If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# 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?
• The error comes from the inaccuracy of the calculator. The area is exactly 1/3.
• How can I integrate expressions like (ax+b)^n, for example 16-(2x+1)^4 ? Are there any videos explaining these?
• 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.
• if we cannot sketch the curve how do we know which curve is on the top and which one is below??
• 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.
• Can I still find the area if I used horizontal rectangles?
• 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.)
• Where did the 2/3 come from when getting the derivative's of square root x and x^2?
• So, it's 3/2 because it's being multiplied 3 times?
• 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.
• 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.
• 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.
• 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?
• 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)