## Antiderivatives

An antiderivative is a differentiable function [latex]F[/latex] whose derivative is equal to [latex]f[/latex] (i.e., [latex]F' = f[/latex]).

### Learning Objectives

Calculate the antiderivative (aka the indefinite integral) for a given function

### Key Takeaways

#### Key Points

- The process of solving for antiderivatives is called antidifferentiation, and its opposite operation is called differentiation, which is the process of finding a derivative.
- Antiderivatives are related to definite integrals through the fundamental theorem of calculus: the definite integral of a function over an interval is equal to the difference between the values of an antiderivative evaluated at the endpoints of the interval.
- The graphs of antiderivatives of a given function are vertical translations of each other, with each graph’s location depending upon the value of constant [latex]C[/latex].

#### Key Terms

**derivative**: a measure of how a function changes as its input changes**definite integral**: the integral of a function between an upper and lower limit

An antiderivative is a differentiable function F whose derivative is equal to [latex]f[/latex] (i.e., [latex]F'=f[/latex]). The process of solving for antiderivatives is called antidifferentiation, and its opposite operation is called differentiation, which is the process of finding a derivative. Antiderivatives are related to definite integrals through the fundamental theorem of calculus: the definite integral of a function over an interval is equal to the difference between the values of an antiderivative evaluated at the endpoints of the interval.

Let’s consider the case of function [latex]F(x) = \frac{x^3}{3}[/latex], which is an antiderivative of [latex]f(x) = x^2[/latex]. As the derivative of a constant is zero, [latex]x^2[/latex] will have an infinite number of antiderivatives, such as [latex]\frac{x^3}{3} + 0[/latex], [latex]\frac{x^3}{3} + 7[/latex], [latex]\frac{x^3}{3} - 42[/latex], [latex]\frac{x^3}{3} + 293[/latex], etc. Therefore, all the antiderivatives of [latex]x^2[/latex] can be obtained by adding the value of [latex]C[/latex] in [latex]F(x) = \frac{x^3}{3} + C[/latex], where [latex]C[/latex] is an arbitrary constant known as the constant of integration. Essentially, the graphs of antiderivatives of a given function are vertical translations of each other, with each graph’s location depending upon the value of [latex]C[/latex].

Antiderivatives are important because they can be used to compute definite integrals with the fundamental theorem of calculus: if [latex]F[/latex] is an antiderivative of the integrable function [latex]f[/latex], and [latex]f[/latex] is continuous over the interval [latex][a, b][/latex], then

[latex]\displaystyle{\int_{a}^{b}f(x)dx = F(b) - F(a)}[/latex]

Because of this rule, each of the infinitely many antiderivatives of a given function [latex]f[/latex] is sometimes called the “general integral” or “indefinite integral” of [latex]f[/latex], and is written using the integral symbol with no bounds:

[latex]\displaystyle{\int f(x)dx}[/latex]

If [latex]F[/latex] is an antiderivative of [latex]f[/latex], and the function [latex]f[/latex] is defined on some interval, then every other antiderivative [latex]G[/latex] of [latex]f[/latex] differs from [latex]F[/latex] by a constant: there exists a number [latex]C[/latex] such that [latex]G(x) = F(x) + C[/latex] for all [latex]x[/latex]. [latex]C[/latex] is called the arbitrary constant of integration. If the domain of [latex]F[/latex] is a disjoint union of two or more intervals, then a different constant of integration may be chosen for each of the intervals. For instance:

[latex]F (x) =\begin{cases} \ -1/x + C_{1} \text{ if } x<0 \\ \ -1/x + C_{2} \text{ if } x>0 \end{cases}[/latex]

is the most general antiderivative of [latex]f(x) = \frac{1}{x^2}[/latex] on its natural domain of:

[latex](-\infty; 0)\bigcup (0; \infty )[/latex].

## Area and Distances

Defined integrals are used in many practical situations that require distance, area, and volume calculations.

### Learning Objectives

Apply integration to calculate problems about the area under a graph, or the distance of an arc

### Key Takeaways

#### Key Points

- The definite integral [latex]\int_{a}^{b}f(x)dx[/latex] is defined informally to be the area of the region in the [latex]xy[/latex]-plane bound by the graph of [latex]f[/latex], the [latex]x[/latex]-axis, and the vertical lines [latex]x = a [/latex] and [latex]x=b[/latex], such that the area above the [latex]x[/latex]-axis adds to the total, and the area below the [latex]x[/latex]-axis subtracts from the total.
- According to the fundamental theorem of calculus: if [latex]f[/latex] is a continuous real-valued function defined on a closed interval [latex][a,b][/latex], then, once an antiderivative [latex]F[/latex] of [latex]f[/latex] is known, the definite integral of [latex]f[/latex] over that interval is given by [latex]\int_{a}^{b}f(x)dx = F(b) - F(a)[/latex].
- When practical approximation does not provide precise enough results for distance, area, and volume calculations, integration must be performed.

