal rectangles?(4 votes)
- As Paul said, integrals are better than rectangles. The other part of your question: Yes, you can integrate with respect to y. Some problems even require that! (Sometimes, area between graphs cannot be expressed easily in integrals with respect to x.)(8 votes)

- Where did the 2/3 come from when getting the derivative's of square root x and x^2?(5 votes)
- So, it's 3/2 because it's being multiplied 3 times?(0 votes)

- Why is it necessary to find the "most positive" of the functions? Would it not work to simply subtract the two integrals and take the absolute value of the final answer? When I look in the hints for the practice sections, you always do a graph to find the "greater" function, but I'm having trouble seeing why that is necessary.(4 votes)
- Is it possible to get a negative number or zero as an answer?(1 vote)
- That depends on the question. If you are simply asking for the area between curves on an interval, then the result will never be negative, and it will only be zero if the curves are identical on that interval. This process requires that you keep track of where each function has a greater value and perform the subtraction in the correct order (or use an absolute value).

However, if you are looking for the net signed area above one curve and below another on an interval, then this can indeed be negative, and it can be zero for non-trivial cases. This is usually the more straightforward calculation, because you can ignore where one function exceeds another and simply perform the integral of a difference.(5 votes)

- y=cosx, lower bound= -pi upper bound = +pi how do i calculate the area here. because sin pi=0 ryt? i can't get an absolute value to that too.(2 votes)
- Someone please explain: Why isn't the constant c included when we're finding area using integration yet when we're solving we have to include it?? Doesn't not including it affect the final answer?(2 votes)
- when we find area we are using definite integration so when we put values then c-c will cancel out.

eg. area of curve y=sinx from o to pi

as this curve is symmetrical we devide in two parts a1 and a2 where ai=a2

a1+a2=total area a

integrating sinxdx from 0 to pi/2

-cosx+c

now putting values

[(o+c)-(1+c)]=1=a1=a2

total area=2a1=2a2=2(!)=2sq. unit(1 vote)

- say the two functions were y=x^2+1 and y=1 when you combine them into one intergral, for example intergral from 0 to 2 of ((x^2+1) - (1)) would you simplify that into the intergral form 0 to 2 of (x^2) or just keep it in its original form(1 vote)

## Video transcript

- [Instructor] We have already covered the notion of area between
a curve and the x-axis using a definite integral. We are now going to then extend this to think about the area between curves. So let's say we care about the region from x equals a to x equals b between y equals f of x
and y is equal to g of x. So that would be this area right over here. So based on what you already know about definite integrals, how would you actually
try to calculate this? Well one natural thing that you might say is well look, if I were to take the integral from a to b of f of x dx, that would give me the entire area below f of x and above the x-axis. And then if I were to subtract from that this area right over here, which is equal to that's the definite integral from a to b of g of x dx. Well then I would net out
with the original area that I cared about. I would net out with this
area right over here. And that indeed would be the case. And we know from our
integration properties that we can rewrite this as the integral from a to b of, let me put some parentheses here, of f of x minus g of x, minus g of x dx. And now I'll make a claim to you, and we'll build a little
bit more intuition for this as we go through this video, but over an integral from a to b where f of x is greater than g of x, like this interval right over here, this is always going to be the case, that the area between the curves is going to be the integral for the x-interval that we
care about, from a to b, of f of x minus g of x. So I know what you're thinking, you're like okay well that
worked when both of them were above the x-axis, but what about the case when f of x is above the x-axis and g of x is below the x-axis? So for example, let's say that we were to
think about this interval right over here. Let's say this is the point c, and that's x equals c, this is x equals d right over here. So what if we wanted to calculate this area that I am shading in right over here? You might say well does
this actually work? Well let's think about now what the integral, let's think about what the integral from c to d of f of x dx represents. Well that would represent
this area right over here. And what would the integral from c to d of g of x dx represent? Well you might say it is this area right over here, but remember, over this interval g of
x is below the x-axis. So this would give you a negative value. But if you wanted this total area, what you could do is take this blue area, which is positive, and then subtract this negative area, and so then you would get
the entire positive area. Well this just amounted to, this is equivalent to the integral from c to d of f of x, of f of x minus g of x again, minus g of x. Let me make it clear, we've
got parentheses there, and then we have our dx. So once again, even over this interval when one of, when f of x was above the x-axis and g of x was below the x-axis, we it still boiled down to the same thing. Well let's take another scenario. Let's take the scenario when they are both below the x-axis. Let's say that we wanted to go from x equals, well I won't
use e since that is a loaded letter in mathematics,
and so is f and g. Well let's just say well
I'm kinda of running out of letters now. Let's say that I am gonna go from I don't know, let's just call this m, and let's call this n right over here. Well n is getting, let's
put n right over here. So what I care about is this area, the area once again below f. We're assuming that we're
looking at intervals where f is greater than g, so below f and greater than g. Will it still amount to this with now the endpoints being m and n? Well let's think about it a little bit. If we were to evaluate that integral from m to n of, I'll just put my dx here, of f of x minus, minus g of x, we already know from
our integral properties, this is going to be equal to the integral from m to n of f of x dx minus the integral from m to n of g of x dx. Now let's think about what
each of these represent. So this yellow integral right over here, that would give this the negative of this area. So that would give a negative value here. But the magnitude of it,
the absolute value of it, would be this area right over there. Now what would just the integral, not even thinking about
the negative sign here, what would the integral of this g of x of this blue integral give? Well that would give this the negative of this entire area. But now we're gonna take
the negative of that, and so this part right over here, this entire part including
this negative sign, would give us, would give us this entire area, the entire area. This would actually give a positive value because we're taking the
negative of a negative. But if with the area that we care about right over here, the area that
we cared about originally, we would want to subtract
out this yellow area. Well this right over here, this yellow integral from, the definite integral
from m to n of f of x dx, that's exactly that. That is the negative of that yellow area. So if you add the blue area, and so the negative of a
negative is gonna be positive, and then this is going to be the negative of the yellow area, you would net out once again to the area that we think about. So in every case we saw, if we're talking about an interval where f of x is greater than g of x, the area between the curves is just the definite
integral over that interval of f of x minus g of x dx.