I can't convince myself why these expressions should be treated as more than one function, thus requiring the use of the chain rule.

not sure if this helps but since x^2 is not x we cannot treat it the same as how we treat x.

I'm having trouble with this equality written in the text: d/dx F(x^3)= F'(x^3).(x^3)' What's the difference between d/dx and ' ?

There is no practical difference, just a difference in the notation. d/dx f(x) = f'(x)

In Practice set 2, we had to find F(x^3), so we took help of F(x). but, problem 2.1 says that F(x) is equal to integral from 0 to x^4, i think it should be F(x^4). please help.

Let 𝐺(𝑥) = ∫cos(𝑡)𝑑𝑡 over the interval [0, 𝑥] ⇒ 𝐺'(𝑥) = cos(𝑥) 𝐹(𝑥) = 𝐺(𝑥⁴) ⇒ 𝐹 '(𝑥) = 𝐺'(𝑥⁴) ∙ 4𝑥³ = 4𝑥³ cos(𝑥⁴)

On the integral sign you go from one point to another. I don't get why this point can be x or something with x. X is not a point on the x axis. It just doesn't make sense. From my understanding, the two points on an integral are two places on the x-axis.

Look more closely. With the Fundamental Theorem of Calculus we are integrating a function of *t* with respect to *t*. The x variable is just the upper limit of the definite integral. x might not be "a point on the x axis", but it can be a point on the t-axis.

I still don't understand, but, so the antiderivative F is same as definite integral. If we find the area under the curve, which is definite integral, we can say that, this function is equal to anti-derivative?

No, the antiderivative is the same as the indefinite integral, which doesn't represent area. An antiderivative is only a function such that its derivative is your original function. The definite integral represents the area under a function between two points. Antiderivatives can help us find definite integrals.

Main content

Course: Integral Calculus (2017 edition) > Unit 3

Lesson 3: Fundamental theorem of calculus: chain rule

Fundamental theorem of calculus review

Google Classroom

Review your knowledge of the fundamental theorem of calculus and use it to solve problems.

What is the fundamental theorem of calculus?

The theorem has two versions.

a) $\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$ ‍

We start with a continuous function

f

and we define a new function for the area under the curve

y = f (t)

F (x) = \int_{a}^{x} f (t) d t

What this version of the theorem says is that the derivative of

F

f

. In other words,

F

is an antiderivative of

f

. Thus, the theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

b) $\int_{a}^{b} f (x) d x = F (b) - F (a)$ ‍

This version gives more direct instructions to finding the area under the curve

y = f (x)

between

x = a

and

x = b

. Simply find an antiderivative

F

and take

F (b) - F (a)

Want to learn more about the fundamental theorem of calculus? Check out this video.

Practice set 1: Applying the theorem

Problem 1.1

g (x) = \int_{1}^{x} \sqrt{2 t + 7} d t

g^{'} (9) =

Want to try more problems like this? Check out this exercise.

Practice set 2: Applying the theorem with chain rule

We can use the theorem in more hairy situations. Let's find, for example, the expression for

\frac{d}{d x} \int_{0}^{x^{3}} \sin (t) d t

. Note that the interval is between

0

and

x^{3}

, not

x

To help us, we define

F (x) = \int_{0}^{x} \sin (t) d t

. According to the fundamental theorem of calculus,

F^{'} (x) = \sin (x)

It follows from our definition that

\int_{0}^{x^{3}} \sin (t) d t

F (x^{3})

, which means that

\frac{d}{d x} \int_{0}^{x^{3}} \sin (t) d t

\frac{d}{d x} F (x^{3})

. Now we can use the chain rule:

\begin{aligned} \frac{d}{d x} \int_{0}^{x^{3}} \sin (t) d t \\ = \frac{d}{d x} F (x^{3}) \\ = F^{'} (x^{3}) \cdot \frac{d}{d x} (x^{3}) \\ = \sin (x^{3}) \cdot 3 x^{2} \end{aligned}

Problem 2.1

F (x) = \int_{0}^{x^{4}} \cos (t) d t

F^{'} (x) =

Want to try more problems like this? Check out this exercise.

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allan1.davis
Posted 8 years ago. Direct link to allan1.davis's post “I can't convince myself w...”
I can't convince myself why these expressions should be treated as more than one function, thus requiring the use of the chain rule.
Button navigates to signup pageComment on allan1.davis's post “I can't convince myself w...”
(17 votes)
Answer
- joshvick8
  Posted 8 years ago. Direct link to joshvick8's post “not sure if this helps bu...”
  not sure if this helps but since x^2 is not x we cannot treat it the same as how we treat x.
  Button navigates to signup page
  (12 votes)
Victor Mayland Nielsen
Posted 7 years ago. Direct link to Victor Mayland Nielsen's post “On the integral sign you ...”
On the integral sign you go from one point to another. I don't get why this point can be x or something with x. X is not a point on the x axis. It just doesn't make sense. From my understanding, the two points on an integral are two places on the x-axis.
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Howard Bradley
  Posted 7 years ago. Direct link to Howard Bradley's post “Look more closely. With t...”
  Look more closely. With the Fundamental Theorem of Calculus we are integrating a function of t with respect to t. The x variable is just the upper limit of the definite integral.
  x might not be "a point on the x axis", but it can be a point on the t-axis.
  Comment on Howard Bradley's post “Look more closely. With t...”
  (6 votes)
John Vanderwalls
Posted 6 years ago. Direct link to John Vanderwalls's post “I'm having trouble with t...”
I'm having trouble with this equality written in the text: d/dx F(x^3)= F'(x^3).(x^3)'