#### Key Terms

**antiderivative**: an indefinite integral**integration**: the operation of finding the region in the [latex]xy[/latex]-plane bound by a given function**definite integral**: the integral of a function between an upper and lower limit

Integration is an important concept in mathematics and—together with its inverse, differentiation —is one of the two main operations in calculus.

Integration is connected with differentiation through the fundamental theorem of calculus: if [latex]f[/latex] is a continuous real-valued function defined on a closed interval [latex][a,b][/latex], then, once an antiderivative F of f is known, the definite integral of [latex]f[/latex] over that interval is given by[latex]\int_{a}^{b}f(x)dx = F(b) - F(a)[/latex]. Definite integrals appear in many practical situations that require distance, area, and volume calculations.

### Area

To start off, consider the curve [latex]y = f(x)[/latex] between [latex]0[/latex] and [latex]x=1[/latex] with [latex]f(x) = \sqrt{x}.[/latex]

We ask, “What is the area under the function [latex]f[/latex], over the interval from [latex]0[/latex] to [latex]1[/latex]? ” and call this (yet unknown) area the integral of [latex]f[/latex]. The notation for this integral will be:

[latex]\displaystyle{\int_{0}^{1} \sqrt{x} dx}[/latex]

As a first approximation, look at the unit square given by the sides [latex]x = 0[/latex] to [latex]x = 1[/latex], [latex]y = f(0) = 0[/latex], and [latex]y = f(1) = 1[/latex]. Its area is exactly [latex]1[/latex]. As it is, the true value of the integral must be somewhat less. Decreasing the width of the approximation rectangles should yield a better result, so we will cross the interval in five steps, using the approximation points [latex]0[/latex], [latex]\frac{1}{5}[/latex], [latex]\frac{2}{5}[/latex], and so on, up to [latex]1[/latex]. Fit a box for each step using the right end height of each curve piece, thus obtaining [latex]\sqrt{\frac{1}{5}}[/latex], [latex]\sqrt{\frac{2}{5}}[/latex], and so on, up to[latex]\sqrt{1} = 1[/latex]. Summing the areas of these rectangles, we get a better approximation for the sought integral, namely:

[latex]\displaystyle{\sqrt{\frac{1}{5}} \left ( \frac{1}{5} - 0 \right ) + \sqrt{\frac{2}{5}} \left ( \frac{2}{5} - \frac{1}{5} \right ) + \cdots + \sqrt{\frac{5}{5}} \left ( \frac{5}{5} - \frac{4}{5} \right ) \approx 0.7497}[/latex]

Notice that we are taking a finite sum of many function values of [latex]f[/latex], multiplied with the differences of two subsequent approximation points. We can easily see that the approximation is still too large. Using more steps produces a closer approximation, but will never be exact: replacing the [latex]5[/latex] subintervals by twelve as depicted, we will get an approximate value for the area of [latex]0.6203[/latex], which is too small. The key idea is the transition from adding a finite number of differences of approximation points multiplied by their respective function values to using an infinite number of fine, or infinitesimal, steps.

Applying the fundamental theorem of calculus to the square root curve, [latex]f(x) = x^{1/2}[/latex], we look at the antiderivative, [latex]F(x) = \frac{2}{3} \cdot x^\frac{3}{2}[/latex], and simply take [latex]F(1) − F(0)[/latex], where [latex]0[/latex] and [latex]1[/latex] are the boundaries of the interval [latex][0,1][/latex]. So the exact value of the area under the curve is computed formally as:

[latex]\displaystyle{\int_{0}^{1}\sqrt{x}dx = \int_{0}^{1}x^{1/2}dx = F(1) - F(0) = \frac{2}{3}}[/latex]

### Distance (Finding arc length by Integrating)

If you know the velocity [latex]v(t) [/latex] of an object as a function of time, you can simply integrate [latex]v(t) [/latex] over time to calculate the distance the object traveled. Since this is equivalent to evaluating the area under the curve [latex]v(t) [/latex], we will not discuss more on this.

