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Finding derivative with fundamental theorem of calculus

AP.CALC:
FUN‑5 (EU)
,
FUN‑5.A (LO)
,
FUN‑5.A.1 (EK)
,
FUN‑5.A.2 (EK)
The Fundamental Theorem of Calculus tells us that the derivative of the definite integral from 𝘢 to 𝘹 of ƒ(𝑡)𝘥𝑡 is ƒ(𝘹), provided that ƒ is continuous. See how this can be used to evaluate the derivative of accumulation functions. Created by Sal Khan.

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  • mr pink red style avatar for user Enya Hsiao
    I don't really understand the role of the lower boundary. It seems to have no effect whatsoever to the result. What changes will take place when we change the lower boundary?
    (93 votes)
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    • blobby green style avatar for user SKapl0929
      The other aspect is that the lower boundary in these has always been a constant number (Pi) when taking the derivative of that boundary it is always zero and therefore irrelevant in the final answer. Check the fourth video in this sequence for more on this.
      (32 votes)
  • leaf blue style avatar for user smarcuseghs
    I don't understand why he needs to put the derivative of x^2 in the final answer. Can someone please help me with this?
    (20 votes)
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    • piceratops ultimate style avatar for user Just Keith
      It is because of the chain rule, as he mentioned. Actually, you ALWAYS have to put the d/dx (of the bound of the definite integral) in the answer. However, in the case where you just have x as the bound, the d/dx = 1. So, you are always putting that derivative in, but in the first example he showed the d/dx was just 1 and didn't affect the final answer.
      (30 votes)
  • piceratops ultimate style avatar for user Michael William Clayton
    What he writes is NOT the second fundamental theorem of calculus. Rather, it is the first fundamental theorem, or the first "part," as some sources prefer. In fact, if you don't believe me, look at the video below (in the AP BC "Integration and accumulation of change" unit) entitled "The Fundamental Theorem of Calculus and definite integrals." In that video Sal labels the first and second theorems correctly, and in fact notes that sometimes they are referred to as "parts" of the Fundamental Theorem.

    I know it's tough to redo videos, but shouldn't you have put one of those pop-ups you have in other videos, where you say things like "what Sal meant to say was ...".
    (23 votes)
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  • blobby green style avatar for user daniel
    Surely this is an improper integral, because cot^2(pi) is not defined. So my question is, does this still matter when applying the fundamental theorem of calculus?
    (15 votes)
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  • blobby green style avatar for user 97.sjross
    If taking the derivative of an integral leaves you with f(x), then would it be right to think of integrals and derivatives as inverse functions?
    (7 votes)
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  • winston default style avatar for user John Shahki
    At , why would we need to apply the chain rule?
    (5 votes)
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  • blobby green style avatar for user Ajeet Dhaliwal
    why is the chain rule used for Sal's second example? Shouldn't it just be cot^2(x^2)?
    (5 votes)
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    • blobby green style avatar for user eajuhnke
      Hey guys, he's using the chain rule because there is essentially a function inside a function... Remember that the chain rule is: If y is a function of u, say y = f(u) and if u is a function of x, say u = g(x), then y = f(u) = f [ g(x) ] For example: if you were asked to find the derivative of (3x^2-5x)^(1/2).... You would let y = u^(1/2) and u = 3x^2 - 5x. Find the derivative of each and multiply them together. So: (1/2)u^(-1/2) * (6x-5) and simplify, but don't forget to replace u with the original u=3x^2-5x! (6x-5) / (2*(3x^2-5x)^(1/2)) Here, we're looking for the derivative of the integral of cot^2(x^2). So, let's apply the chain rule. Let F'(x^2) = cot^2(u) and let u=x^2... Find the derivative of each and multiply together (the derivative of an integral is itself): cot^2(u) * 2x... replacing u and you have cot^2(x^2)*2x
      (5 votes)
  • ohnoes default style avatar for user charlestang06
    When Sal Khan said it was the second fundamental theorem of calculus, did he mean the first?
    (5 votes)
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  • blobby green style avatar for user weirdmind1
    I am having some problems conceptualizing the fundamental theorem of calculus,first thing i want to ask is this :when we take the derivative of an integral of some functions we get back to the original function because derivatives and integrals are inverse operators,right ? Second thing :after we take the derivative we get the original function at x but isnt f(t) the same function at x ,what is the distinction here ?i hope i explained my self ,sorry for my rude english.Thank you all
    (2 votes)
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  • piceratops ultimate style avatar for user Gladwin
    why are we finding the derivative of Capital F(x) ?
    (4 votes)
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Video transcript

- [Instructor] Let's say that we have the function g of x, and it is equal to the definite integral from 19 to x of the cube root of t dt. And what I'm curious about finding or trying to figure out is, what is g prime of 27? What is that equal to? Pause this video and try to think about it, and I'll give you a little bit of a hint. Think about the second fundamental theorem of calculus. All right, now let's work on this together. So we wanna figure out what g prime, we could try to figure out what g prime of x is, and then evaluate that at 27, and the best way that I can think about doing that is by taking the derivative of both sides of this equation. So let's take the derivative of both sides of that equation. So the left-hand side, we'll take the derivative with respect to x of g of x, and the right-hand side, the derivative with respect to x of all of this business. Now, the left-hand side is pretty straight forward. The derivative with respect to x of g of x, that's just going to be g prime of x, but what is the right-hand side going to be equal to? Well, that's where the second fundamental theorem of calculus is useful. I'll write it right over here. Second fundamental, I'll abbreviate a little bit, theorem of calculus. It tells us, let's say we have some function capital F of x, and it's equal to the definite integral from a, sum constant a to x of lowercase f of t dt. The second fundamental theorem of calculus tells us that if our lowercase f, if lowercase f is continuous on the interval from a to x, so I'll write it this way, on the closed interval from a to x, then the derivative of our capital f of x, so capital F prime of x is just going to be equal to our inner function f evaluated at x instead of t is going to become lowercase f of x. Now, I know when you first saw this, you thought that, "Hey, this might be some cryptic thing "that you might not use too often." Well, we're gonna see that it's actually very, very useful and even in the future, and some of you might already know, there's multiple ways to try to think about a definite integral like this, and you'll learn it in the future. But this can be extremely simplifying, especially if you have a hairy definite integral like this, and so this just tells us, hey, look, the derivative with respect to x of all of this business, first we have to check that our inner function, which would be analogous to our lowercase f here, is this continuous on the interval from 19 to x? Well, no matter what x is, this is going to be continuous over that interval, because this is continuous for all x's, and so we meet this first condition or our major condition, and so then we can just say, all right, then the derivative of all of this is just going to be this inner function replacing t with x. So we're going to get the cube root, instead of the cube root of t, you're gonna get the cube root of x. And so we can go back to our original question, what is g prime of 27 going to be equal to? Well, it's going to be equal to the cube root of 27, which is of course equal to three, and we're done.