  Calculus

# Understanding the Fundamental Theorem of Calculus

04.10.2022 • 6 min read

## Rachel McLean

Subject Matter Expert

In this article, we’ll discuss the meaning of the Fundamental Theorem of Calculus. Then, we’ll break down the First Fundamental Theorem of Calculus and the Second Fundamental Theorem of Calculus. Finally, we’ll practice with some examples.

## What Is the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus helps to define the relationship between integration and differentiation.

The theorem has two parts:

1. The First Fundamental Theorem of Calculus

2. The Second Fundamental Theorem of Calculus

The First Fundamental Theorem of Calculus reveals that integration is the inverse process of differentiation, while the Second Fundamental Theorem of Calculus illuminates the relationship between the integral and the antiderivative function. You might also hear this theorem referred to as the “FTC.”

In this lesson clip, Dr. Hannah Fry shares what this theorem actually means, how it’s calculated, and what it can further allow us to do.

Integration and differentiation were initially developed separately because we did not suspect that they were related. The Fundamental Theorem of Calculus became the connective tissue that linked these essential operations.

The Fundamental Theorem of Calculus was first articulated in 1668 by Scottish mathematician James Gregory. In the later 17th century, English mathematicians Isaac Barrow and Isaac Newton and German mathematician Gottfried Wilhelm Leibniz separately developed the Fundamental Theorem of Calculus. In 1823, French mathematician Austin-Louis Cauchy rigorously proved the theorem.

## What Is the First Fundamental Theorem of Calculus?

The First Fundamental Theorem of Calculus shows that integration and differentiation are inverse operations. We also refer to it as the Fundamental Theorem of Differential Calculus.

Let $f$ be a continuous function on the interval $[a, b]$. Let the function $F(x)$ be defined by

$F(x) = \int_a^x f(t)\,dt$

Then $F$ is continuous on $[a, b]$ and differentiable on $(a, b)$, and $F’(x) = f(x)$ on $(a, b)$.

The First Fundamental Theorem of Calculus formula explains why $F(x)$ is called the antiderivative function of $f$ on $[a, b]$. Taking the derivative of $F(x)= \int_a^x f(t)\,dt$ gives back the original function $f(x)$. To better understand the relationship between a function $f$ and its antiderivative, it can be helpful to ask yourself, “What function $F(x)$ gives back $f(x)$ when differentiated?”

Dr. Tim Chartier explains more about antiderivatives in this lecture clip:

We can rewrite the theorem using Leibniz’s notation like this:

$F'(x) = \frac{d}{dx} \int_a^x f(t)\,dt = f(x)$

This theorem guarantees that any continuous function has an antiderivative function.

Let’s consider two different examples. Consider the function $F(x) = \int_3^x (6t^4 + \sin{(t)} \, dt$. Suppose we want to find $F(x)$. Note that $6t^4 + \sin{(t)}$ is continuous. Then, by the First Fundamental Theorem of Calculus, we have:

$F'(x) = \frac{d}{dx} \int_a^x f(t)\,dt = f(x)$
$F’(x) = \frac{d}{dx} \int_3^x (6t^4 + \sin{(t)}) \, dt = 6x^4 + \sin{(x)}$

Many problems will require you to use the First Fundamental Theorem of Calculus with the Chain Rule for derivatives. The Chain Rule states that $\frac{d}{dx}f(g(x)) = f’(g(x))g’(x)$.

Then, since $\frac{d}{dx} \int_a^x f(t)\,dt = f(x)$ by the First Fundamental Theorem of Calculus, this means that $\frac{d}{dx} \int_a^{g(x)} f(t)\,dt = f(g(x))g’(x)$.

For example, consider the function $F(x) = \int_0^{\tan{(x)}} e^t \, dt$. Notice that the upper bound of the integral is $\tan{(x)}$ instead of $x$. Suppose we want to find $F’(x)$. Then, by the First Fundamental Theorem of Calculus and the tangent function trigonometry rule for derivatives, we have:

$F’(x) = \frac{d}{dx} \int_0^{\tan{(x)}} e^t \, dt$
$F’(x) = e^{\tan{(x)}} \cdot \frac{d}{dx}\tan{(x)}$
$F’(x) = e^{\tan{(x)}} \cdot \sec ^2 (x)$

## What Is the Second Fundamental Theorem of Calculus?

The Second Fundamental Theorem of Calculus clarifies the relationship between the integral and the antiderivative function. It is also referred to as the Fundamental Theorem of Integral Calculus.

Let $f$ be continuous on $[a, b]$ and let $F$ be the antiderivative of $f$. Then:

$\int_{a}^{b} f(x)\,dx = F(x)\Big|_a^b = F(b) - F(a)$

The above statement means that we can evaluate the definite integral of a function on $[a, b]$ by taking the difference between the indefinite integral of the function evaluated at $a$ and the indefinite integral of the function evaluated at $b$. This gives us the area bounded by the curve of $f$, the x-axis, and the lines $x = a$ and $x = b$.

The Second Fundamental Theorem of Calculus allows us to calculate definite integrals without Riemann sums.