However, you can also use integrals to calculate length—for example, the length of an arc described by a function [latex]y = f(x)[/latex]. Consider an infinitesimal part of the curve [latex]ds[/latex] on the curve (or consider this as a limit in which the change in [latex]s[/latex] approaches [latex]ds[/latex]). According to Pythagoras’s theorem, [latex]ds^2=dx^2+dy^2[/latex], from which we can determine:

[latex]\displaystyle{\frac{ds^2}{dx^2}=1+\frac{dy^2}{dx^2}}[/latex]

[latex]\displaystyle{ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx}[/latex]

[latex]\displaystyle{s = \int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }\, dx}[/latex]

Equivalently, if a curve is defined parametrically by [latex]x = X(t)[/latex] and [latex]y = Y(t)[/latex], we get:

[latex]\displaystyle{ds = \sqrt{ \left( \frac{dx}{dt} \right)^2+ \left(\frac{dy}{dt} \right)^2} \cdot dt}[/latex]

then its arc length between [latex]t = a[/latex] and [latex]t = b[/latex] is:

[latex]\displaystyle{s = \int_{a}^{b} \sqrt { [X'(t)]^2 + [Y'(t)]^2 }\, dt}[/latex]

### Example

For the following curve described by the parameter [latex]t[/latex]:

[latex]\begin{cases} y = t^5 \\ x = t^3 \end{cases}[/latex]

the arc length integral for values of [latex]t[/latex] from [latex]-1[/latex] to [latex]1[/latex] is:

[latex]\displaystyle{\int_{-1}^1 \sqrt{(3t^2)^2 + (5t^4)^2}\,dt}[/latex]

[latex]\displaystyle{= \int_{-1}^1 \sqrt{9t^4 + 25t^8}\,dt}[/latex]

## The Definite Integral

A definite integral is the area of the region in the [latex]xy[/latex]-plane bound by the graph of [latex]f[/latex], the [latex]x[/latex]-axis, and the vertical lines [latex]x=a[/latex] and [latex]x=b[/latex].

### Learning Objectives

Compute the definite integral of a function over a set interval

### Key Takeaways

#### Key Points

- Integration is an important concept in mathematics and—together with its inverse, differentiation —is one of the two main operations in calculus.
- Integration is connected with differentiation through the fundamental theorem of calculus: if [latex]f[/latex] is a continuous real-valued function defined on a closed interval [latex][a, b][/latex], then, once an antiderivative [latex]F[/latex] of [latex]f[/latex] is known, the definite integral of [latex]f[/latex] over that interval is given by [latex]\int_{a}^{b}f(x)dx = F(b) - F(a)[/latex].
- Definite integrals appear in many practical situations, and their actual calculation is important in the type of precision engineering (of any discipline) that requires exact and rigorous values.

#### Key Terms

**definite integral**: the integral of a function between an upper and lower limit**integration**: the operation of finding the region in the [latex]xy[/latex]-plane bound by the function**antiderivative**: an indefinite integral

Integration is an important concept in mathematics and—together with its inverse, differentiation—is one of the two main operations in calculus. Given a function [latex]f[/latex] of a real variable [latex]x[/latex] and an interval [latex][a, b][/latex] of the real line, the definite integral [latex]\int_{a}^{b}f(x)dx[/latex] is defined informally to be the area of the region in the [latex]xy[/latex]-plane bound by the graph of [latex]f[/latex], the [latex]x[/latex]-axis, and the vertical lines [latex]x = a[/latex] and [latex]x=b[/latex], such that the area above the [latex]x[/latex]-axis adds to the total, and that the area below the [latex]x[/latex]-axis subtracts from the total. Integrals such as these are termed definite integrals.

The principles of integration were formulated independently by Isaac Newton and Gottfried Leibniz in the late 17th century. Through the fundamental theorem of calculus, which they independently developed, integration is connected with differentiation: if [latex]f[/latex] is a continuous real-valued function defined on a closed interval [latex][a, b][/latex], then, once an antiderivative [latex]F[/latex] of [latex]f[/latex] is known, the definite integral of [latex]f[/latex] over that interval is given by

[latex]\displaystyle{\int_{a}^{b}f(x)dx = F(b) - F(a)}[/latex]

Definite integrals appear in many practical situations. If a swimming pool is rectangular with a flat bottom, then from its length, width, and depth we can easily determine the volume of water it can contain (to fill it), the area of its surface (to cover it), and the length of its edge (to rope it). But if it is oval with a rounded bottom, all of these quantities call for integrals. Practical approximations may suffice for such trivial examples, but precision engineering (of any discipline) requires exact and rigorous values for these elements.

