Calculus

# Definite Integrals: What Are They and How to Calculate Them

## Rachel McLean

Subject Matter Expert

Knowing how to find definite integrals is an essential skill in calculus. In this article, we’ll learn the definition of definite integrals, how to evaluate definite integrals, and practice with some examples.

## Defining Definite Integrals

What is a definite integral? Definite integrals are used to calculate the area between a curve and the x-axis on a specific interval. (If you need to review, see our beginner's guide to integrals).

If we want to evaluate the definite integral of a real-valued function $f$ with respect to $x$ on the interval [a, b], where $a$ and $b$ are real numbers and $a \leq b$, we use the following notation:

$\int_{a}^{b} f(x)dx = A$

In this notation, the curved integral sign $\int$ indicates the operation of taking an integral. The rest of this notation is composed of three parts:

• The integrand $f(x)$

• The integral bounds $a$ and $b$, where $a$ is the lower bound and $b$ is the upper bound. These are also referred to as limits.

• The differential $dx$, which tells us that we are integrating $f$ with respect to the variable $x$.

Altogether, this notation represents the area enclosed by $f(x)$, the x-axis, and the lines $x=a$ and $x=b$. Graphically, we can visualize $\int_{a}^{b} f(x)dx$ as something like this:

## Definite Integrals vs. Indefinite Integrals

Before we learn exactly how to solve definite integrals, it’s important to understand the difference between definite and indefinite integrals.

Definite integrals find the area between a function’s curve and the x-axis on a specific interval, while indefinite integrals find the antiderivative of a function. Finding the indefinite integral and finding the definite integral are operations that output different things.

Calculating the indefinite integral takes in one function, and outputs another function: the antiderivative function of $f(x)$, notated by $F(x)$.

This output function is accompanied by an arbitrary constant C and does not involve lower and upper boundaries. By contrast, calculating the definite integral always outputs a real number, which represents the area under the curve on a specific interval. You can see the difference in their notations below:

• The indefinite integral $\int f(x)dx = F(x) + C$

• The definite integral $\int_{a}^{b} f(x)dx = A$, for some real number A.

Given $f(x)$, the indefinite integral answers the question, “What function, when differentiated, gives us $f(x)$?” The indefinite integral gives us a family of functions $F$ since infinite functions will satisfy this question. Thus, the indefinite integral gives us an “indefinite” answer. The definite integral gives us a real number — a unique “definite” answer.

You can learn more about the difference with this lesson sample on indefinite integrals by one of our instructors Dr. Hannah Fry.

## How to Calculate Definite Integrals

To find the definite integral of a function, we can use the Fundamental Theorem of Calculus, which states: If $f$ is continuous and $F$ is an antiderivative of $f$, then $\int_{a}^{b} f(x)dx = [F(x)]^b_a = F(b) - F(a)$.

This means that to find the definite integral of a function on the interval [a, b], we simply take the difference between the indefinite integral of the function evaluated at $a$ and the indefinite integral of the function evaluated at $b$.

We can break this process down into four steps:

1. Find the indefinite integral $F(x)$. You can use the Rules of Integration that you learned with indefinite integrals to help with this part.

2. Find $F(b)$. This is found by plugging the upper bound $b$ into the indefinite integral found in Step 1.

3. Find $F(a)$. This is found by plugging the lower bound $a$ into the indefinite integral found in Step 1.

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

Let’s do one example together. Let’s calculate the definite integral of the function $f(x) = 4x^3-2x$ on the interval [1, 2].

We'll follow the four steps given above.

Step 1:

$\int (4x^3-2x) dx = x^4 - x^2 = F(x)$

Step 2:

$F(2) = 2^4-2^2 = 16-4 = 12$

Step 3:

$F(1) = 1^4-1^2 = 1-1 = 0$

Step 4:

$F(2)-F(1) = 12 - 0 = 12$

Thus, $\int_{1}^{2} (4x^3-2x) dx = 12$.

## Properties of Definite Integrals and Key Equations

Let’s review some of the key properties of definite integrals. These will be useful for solving more complex integral problems. In the following properties, assume that $f$ and $g$ are continuous functions, and let $k$ be a constant.

