# The Fundamental Theorem of Calculus

So far, we know that calculating the integral (or antiderivative) of a function gives a general function that represents the area under the curve of that function. What if we want to determine the exact area under a function on a given interval? We use the Fundamental Theorem of Calculus (FTC) to do this!

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The FTC enables us to solve slightly more complex problems that are otherwise possible. For example, it's easy to determine who will travel farther in 30 minutes if you're going 10 miles per hour and your friend is moving at 12 miles per hour. But what if your velocities are not constant and are modeled by an equation that changes with respect to time? How would you answer this? Luckily, we have the FTC to allow us to do this. It can be split up into 2 parts. Let's see how it works!

## The first part of the Fundamental Theorem of Calculus

If the function f is continuous on [a, b], then the function g, defined by

$g\left(x\right)={\int }_{a}^{x}f\left(t\right)dt$ where $a\le x\le b$,

is continuous on [a, b] and differentiable on (a, b), and $g\text{'}\left(x\right)=f\left(x\right)$.

Put into words, the derivative of a definite integral with respect to the upper limit equals the integrand evaluated at the upper limit. Integration and differentiation are inverses!

The derivative of the integral of the function In layman's terms in equation form:

$\int f\text{'}\left(x\right)dx=f\left(x\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{d}{dx}\left(\int f\left(x\right)dx\right)=f\left(x\right)$

The derivative of the integral of a function results in the function itself! So, all part 1 of the FTC is saying is that integration and differentiation are inverses of each other - they undo each other!

## The second part of the Fundamental Theorem of Calculus

This part is informally referred to as "the evaluating part of the theorem".

If the function f is continuous on [a, b], then

${\int }_{a}^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right)$

where F is the antiderivative of f, or a function such that $F\text{'}=f$.

Essentially, Part 2 is the inverse of Part 1. If the antiderivative F of f is already known, we can evaluate the integral by subtracting the values of F at the endpoints of the interval of integration.

## Proof of the Fundamental Theorem of Calculus

### Part 1

First, we need to start off with a few pre-requirements. We need:

1. A function f that is continuous on [a,b]
2. The integral $F\left(x\right)={\int }_{a}^{x}f\left(t\right)dt$continuous on [a,b] and differentiable on (a,b)

Remember, we have to prove that the derivative of the integral on the function f is the function f itself. In other words

$\frac{d}{dx}{\int }_{a}^{x}f\left(t\right)dt=f\left(x\right)$

Let x and $x+h$ be in the interval (a, b). Then, by the definition of the derivative, we have

$\begin{array}{rcl}F\text{'}\left(x\right)& =& \underset{h\to 0}{\mathrm{lim}}\frac{F\left(x+h\right)-F\left(x\right)}{h}\\ & =& \underset{h\to 0}{\mathrm{lim}}\frac{{\int }_{a}^{x+h}f\left(t\right)dt-{\int }_{a}^{x}f\left(t\right)dt}{h}\\ & =& \underset{h\to 0}{\mathrm{lim}}\frac{\left({\int }_{a}^{x}f\left(t\right)dt+{\int }_{x}^{x+h}f\left(t\right)dt\right)-{\int }_{a}^{x}f\left(t\right)dt}{h}\\ & =& \underset{h\to 0}{\mathrm{lim}}\frac{{\int }_{x}^{x+h}f\left(t\right)dt}{h}\\ & & \end{array}$

Factoring out h leaves us with

$F\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1}{h}\left({\int }_{x}^{x+h}f\left(t\right)dt\right)\right)$

For this next part, we need to use our Mean Value Theorem for Integrals muscles, found here! The area under any two points of a curve, let's say between points a and b, is calculated as

$Areaunder2points,aandb=width×averageheightofthefunctionon\left[a,b\right]$

Rewriting the equation above In math notation according to the diagram below, we have