For example, consider the curve [latex]y = f(x)[/latex] between 0 and *x* = 1 with [latex]f(x) = \sqrt{x}.[/latex]

We ask, “What is the area under the function [latex]f[/latex], over the interval from 0 to 1? ” and call this (yet unknown) area the integral of [latex]f[/latex]. The notation for this integral will be:

[latex]\int_{0}^{1} \sqrt{x} dx[/latex]

As a first approximation, look at the unit square given by the sides [latex]x = 0[/latex] to [latex]x = 1[/latex], [latex]y = f(0) = 0[/latex], and [latex]y = f(1) = 1[/latex]. Its area is exactly [latex]1[/latex]. As it is, the true value of the integral must be somewhat less. Decreasing the width of the approximation rectangles should yield a better result, so we will cross the interval in five steps, using the approximation points [latex]0[/latex], [latex]\frac{1}{5}[/latex], [latex]\frac{2}{5}[/latex], and so on, up to [latex]1[/latex]. Fit a box for each step using the right end height of each curve piece, thus obtaining [latex]\sqrt{\frac{1}{5}}[/latex], [latex]\sqrt{\frac{2}{5}}[/latex], and so on, up to[latex]\sqrt{1} = 1[/latex]. Summing the areas of these rectangles, we get a better approximation for the sought integral, namely:

[latex]\displaystyle{\sqrt{\frac{1}{5}} \left ( \frac{1}{5} - 0 \right ) + \sqrt{\frac{2}{5}} \left ( \frac{2}{5} - \frac{1}{5} \right ) + \cdots + \sqrt{\frac{5}{5}} \left ( \frac{5}{5} - \frac{4}{5} \right ) \approx 0.7497}[/latex]

Notice that we are taking a finite sum of many function values of [latex]f[/latex], multiplied with the differences of two subsequent approximation points. We can easily see that the approximation is still too large. Using more steps produces a closer approximation, but will never be exact: replacing the [latex]5[/latex] subintervals by twelve as depicted, we will get an approximate value for the area of [latex]0.6203[/latex], which is too small. The key idea is the transition from adding a finite number of differences of approximation points multiplied by their respective function values to using an infinite number of fine, or infinitesimal, steps.

As for the actual calculation of integrals, the fundamental theorem of calculus, due to Newton and Leibniz, is the fundamental link between the operations of differentiating and integrating. Applied to the square root curve, [latex]f(x) = x^{1/2}[/latex], the theorem says to look at the antiderivative:

[latex]\displaystyle{F(x) = \frac{2}{3} \cdot x^\frac{3}{2}}[/latex]

and simply take [latex]F(1) − F(0)[/latex], where [latex]0[/latex] and [latex]1[/latex] are the boundaries of the interval [latex][0,1][/latex]. So the exact value of the area under the curve is computed formally as:

[latex]\displaystyle{\int_{0}^{1}\sqrt{x}dx = \int_{0}^{1}x^{1/2}dx = F(1) - F(0) = \frac{2}{3}}[/latex]

## The Fundamental Theorem of Calculus

The fundamental theorem of calculus is a theorem that links the concept of the derivative of a function to the concept of the integral.

### Learning Objectives

Define the first and second fundamental theorems of calculus

### Key Takeaways

#### Key Points

- The first part of the theorem shows that an indefinite integration can be reversed by differentiation.
- The second part allows one to compute the definite integral of a function by using any one of its infinitely many antiderivatives.
- The second part of the theorem has invaluable practical applications because it markedly simplifies the computation of definite integrals.

#### Key Terms

**antiderivative**: an indefinite integral**derivative**: a measure of how a function changes as its input changes**definite integral**: the integral of a function between an upper and lower limit

The fundamental theorem of calculus is a theorem that links the concept of the derivative of a function to the concept of the integral.

There are two parts to the theorem. Loosely put, the first part deals with the derivative of an antiderivative, while the second part deals with the relationship between antiderivatives and definite integrals.

The first part of the theorem, sometimes called the first fundamental theorem of calculus, shows that an indefinite integration can be reversed by differentiation. This part of the theorem is also important because it guarantees the existence of antiderivatives for continuous functions.