What's the difference between d/dx and ' ?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Mark Geary
  Posted 6 years ago. Direct link to Mark Geary's post “There is no practical dif...”
  There is no practical difference, just a difference in the notation.
  d/dx f(x) = f'(x)
  Button navigates to signup page
  (4 votes)
bgautam0707
Posted 7 years ago. Direct link to bgautam0707's post “In Practice set 2, we had...”
In Practice set 2, we had to find F(x^3), so we took help of F(x).
but, problem 2.1 says that F(x) is equal to integral from 0 to x^4, i think it should be F(x^4).
please help.
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Jerry Nilsson
  Posted 7 years ago. Direct link to Jerry Nilsson's post “Let 𝐺(𝑥) = ∫cos(𝑡)𝑑𝑡...”
  Let 𝐺(𝑥) = ∫cos(𝑡)𝑑𝑡 over the interval [0, 𝑥] ⇒ 𝐺'(𝑥) = cos(𝑥)
  𝐹(𝑥) = 𝐺(𝑥⁴) ⇒ 𝐹 '(𝑥) = 𝐺'(𝑥⁴) ∙ 4𝑥³ = 4𝑥³ cos(𝑥⁴)
  Comment on Jerry Nilsson's post “Let 𝐺(𝑥) = ∫cos(𝑡)𝑑𝑡...”
  (2 votes)
beilingliu77
Posted 6 years ago. Direct link to beilingliu77's post “I still don't understand,...”
I still don't understand, but, so the antiderivative F is same as definite integral. If we find the area under the curve, which is definite integral, we can say that, this function is equal to anti-derivative?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Gavin Yu
  Posted 6 years ago. Direct link to Gavin Yu's post “No, the antiderivative is...”
  No, the antiderivative is the same as the indefinite integral, which doesn't represent area. An antiderivative is only a function such that its derivative is your original function.
  
  The definite integral represents the area under a function between two points. Antiderivatives can help us find definite integrals.
  Button navigates to signup page
  (3 votes)
ximena.tamariz
Posted 4 years ago. Direct link to ximena.tamariz's post “I think I understand the ...”
I think I understand the theorem when the lower limit is zero and the upper limit is x, but what do you do when the lower limit is 2 and the upper limit is x. A specific problem I have is: d/dx(∫(upper limit x, lower limit 2)sin(t^4)dt)
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(2 votes)
Answer
Evana Holevas
Posted 5 years ago. Direct link to Evana Holevas's post “What is the difference be...”
What is the difference between the FTC and the Antiderivative Method? I'm working on a problem that wants me to use both formulas and it says the functions will be different however I'm having trouble remembering. Would someone mind explaining?
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(2 votes)
Answer
beilingliu77
Posted 6 years ago. Direct link to beilingliu77's post “Using, problem 2.1 as exa...”
Using, problem 2.1 as example
when using the chain rule, why does F'(x) stay the same as COS(X^4)*4X^3?
I thought the chain rule should be
f(x)=cos(x)
f'(x)=-sin(x)
g(x)=x^4
g'(x)=4x^3

apply chain rule here, should be f'(g(x))*g'(x)
which should be f'(x^4)*4x^3
-sin(x^4)*4x^3

or am I confused with something? Thanks for the help
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(1 vote)
Answer
- Mateusz Jastrzębski
  Posted 6 years ago. Direct link to Mateusz Jastrzębski's post “Your conclusions are alri...”
  Your conclusions are alright but you're not solving for what's being asked.
  
  The definite integral equals F(x)=Integral(f(t)) from 0 to x^4. Now, if you take the derivative of this integral you get f(x^4) times d/dx(x^4). You don't differentiate the f(t) because it is in fact your original function before integration.
  
  Fundamental Theorem of Calculus is tricky to understand but once you know it by heart it'll never leave you. If you're struggling to get a good grasp on this fundamental concept try this: check out this video - https://www.khanacademy.org/math/ap-calculus-ab/ab-antiderivatives-ftc/ab-fundamental-theorem-of-calc/v/fundamental-theorem-of-calculus - and consider the following: it is a very subtle and simple idea:
  
  On a diagram, there's a yellow line with an x label. What is this yellow line? It's f(x). What does its value represent? It represents the rate of change of F around that point x and naturally it means it is the derivative of the integral F. The slope of F at point x is equal to f(x)
  Button navigates to signup page
  (1 vote)
20leunge
Posted 6 years ago. Direct link to 20leunge's post “So is the derivative the ...”
So is the derivative the inverse of the integral?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Mark Geary
  Posted 6 years ago. Direct link to Mark Geary's post “Yes, and the integral is ...”
  Yes, and the integral is the inverse of the derivative.
  Button navigates to signup page
  (1 vote)
ddan281847
Posted 6 years ago. Direct link to ddan281847's post “I don't really understand...”
I don't really understand how a derivative of this is just the function of the equation that was used. But the difference is if x matters since the part being cut is constantly being increase as x increases and the functions derivative is the same as the function being measured?
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(1 vote)
Answer
- ThePencilThief
  Posted 6 years ago. Direct link to ThePencilThief's post “The reason why the deriva...”
  The reason why the derivative is the function of the eqution is because the FTC is defined in such a way, F'(x)=f(x), two separate functions linked in that way.
  An integral can also be stated as an anti-derivative, so the opposite of a derivative. If you take the derivative of an integral, it is like taking the square root of x squared, they cancel out, and you are left with x. More can be seen as we take the limits of integrals as x approaches infinite, but we are not yet there in these videos. Take time to really get a good foundation in the FTC because it is VITAL to calculus understanding. I hope I answered your question properly, have a good day!
  Comment on ThePencilThief's post “The reason why the deriva...”
  (1 vote)

Integral Calculus (2017 edition)