${\int }_{a}^{b}f\left(x\right)dx=f\left(c\right)\left(b-a\right)$

The area under the curve is equal to the width of the interval multiplied by the average value of the curve - StudySmarter Originals

Rewriting the formula above, subbing in x for a, $x+h$ for b, and h for $b-a$, we have

$\begin{array}{rcl}f\left(c\right)·h& =& {\int }_{x}^{x+h}f\left(t\right)dt\\ f\left(c\right)& =& \frac{1}{h}{\int }_{x}^{x+h}f\left(t\right)dt\end{array}$

What happens as $h\to 0$? Two things happen:

1. $x+h\to 0$, and so
2. $c\to x$

Mathematically:

1. $\underset{h\to 0}{\mathrm{lim}}f\left(c\right)=f\left(x\right)$

Ok, let's review what we have determined so far. First, we know that

1. $F\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1}{h}\left({\int }_{x}^{x+h}f\left(t\right)dt\right)\right)$
2. $\begin{array}{rcl}f\left(c\right)& =& \frac{1}{h}{\int }_{x}^{x+h}f\left(t\right)dt\end{array}$

Subbing equation II into equation I we get;

$F\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(c\right)$

But wait, equation 1 above tells us what $\underset{h\to 0}{\mathrm{lim}}f\left(c\right)$ is! It's $f\left(x\right)$! So

$\begin{array}{rcl}F\text{'}\left(x\right)& =& \underset{h\to 0}{\mathrm{lim}}f\left(c\right)\\ & =& f\left(x\right)\end{array}$

From the start of this proof, we saw that

$F\left(x\right)={\int }_{a}^{x}f\left(t\right)dt$

Therefore, the derivative of $F\left(x\right),F\text{'}\left(x\right),$ is

$F\text{'}\left(x\right)=\frac{d}{dx}{\int }_{a}^{x}f\left(t\right)dt$

Finally, we have

$\begin{array}{rcl}F\text{'}\left(x\right)& =& \frac{d}{dx}{\int }_{a}^{x}f\left(t\right)dt\\ & =& f\left(x\right)\end{array}$

### Part 2

Let $g\left(x\right)={\int }_{a}^{x}f\left(t\right)dt$. Part 1 of FTC says that

$g\text{'}\left(x\right)=f\left(x\right)$

If F is the antiderivative of f on the interval $\left[a,b\right]$, then

$F\left(x\right)=g\left(x\right)+C$

where C is a constant.

Let's say we want to find the area under the curve between the same point. This area should be zero since there is no width! So the area at this point, let's say $x=a$, is calculated as

$\begin{array}{rcl}g\left(a\right)& =& {\int }_{a}^{a}f\left(t\right)dt\\ & =& 0\end{array}$

Using the fact that $F\left(x\right)=g\left(x\right)+C$, subbing in $x=a$ and $x=b$, and using the fact that $g\left(a\right)=0$ gives us

$\begin{array}{rcl}F\left(b\right)-F\left(a\right)& =& \left[g\left(b\right)+C\right]-\left[g\left(a\right)+C\right]\\ & =& g\left(b\right)-g\left(a\right)\\ & =& g\left(b\right)\\ & =& {\int }_{a}^{b}f\left(t\right)dt\end{array}$

## Applications of the Fundamental Theorem of Calculus

Calculus is an extremely powerful tool for evaluating integrals; it allows us to evaluate integrals without approximations or geometry. The main application of the FTC is finding exact integral answers. However, using the FTC, we can also find and study antiderivatives more abstractly.

As previously mentioned, the FTC helps us to solve more complex problems. Believe it or not, integration and the Fundamental Theorem of Calculus can help us calculate the center of mass, which is essentially a balancing point. If you stand and try to balance on one leg, you have to adjust your center of mass to balance on one leg rather than two. In Calculus, the center of mass of a graphical region is the point where the region is perfectly horizontally balanced. The FTC can help us to solve for the center of mass.