Let [latex]f[/latex] be a continuous real-valued function defined on a closed interval [latex][a,b][/latex]. Let [latex]F[/latex] be the function defined, for all [latex]x[/latex] in [latex][a,b][/latex], by:

[latex]\displaystyle{F(x) = \int_{a}^{b}f(t)dt}[/latex]

Now, [latex]F[/latex] is continuous on [latex][a,b][/latex], differentiable on the open interval [latex]( a,b )[/latex], and [latex]F'(x) = f(x)[/latex] for all [latex]x[/latex] in [latex]( a,b )[/latex].

The second part, sometimes called the second fundamental theorem of calculus, allows one to compute the definite integral of a function by using any one of its infinitely many antiderivatives. This part of the theorem has invaluable practical applications, because it markedly simplifies the computation of definite integrals.

Let [latex]f[/latex] and [latex]F[/latex] be real-valued functions defined on a closed interval [latex][a,b][/latex] such that the derivative of [latex]F[/latex] is [latex]f[/latex]. That is, [latex]f[/latex] and [latex]F[/latex] are functions such that, for all [latex]x[/latex] in [latex][a,b][/latex], [latex]F'(x) = f(x)[/latex]. If [latex]f[/latex] is a Riemann integrable on [latex][a,b][/latex], then:

[latex]\displaystyle{\int_{a}^{b}f(x)dx = F(b) - F(a)}[/latex]

The first published statement and proof of a restricted version of the fundamental theorem was by James Gregory (1638–1675). Isaac Barrow (1630–1677) proved a more generalized version of the theorem, while Barrow’s student Isaac Newton (1643–1727) completed the development of the surrounding mathematical theory. Gottfried Leibniz (1646–1716) systematized the knowledge into a calculus for infinitesimal quantities and introduced the notation used today.

## Indefinite Integrals and the Net Change Theorem

An indefinite integral is defined as [latex]\int f(x)dx = F(x)+ C[/latex], where [latex]F[/latex] satisfies [latex]F'(x) = f(x)[/latex] and where [latex]C[/latex] is any constant.

### Learning Objectives

Apply the basic properties of indefinite integrals, including the constant, sum, and difference rules

### Key Takeaways

#### Key Points

- The constant rule for indefinite integrals: [latex]\int cf(x)dx = c\int f(x)dx[/latex]
- The sum rule for indefinite integrals: [latex]\int (f(x)+ g(x)) dx = \int f(x)dx + \int g(x)dx[/latex]
- The difference rule for indefinite integrals: [latex]\int (f(x)- g(x)) dx = \int f(x)dx - \int g(x)dx[/latex]
- The integral of a rate of change is the net change (displacement for position functions ): [latex]\int_{a}^{b} f(x)dx = f(b) - f(a)[/latex]

#### Key Terms

**antiderivative**: an indefinite integral**definite integral**: the integral of a function between an upper and lower limit**integral**: also sometimes called antiderivative; the limit of the sums computed in a process in which the domain of a function is divided into small subsets and a possibly nominal value of the function on each subset is multiplied by the measure of that subset, all these products then being summed

### Indefinite Integrals and Antiderivatives

As you remember from the atoms on antiderivatives, [latex]F[/latex] is said to be an antiderivative of [latex]f[/latex] if [latex]F'(x) = f(x)[/latex]. However, [latex]F[/latex] is not the only antiderivative. We can add any constant [latex]C[/latex] to [latex]F[/latex] without changing the derivative. With this in mind, we define the indefinite integral as follows: [latex]\int f(x)dx = F(x)+ C[/latex], where [latex]F[/latex] satisfies [latex]F'(x) = f(x)[/latex] and [latex]C[/latex] is any constant.

[latex]f(x)[/latex], the function being integrated, is known as the integrand. Note that the indefinite integral yields a family of functions.

For example, the function [latex]F(x) = \frac{x^3}{3}[/latex] is an antiderivative of [latex]f(x) = x^2[/latex]. Since the derivative of a constant is zero, [latex]x^2[/latex] will have an infinite number of antiderivatives, such as [latex]\left ( \frac{x^3}{3} \right ) + 0[/latex], [latex]\left ( \frac{x^3}{3} \right ) + 7[/latex], [latex]\left ( \frac{x^3}{3} \right ) - 42[/latex],[latex]\left ( \frac{x^3}{3} \right ) + 293[/latex], etc. Therefore, all the antiderivatives of [latex]x^2[/latex] can be obtained by changing the value of [latex]C[/latex] in [latex]F(x) = \left ( \frac{x^3}{3} \right ) + C[/latex], where [latex]C[/latex] is an arbitrary constant known as the constant of integration. Essentially, the graphs of antiderivatives of a given function are vertical translations of each other, with each graph’s location depending upon the value of [latex]C[/latex].