If you need a review of indefinite integrals using the fundamental theorem as well as the basic rules of integration, Dr. Hannah Fry explains all this in the following lesson clip below:

### 4 Steps for Evaluating Definite Integrals

Here are four steps to evaluating definite integrals using the Second Fundamental Theorem of Calculus:

Determine the indefinite integral $F(x)$ using rules of integration.

Find $F(b)$ by plugging $b$ into $F(x)$.

Find $F(a)$ by plugging $a$ into $F(x)$.

Take the difference $F(b) - F(a)$.

## 6 Example Exercises of Fundamental Theorem of Calculus

Now, let’s look at some examples of how to use the Fundamental Theorem of Calculus.

### Example 1

Let $k(x) = \int_2^x (5^t+7) \, dt$. Find $k’(x)$.

### Solution

Note that $5^t+7$ is continuous. Let $f(t) = 5^t+7$. Then, by the First Fundamental Theorem of Calculus, we know that

$F'(x) = \frac{d}{dx} \int_a^x f(t)\,dt = f(x)$
$k’(x) = \frac{d}{dx} \int_2^x (5^t+7)\,dt = 5^x + 7$

Thus $k’(x) = 5^x + 7$.

### Example 2

Let $F(x) = \int_1^{x^2} \sin{(t)}\, dt$. Find $F’(x)$.

### Solution

In this problem, we’ll need to combine the First Fundamental Theorem of Calculus and the Chain Rule for derivatives. Then, using the power rule for derivatives, we have:

$F’(x) = \frac{d}{dx} \int_1^{x^2}\sin{(t)}\, dt$
$F’(x) = \sin{(x^2)} \cdot \frac{d}{dx}(x^2)$
$F’(x) = 2x\sin{(x^2)}$

Thus $F’(x) = 2x\sin{(x^2)}$.

### Example 3

Let $F(x) = \int_1^{\sqrt{x}} e^t \, dt$. Find $F’(x)$.

### Solution

In this problem, we’ll need to combine the First Fundamental Theorem of Calculus and the Chain Rule for derivatives. Then, using the power rule for derivatives, we have:

$F’(x) = \frac{d}{dx} \int_1^{\sqrt{x}} e^t \, dt$
$F’(x) = e^{\sqrt{x}} \cdot \frac{d}{dx} \sqrt{x}$
$F’(x) = e^{\sqrt{x}} \cdot \frac{d}{dx} x^{1}{2}$
$F’(x) = e^{\sqrt{x}} \cdot \frac{1}{2}x^{\frac{-1}{2}}$
$F’(x) = \frac{e^{\sqrt{x}}}{2\sqrt{x}}$

Thus $F’(x) = \frac{e^{\sqrt{x}}}{2\sqrt{x}}$.

### Example 4

Evaluate $\int_1^2 (3x^2+2) \, dx$.

### Solution

For this problem, we can use the Second Fundamental Theorem of Calculus. First, we must find the antiderivative function $F(x)$. Using the power rule for integrals, we have:

$F(x) = \int (3x^2 + 2) \, dx = \frac{3x^3}{3} + 2x = x^3 + 2x$

Now, we can evaluate $F(x)$ at $x = 1$ and $x = 2$ and take their difference to find $\int_1^2 (3x^2+2) \, dx$.

$\int_1^2 (3x^2+2) \, dx = x^3 + 2x \Big|_1^2$
$= (8+4) - (1+2)$
$= 12 - 3$
$=9$

Thus $\int_1^2 (3x^2+2) \, dx = 9$.

### Example 5

Evaluate $\int_1^3 \frac{dx}{x}$.

### Solution

For this problem, we can use the Second Fundamental Theorem of Calculus. First, we must find the antiderivative function $F(x)$. Using the reciprocal rule for integrals, we have:

$F(x) = \int \frac{dx}{x} = \ln|x|$

Now, we can evaluate $F(x)$ at $x = 1$ and $x = 3$ and take their difference to find $\int_1^3 \frac{dx}{x}$.

$\int_1^3 \frac{dx}{x} = \ln|x| \Big|_1^3$
$= \ln3 - \ln1$
$\approx 1.099 - 0$
$\approx 1.099$

Thus $\int_1^3 \frac{dx}{x} \approx 1.099$.

### Example 6

Evaluate $\int_0^{\frac{3\pi}{2}} \sin{(x)} \, dx$.

### Solution

For this problem, we can use the Second Fundamental Theorem of Calculus. First, we must find the antiderivative function $F(x)$. Using the sine function trigonometry rule for integrals, we have:

$F(x) = \int \sin{(x)} \, dx = -\cos{(x)}$

Now, we can evaluate $F(x)$ at $x = 0$ and $x = \frac{3\pi}{2}$ and take their difference to find $\int_0^{\frac{3\pi}{2}} \sin{(x)} \, dx$.

$\int_0^{\frac{3\pi}{2}} \sin{(x)} \, dx = -\cos{(x)} \Big|_0^{ \frac{3\pi}{2}}$
$= -\cos{( \frac{3\pi}{2})} - (-\cos{(0)})$
$= 0 - (-1)$
$=1$

Thus $\int_0^{\frac{3\pi}{2}} \sin{(x)} \, dx = 1$.

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