## Examples of the Fundamental Theorem of Calculus

Let's first do 3 examples using FTC part 1!

### Example 1

Find the derivative of $g\left(x\right)={\int }_{1}^{x}\frac{1}{{t}^{2}+1}dt$ and evaluate $g\text{'}\left(10\right)$.

#### Step 1: Ensure g(x) is continuous

$\frac{1}{{t}^{2}+1}$ is continuous on the interval $\left[1,x\right]$. Plugging in any number between 1 and infinity will not produce any discontinuities.

#### Step 2: Take the derivative of both sides with respect to x

$\frac{d}{dx}\left[g\left(x\right)\right]=\frac{d}{dx}\left[{\int }_{1}^{x}\frac{1}{{t}^{2}+1}dt\right]$

#### Step 3: Apply the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus, all we need to do is replace the t variable of the inner function with x such that

$g\text{'}\left(x\right)=\frac{1}{{x}^{2}+1}$

#### Step 4: Plug in x = 10 to g'(x)

Now that we have g'(x), we simply plug in 10 to find g'(10).

$g\text{'}\left(10\right)=\frac{1}{{10}^{2}+1}=\frac{1}{101}$

### Example 2

Now that we better understand the Fundamental Theorem of Calculus, let's move on to a more complex example involving The Chain Rule. If you haven't gone over the chain rule, please check out our article before proceeding!

Find the derivative of $g\left(x\right)={\int }_{1}^{{x}^{2}}\mathrm{cos}\left(r\right)dr$.

#### Step 1: Ensure g(x) is continuous

We know that the cosine function has no discontinuities.

#### Step 2: Substitute u(x) in for the upper limit function

We let $u\left(x\right)={x}^{2}$, then

$g\left(x\right)={\int }_{1}^{u\left(x\right)}\mathrm{cos}\left(r\right)dr$

#### Step 3: Take the derivative of both sides with respect to x

$\frac{d}{dx}\left[g\left(x\right)\right]=\frac{d}{dx}\left[{\int }_{1}^{u\left(x\right)}\mathrm{cos}\left(r\right)dr\right]$

#### Step 3: Apply the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus, all we need to do is replace the r variable of the inner function with u(x) such that

$g\text{'}\left(x\right)=\mathrm{cos}\left(u\left(x\right)\right)$

#### Step 4: Apply the Chain Rule

We must also apply the Chain Rule, so

$g\text{'}\left(x\right)=\mathrm{cos}\left(u\left(x\right)\right)\frac{du}{dx}=\mathrm{cos}\left({x}^{2}\right)\left[\frac{d}{dx}\left({x}^{2}\right)\right]=2x·\mathrm{cos}\left({x}^{2}\right)$

### Example 3

Let's try an example with two variable limits of integration.

Evaluate

$g\left(t\right)={\int }_{t}^{{t}^{3}}{x}^{4}dx$

#### Step 1: Ensure g(t) is continuous

There is no such number that will produce a discontinuity when plugged in.

#### Step 2: Split the integral in two

Since both limits of integration are variables, we must split the integral in two and use our definite integral rules to match the form of the Fundamental Theorem of Calculus.

$\begin{array}{rcl}g\left(t\right)& =& {\int }_{t}^{{t}^{3}}{x}^{4}dx\\ & =& {\int }_{t}^{0}{x}^{4}dx+{\int }_{0}^{{t}^{3}}{x}^{4}dx\\ & =& -{\int }_{0}^{t}{x}^{4}dx+{\int }_{0}^{{t}^{3}}{x}^{4}dx\end{array}$

#### Step 3: Substitute u(t) for the upper limit

We let $u\left(t\right)={t}^{3}$, then

$g\left(t\right)=-{\int }_{0}^{t}{x}^{4}dx+{\int }_{0}^{u\left(t\right)}{x}^{4}dx$