Indefinite integrals exhibit the following basic properties.

### The Constant Rule for Indefinite Integrals

[latex]\int cf(x)dx = c\int f(x)dx[/latex]

### The Sum Rule for Indefinite Integrals

[latex]\int (f(x)+ g(x)) dx = \int f(x)dx + \int g(x)dx[/latex]

### The Difference Rule for Indefinite Integrals

[latex]\int (f(x)- g(x)) dx = \int f(x)dx - \int g(x)dx[/latex]

### Definite Integrals and the Net Change Theorem

Integrating over a specified domain yields what is called a ” definite integral ” (in that the domain is defined). Integrating over a domain [latex]D[/latex] is written as [latex]\int_{a}^{b} f(x)dx[/latex] if the domain is an interval [latex][a, b][/latex] of [latex]x[/latex].

Such a problem can be solved using the net change theorem, which states that the integral of a rate of change is the net change (displacement for position functions):

[latex]\displaystyle{\int_{a}^{b} f(x)dx = f(b) - f(a)}[/latex]

Basically, the theorem states that the integral of or [latex]F'[/latex] from [latex]a[/latex] to [latex]b[/latex] is the area between [latex]a[/latex] and [latex]b[/latex], or the difference in area from the position of [latex]f(a)[/latex] to the position of [latex]f(b)[/latex]. This can be applied to find values such as volume, concentration, density, population, cost, and velocity.

## The Substitution Rule

Integration by substitution is an important tool for mathematicians used to find integrals and antiderivatives.

### Learning Objectives

Use [latex]u[/latex]-substitution (the substitution rule) to find the antiderivative of more complex functions

### Key Takeaways

#### Key Points

- The substitution [latex]x = g(t)[/latex] yields [latex]\frac{dx}{dt} = g'(t)[/latex] and therefore, formally, [latex]dx = g'(t)dt[/latex], which is the required substitution for [latex]dx[/latex].
- [latex]u[/latex]-substitution (also called [latex]w[/latex]-substitution) is used to simplify a given integral.
- Substitution can be used to determine antiderivatives.

#### Key Terms

**integration**: the operation of finding the region in the [latex]x[/latex]–[latex]y[/latex] plane bound by the function**antiderivative**: an indefinite integral

Integration by substitution, also known as [latex]u[/latex]-substitution, is a method for finding integrals. Using the fundamental theorem of calculus often requires finding an antiderivative. For this and other reasons, integration by substitution is an important tool for mathematicians. It is the counterpart to the chain rule of differentiation.

Let [latex]I\subseteq \mathbb{R}[/latex] be an interval and [latex]g:[a, b] \rightarrow 1[/latex] be a continuously differentiable function. Suppose that [latex]f:I \rightarrow \mathbb{R}[/latex] is a continuous function. Then:

[latex]\displaystyle{\int_{g(a)}^{g(b)}f(x)dx = \int_{a}^{b}f(g(t))g'(t)dt}[/latex]

Using Leibniz notation, the substitution [latex]x = g(t)[/latex] yields:

[latex]\displaystyle{\frac{dx}{dt} = g'(t)}[/latex]

and therefore, formally:

[latex]dx = g'(t)dt[/latex]

which is the required substitution for [latex]dx[/latex].

The formula is used to transform one integral into another integral that is easier to compute. Thus, the formula can be used from left to right or from right to left in order to simplify a given integral. When used in the latter manner, it is often known as [latex]u[/latex]-substitution (or [latex]w[/latex]-substitution).

For example, consider the following integral:

[latex]\displaystyle{\int_0^2 x \cos(x^2+1) dx}[/latex]

If we make the substitution [latex]u = x^2 + 1[/latex], we obtain [latex]du = 2x dx[/latex] and

[latex]\begin{align} \displaystyle{\int_{x=0}^{x=2}x \cos(x^{2}+1)} &= \displaystyle{ \frac{1}{2}\int_{u=1}^{u=5} \cos(u)du} \\ &= \frac{1}{2} \left (\sin(5) - \sin(1) \right ) \end{align}[/latex]

It is important to note that since the lower limit [latex]x = 0[/latex] was replaced with [latex]u = 0^2 + 1 = 1[/latex], and the upper limit [latex]x=2[/latex] replaced with [latex]u = 2^2 + 1 = 5[/latex], a transformation back into terms of [latex]x[/latex] was unnecessary.

