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## Class 12 math (India)

### Course: Class 12 math (India)>Unit 10

Lesson 3: Fundamental theorem of calculus: chain rule

# Fundamental theorem of calculus review

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) $\frac{d}{dx}{\int }_{a}^{x}f\left(t\right)\phantom{\rule{0.167em}{0ex}}dt=f\left(x\right)$‍

We start with a continuous function $f$ and we define a new function for the area under the curve $y=f\left(t\right)$:
$F\left(x\right)={\int }_{a}^{x}f\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$
What this version of the theorem says is that the derivative of $F$ is $f$. 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) ${\int }_{a}^{b}\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}f\left(x\right)dx=F\left(b\right)\phantom{\rule{-0.167em}{0ex}}-\phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}F\left(a\right)$‍

This version gives more direct instructions to finding the area under the curve $y=f\left(x\right)$ between $x=a$ and $x=b$. Simply find an antiderivative $F$ and take $F\left(b\right)-F\left(a\right)$.

## Practice set 1: Applying the theorem

Problem 1.1
$g\left(x\right)={\int }_{\phantom{\rule{0.167em}{0ex}}1}^{\phantom{\rule{0.167em}{0ex}}x}\sqrt{2t+7}\phantom{\rule{0.167em}{0ex}}dt\phantom{\rule{0.167em}{0ex}}$
$g{\phantom{\rule{0.167em}{0ex}}}^{\prime }\left(9\right)\phantom{\rule{0.167em}{0ex}}=$

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 $\frac{d}{dx}{\int }_{0}^{{x}^{3}}\mathrm{sin}\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$. Note that the interval is between $0$ and ${x}^{3}$, not $x$.
To help us, we define $F\left(x\right)={\int }_{0}^{x}\mathrm{sin}\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$. According to the fundamental theorem of calculus, ${F}^{\prime }\left(x\right)=\mathrm{sin}\left(x\right)$.
It follows from our definition that ${\int }_{0}^{{x}^{3}}\mathrm{sin}\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$ is $F\left({x}^{3}\right)$, which means that $\frac{d}{dx}{\int }_{0}^{{x}^{3}}\mathrm{sin}\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$ is $\frac{d}{dx}F\left({x}^{3}\right)$. Now we can use the chain rule:
$\begin{array}{rl}& \phantom{=}\frac{d}{dx}{\int }_{0}^{{x}^{3}}\mathrm{sin}\left(t\right)\phantom{\rule{0.167em}{0ex}}dt\\ \\ & =\frac{d}{dx}F\left({x}^{3}\right)\\ \\ & ={F}^{\prime }\left({x}^{3}\right)\cdot \frac{d}{dx}\left({x}^{3}\right)\\ \\ & =\mathrm{sin}\left({x}^{3}\right)\cdot 3{x}^{2}\end{array}$
Problem 2.1
$F\left(x\right)={\int }_{0}^{{x}^{4}}\mathrm{cos}\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$
${F}^{\prime }\left(x\right)=$

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

## Want to join the conversation?

• 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.
• 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'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).
• 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)
• 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.
• 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)
• 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?
• 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
(1 vote)
• 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)
(1 vote)
• So is the derivative the inverse of the integral?
(1 vote)
• Yes, and the integral is the inverse of the derivative.
(1 vote)
• 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?
(1 vote)
• 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!
(1 vote)