# Integrals

## 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

## 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

### 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-display]\displaystyle{\int_{0}^{1} \sqrt{x} dx}[/latex-display] 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-display]\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-display] 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-display]\displaystyle{\int_{0}^{1}\sqrt{x}dx = \int_{0}^{1}x^{1/2}dx = F(1) - F(0) = \frac{2}{3}}[/latex-display]### 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-display]\displaystyle{\frac{ds^2}{dx^2}=1+\frac{dy^2}{dx^2}}[/latex-display] [latex-display]\displaystyle{ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx}[/latex-display] [latex-display]\displaystyle{s = \int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }\, dx}[/latex-display] Equivalently, if a curve is defined parametrically by [latex]x = X(t)[/latex] and [latex]y = Y(t)[/latex], we get: [latex-display]\displaystyle{ds = \sqrt{ \left( \frac{dx}{dt} \right)^2+ \left(\frac{dy}{dt} \right)^2} \cdot dt}[/latex-display] then its arc length between [latex]t = a[/latex] and [latex]t = b[/latex] is: [latex-display]\displaystyle{s = \int_{a}^{b} \sqrt { [X'(t)]^2 + [Y'(t)]^2 }\, dt}[/latex-display]### Example

For the following curve described by the parameter [latex]t[/latex]: [latex-display]\begin{cases} y = t^5 \\ x = t^3 \end{cases}[/latex-display] the arc length integral for values of [latex]t[/latex] from [latex]-1[/latex] to [latex]1[/latex] is: [latex-display]\displaystyle{\int_{-1}^1 \sqrt{(3t^2)^2 + (5t^4)^2}\,dt}[/latex-display] [latex-display]\displaystyle{= \int_{-1}^1 \sqrt{9t^4 + 25t^8}\,dt}[/latex-display]## 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

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

## 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

## 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-display]\int cf(x)dx = c\int f(x)dx[/latex-display]### The Sum Rule for Indefinite Integrals

[latex-display]\int (f(x)+ g(x)) dx = \int f(x)dx + \int g(x)dx[/latex-display]### The Difference Rule for Indefinite Integrals

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

## 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

_{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

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