Substitution can be used to determine antiderivatives if one chooses a relation between [latex]x[/latex] and [latex]u[/latex], determines the corresponding relation between [latex]dx[/latex] and [latex]du[/latex] by differentiating, and performs the substitutions. An antiderivative for the substituted function can hopefully be determined; the original substitution between [latex]u[/latex] and [latex]x[/latex] is then undone.

Similar to our first example above, we can determine the following antiderivative with this method:

[latex]\displaystyle{\int x \cos(x^{2}+1)}[/latex]

[latex]\displaystyle{= \frac{1}{2}\int 2x\cos(x^{2}+1)dx}[/latex]

[latex]\displaystyle{= \frac{1}{2}\int \cos(u)du}[/latex]

[latex]\displaystyle{ = \frac{1}{2}\sin(u) + C}[/latex]

[latex]\displaystyle{ = \frac{1}{2} \sin(x^{2}+1)+C}[/latex]

where [latex]C[/latex] is an arbitrary constant of integration. Note that there were no integral boundaries to transform, but in the last step we had to revert the original substitution, [latex]u = x^2 + 1[/latex].

## Further Transcendental Functions

A transcendental function is a function that is not algebraic.

### Learning Objectives

Identify a transcendental function as one that cannot be expressed as the finite sequence of an algebraic operation

### Key Takeaways

#### Key Points

- Transcendental functions cannot be expressed as a solution of a polynomial equation whose coefficients are themselves polynomials with rational coefficients.
- Examples of transcendental functions include the exponential function, the logarithm, and the trigonometric functions.
- Transcendental functions can be an easy-to-spot source of dimensional errors.

#### Key Terms

**polynomial**: an expression consisting of a sum of a finite number of terms, each term being the product of a constant coefficient and one or more variables raised to a non-negative integer power**trigonometric function**: any function of an angle expressed as the ratio of two of the sides of a right triangle that has that angle, or various other functions that subtract 1 from this value or subtract this value from 1 (such as the versed sine)**exponential function**: any function in which an independent variable is in the form of an exponent; they are the inverse functions of logarithms

A transcendental function is a function that is not algebraic. Such a function cannot be expressed as a solution of a polynomial equation whose coefficients are themselves polynomials with rational coefficients. Examples of transcendental functions include the exponential function, the logarithm, and the trigonometric functions.

Formally, an analytic function [latex]ƒ(z)[/latex] of the real or complex variables [latex]z_1, \cdots,z_n[/latex] is transcendental if [latex]z_1, \cdots,z_n[/latex], [latex]ƒ(z)[/latex] are algebraically independent, i.e., if [latex]ƒ[/latex] is transcendental over the field [latex]C(z_1, \cdots,z_n)[/latex].

A transcendental function is a function that “transcends” algebra in the sense that it cannot be expressed in terms of a finite sequence of the algebraic operations of addition, multiplication, power, and root extraction.

The following functions are transcendental:

[latex]f_1(x) = x^{\Pi }[/latex]

[latex]f_2(x) = c^{x }, x \neq 0, 1[/latex]

[latex]f_3(x) = x^{x}[/latex]

[latex]f_4(x) = x^{\frac{1}{x}}[/latex]

[latex]f_5(x) = \log_{c}(x)[/latex]

[latex]f_6(x) = \sin(x)[/latex]

Note that, for [latex]ƒ_2[/latex] in particular, if we set [latex]c[/latex] equal to [latex]e[/latex], the base of the natural logarithm, then we find that [latex]e^x[/latex] is a transcendental function. Similarly, if we set [latex]c[/latex] equal to [latex]e[/latex] in ƒ_{5}, then we find that [latex]\ln(x)[/latex], the natural logarithm, is a transcendental function.

In dimensional analysis, transcendental functions are notable because they make sense only when their argument is dimensionless (possibly after algebraic reduction). Because of this, transcendental functions can be an easy-to-spot source of dimensional errors. For example, [latex]\log(5 \text{ meters})[/latex] is a nonsensical expression, unlike [latex]\log \left ( \frac{5 \text{ meters}}{3 \text{ meters}} \right )[/latex] or [latex]\log(3)\text{ meters}[/latex]. One could attempt to apply a logarithmic identity to get [latex]\log(10) + \log(m)[/latex], which highlights the problem: applying a non-algebraic operation to a dimension creates meaningless results.

## Numerical Integration

Numerical integration is a method of approximating the value of a definite integral.

### Learning Objectives

Solve for the definite integral of a continuous function over a closed interval

### Key Takeaways

#### Key Points

- Numerical methods of approximation are useful when the integral is difficult to solve by hand.
- Some methods of numerical integration include the rectangle rule and the trapezoidal rule.
- Since computers cannot solve integrals by hand and are very fast at arithmetic, they use numerical methods to solve integrals.

