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## Class 12 math (India)

### Unit 10: Lesson 2

Fundamental theorem of calculus- The fundamental theorem of calculus and accumulation functions
- Finding derivative with fundamental theorem of calculus
- Finding derivative with fundamental theorem of calculus: x is on both bounds
- Finding derivative with fundamental theorem of calculus
- Proof of fundamental theorem of calculus

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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.

## Want to join the conversation?

- 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)
- 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)

- 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)
- 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)

- 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)- Actually, there is no specific FIRST versus SECOND part of the theorem. Textbooks place them in different order depending on the curriculum for that course. So maybe this order is different from the order that you earned, but it is not objectively incorrect.(1 vote)

- 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)
- As long as the integral is convergent, it doesn't matter. But you have to check whether the integral is convergent...(11 votes)

- 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)
- Yes. That result is known as the Fundamental Theorem of Calculus(0 votes)

- At2:35, why would we need to apply the chain rule?(5 votes)
- Because you have a function of a function. Your inner function is
`x²`

, while your outer function is the whole integral. In order to differentiate a function of a function, you have to use the chain rule.(10 votes)

- why is the chain rule used for Sal's second example? Shouldn't it just be cot^2(x^2)?(5 votes)
- 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)

- When Sal Khan said it was the second fundamental theorem of calculus, did he mean the first?(5 votes)
- 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)
- why are we finding the derivative of Capital F(x) ?(4 votes)
- When he uses the capital F(x) it's the integral of lowercase f(x).

So the derivative of F(x) is f(x).(2 votes)

## 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.