#### Step 4: Take the derivative of both sides with respect to t

$\frac{d}{dt}\left[g\left(t\right)\right]=\frac{d}{dt}\left[-{\int }_{0}^{t}{x}^{4}dx+{\int }_{0}^{u\left(t\right)}{x}^{4}dx\right]$

#### Step 4: Apply the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus, all we need to do is replace the x variable of the inner function with t and u(x) such that

$g\text{'}\left(t\right)=-{t}^{4}+\left(u{\left(t\right)\right)}^{4}$

Again, we must also apply the Chain Rule, so

$\begin{array}{rcl}g\text{'}\left(t\right)& =& -{t}^{4}+{\left({t}^{3}\right)}^{4}\frac{du}{dx}\\ & =& -{t}^{4}+{\left({t}^{3}\right)}^{4}\left[\frac{d}{dx}\left({t}^{3}\right)\right]\\ & =& -{t}^{4}+{\left({t}^{3}\right)}^{4}\left(3{t}^{2}\right)\\ & =& -{t}^{4}+3{t}^{14}\end{array}$

### Example 4

Now let's try an example using part 2 of the FTC!

Use the Fundamental Theorem of Calculus Part 2 to evaluate

${\int }_{0}^{5}4-{x}^{2}dx$

#### Step 1: Ensure that the function is continuous over the interval of integration

Since the function inside the integrand is a polynomial, we know that it is continuous over the entire interval.

#### Step 2: Find the antiderivative of the function

The antiderivative of the function is

$\int 4-{x}^{2}dx=4x-\frac{{x}^{3}}{3}$

#### Step 3: Apply the Fundamental Theorem of Calculus

To apply part 2 of the FTC, we simply plug in $x=5$ and $x=0$ into the antiderivative and subtract.

${\int }_{0}^{5}4-{x}^{2}dx=\left[4\left(5\right)-\frac{{5}^{3}}{3}\right]-\left[4\left(0\right)-\frac{{0}^{3}}{3}\right]\phantom{\rule{0ex}{0ex}}=\frac{-65}{3}$

## The Fundamental Theorem of Calculus - Key takeaways

• The Fundamental Theorem of Calculus relates integrals to derivatives
• The FTC Part 1 states that if the function f is continuous on [a, b], then the function g is defined by$g\left(x\right)={\int }_{a}^{x}f\left(t\right)dt$ where $a\le x\le b$is continuous on [a, b] and differentiable on (a, b), and $g\text{'}\left(x\right)=f\left(x\right)$
• The FTC Part 2 states that if the function f is continuous on [a, b], then${\int }_{a}^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right)$where F is the antiderivative of f, or a function such that $F\text{'}=f$
• The Fundamental Theorem of Calculus gives us a way to evaluate integrals exactly without geometry

#### Flashcards in The Fundamental Theorem of Calculus 57

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What is the fundamental theorem of calculus?

The Fundamental Theorem of Calculus shows the relationship between the integral and the derivative. Essentially, the theorem states that the derivative of a definite integral with respect to the upper limit is the same as the integrand evaluated at the upper limit.

How to use the fundamental theorem of calculus?

You can use the formula for part 2 of the theorem to calculate the area under the curve between 2 points.

When does the fundamental theorem of calculus not apply?

The function must be continuous over the interval of integration.

How to use the first fundamental theorem of calculus?

Taking the derivative of an integral gives the original function that was being integrated.

What is an example of the fundamental theorem of calculus?

The derivative of the integral from 0 to x of the function f(t)=t is x.

## Test your knowledge with multiple choice flashcards

The Evaluation Theorem is another way of referring to the evaluation part of ____.

The value of the definite integral$\int_a^b f(x)\,\mathrm{d}x$gives the ____ $$x=a$$ and $$x=b$$.

The Evaluation Theorem tells you that$\int_a^b f(t)\,\mathrm{d}t = F(b)-F(a).$The function $$F$$ is a(n) ____.

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