#### Key Terms

**approximation**: An imprecise solution or result that is adequate for a defined purpose.**definite integral**: the integral of a function between an upper and lower limit

Given a function [latex]f[/latex] of a real variable [latex]x[/latex] and an interval of the real line, the definite integral [latex]\int_{a}^{b}f(x)dx [/latex] is defined informally to be the area of the region in the [latex]xy[/latex]-plane bound by the graph of [latex]f[/latex], the [latex]x[/latex]-axis, and the vertical lines [latex]x=a[/latex] and [latex]x=b[/latex], such that the area above the [latex]x[/latex]-axis adds to the total, and that the area below the [latex]x[/latex]-axis subtracts from the total. These integrals are termed “definite integrals.”

Numerical integration constitutes a broad family of algorithms for calculating the numerical value of a definite integral.

There are several reasons for carrying out numerical integration.

The integrand [latex]f(x)[/latex] may be known only at certain points, such as obtained by sampling. Some embedded systems and other computer applications may need numerical integration for this reason.

A formula for the integrand may be known, but it may be difficult or impossible to find an antiderivative which is an elementary function. An example of such an integrand is [latex]f(x) = \exp(x^2)[/latex], the antiderivative of which (the error function, times a constant) cannot be written in elementary form. It may be possible to find an antiderivative symbolically, but it may be easier to compute a numerical approximation than to compute the antiderivative. That may be the case if the antiderivative is given as an infinite series or product, or if its evaluation requires a special function which is not available.

Numerical integration methods can generally be described as combining evaluations of the integrand to get an approximation to the integral.

The integrand is evaluated at a finite set of points called integration points and a weighted sum of these values is used to approximate the integral. The integration points and weights depend on the specific method used and the accuracy required from the approximation. An important part of the analysis of any numerical integration method is to study the behavior of the approximation error as a function of the number of integrand evaluations. A method which yields a small error for a small number of evaluations is usually considered superior. Reducing the number of evaluations of the integrand reduces the number of arithmetic operations involved, and therefore reduces the total round-off error. Also, each evaluation takes time, and the integrand may be arbitrarily complicated. A ‘brute force ‘ kind of numerical integration can be done, if the integrand is reasonably well-behaved (i.e. piecewise continuous and of bounded variation), by evaluating the integrand with very small increments.

The simplest method to numerically solve integrals is to use the midpoint or rectangle rule. Draw rectangles which pass through the following point:

[latex]\displaystyle{\left ( \frac{(a+b)}{2}, f \left ( \frac{(a+b)}{2} \right ) \right )}[/latex]

The way to do this is as follows:

[latex]\displaystyle{∫_b^af(x)dx \approx (b−a)f \left ( \frac{a+b}{2} \right ) }[/latex]

The area can then be approximated by adding up the areas of the rectangles. Notice that the smaller the rectangles are made, the more accurate the approximation.

The trapezoidal rule is a more accurate method of numerical integration. Instead of taking [latex]f(x)[/latex] to be a constant around a chosen [latex]x[/latex], [latex]f(x)[/latex] is approximated as having a constant slope around [latex]x[/latex], where the slope is the average between the chosen points.

[latex]\begin{align}\int_{a}^{b} f(x) dx &\approx(b - a) f(a) + f(b) 2\int_a^b f(x)dx\\ & \approx (b-a) \frac{f(a) + f(b)}{2}\end{align}[/latex]

For either one of these rules, we can make a more accurate approximation by breaking up the interval [latex][a, b][/latex] into some number [latex]n[/latex] of subintervals, computing an approximation for each subinterval, then adding up all the results. This is called a composite rule, extended rule, or iterated rule. For example, the composite trapezoidal rule can be stated as:

[latex]\displaystyle{\int_a^b f(x)\,dx \approx \frac{b-a}{n} \left( {f(a) \over 2} + \sum_{k=1}^{n-1} \left( f \left( a+k \frac{b-a}{n} \right) \right) + {f(b) \over 2} \right)}[/latex]

where the subintervals have the form [latex][k h, (k+1) h][/latex], with [latex]h = \frac{(b−a)}{n}[/latex] and [latex]k = 0, 1, 2, \cdots, n−1[/latex].

Since computers are able to do many arithmetic operations in a small amount of time, they use numerical integration to approximate the values of integrals rather than solving them the way a person would.