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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?
• 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.
• 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?
• 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.
• 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 ...".
• 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?
• As long as the integral is convergent, it doesn't matter. But you have to check whether the integral is convergent...
• 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?
• Yes. That result is known as the Fundamental Theorem of Calculus
• At , why would we need to apply the chain rule?
• 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.
• why is the chain rule used for Sal's second example? Shouldn't it just be cot^2(x^2)?
• 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