### Zero-Length Interval Rule

When a=b, the interval has length 0, and so the definite integral definite integral of a function on [a, b] is 0.

$\int_{a}^{a} f(x)dx = 0$

### Reverse Bounds Rule

To find the definite integral of a function on [a, b] where $a > b$, we can simply reverse the sign of $\int_{b}^{a} f(x)dx$.

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

If $a$, $b$, and $c$ are real numbers on a closed interval, then $\int_{a}^{c} f(x)dx$ can be found by adding integrals as follows:

$\int_{a}^{c} f(x)dx = \int_{a}^{b} f(x)dx + \int_{b}^{c} f(x)dx$

### Constant Multiplier Rule

You can pull constants outside of an integral.

$\int_{a}^{b} kf(x)dx = k\int_{a}^{b} f(x)dx$

### Sum and Difference Rule

The integral of the sum or difference of two functions is the sum or difference of their integrals.

$\int_{a}^{b} [f(x) \pm g(x)]dx = \int_{a}^{b} f(x)dx \pm \int_{a}^{b} g(x)dx$

### Integral of a Constant Rule

The integral of a constant over [a, b] is equal to the constant multiplied by the difference $b-a$.

$\int_{a}^{b} kdx = k(b-a)$

### Comparison Properties of Definite Integrals

• If $f(x) \geq 0$ on [a, b], then $\int_{a}^{b} f(x)dx \geq 0$.

• If $f(x) \leq 0$ on [a, b], then $\int_{a}^{b} f(x)dx \leq 0$.

• If $f(x) \geq g(x)$ on [a, b], then $\int_{a}^{b} f(x)dx \geq \int_{a}^{b} g(x)dx$.

### Average Value of a Function

The average value of a function on [a, b] is defined by:

$f_{avg}=\frac{1}{b-a}\int_{a}^{b} f(x)dx=\frac{F(b)-F(a)}{b-a}$

### The Mean Value Theorem

This theorem tells us that there’s at least one point c inside the open interval (a,b) at which $f(c)$ will be equal to the average value of the function over [a, b]. That is, there exists a $c$ on (a, b) such that:

$f(c) = \frac{1}{b-a}\int_{a}^{b} f(x)dx$

or equivalently

$f(c)(b-a) = \int_{a}^{b} f(x)dx$

## 3 Practice Exercises and Solutions

Here are three exercises for you to practice how to do a definite integral and their solutions.

### Exercise 1

Calculate the definite integral of the function $f(x) = \cos{(x)}$ on the interval $[0, \frac{\pi}{2}]$.

### Solution:

$\int_{0}^{\frac{\pi}{2}} \cos{(x)}dx = [\sin{(x)}]^{\frac{\pi}{2}}_0$
$= \sin{(\frac{\pi}{2})} - \sin{(0)}$
$= 1 - 0$
$=1$

### Exercise 2

Determine the average value of $f(x) = 12x^2-2x$ on $[2, 4]$.

### Solution:

$f_{avg}=\frac{1}{4-2}\int_{2}^{4} (12x^2-2x)dx$
$=\frac{1}{2}\left[\frac{12x^3}{3}-\frac{2x^2}{2}\right]^4_2$
$=\frac{1}{2}\left[4x^3-x^2\right]^4_2$
$=\frac{1}{2}((4 \cdot 4^3-4^2)-(4 \cdot 2^3-2^2))$
$=\frac{1}{2}((256-16)-(32-4))$
$=\frac{1}{2}(240-28)$
$=\frac{1}{2}(212)$
$=106$

### Exercise 3

Given that $\int_{3}^{10} f(x)dx = 17$ and $\int_{7}^{10} f(x)dx=9$, evaluate $\int_{3}^{7} f(x)dx$.

### Solution:

$\int_{3}^{10} f(x)dx = \int_{3}^{7} f(x)dx + \int_{7}^{10} f(x)dx$
$\int_{3}^{7} f(x)dx = \int_{3}^{10} f(x)dx - \int_{7}^{10} f(x)dx$
$\int_{3}^{7} f(x)dx = 17 - 9$
$\int_{3}^{7} f(x)dx = 8$

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