Integration

Introducing Integration

Integration is the second major branch of calculus. Just as subtraction is the inverse of addition, integration is the inverse process of differentiation. It has two fundamental interpretations that, remarkably, turn out to be the same concept.

1. Integration as an Antiderivative

This is the simplest way to think about integration. If differentiation is the process of finding the slope (rate of change), then integration is the process of finding the original function given its slope.

If $\frac{d}{dx}F(x) = f(x)$, then we say that $F(x)$ is the antiderivative or integral of $f(x)$. We write this as:

$$\int f(x) \,dx = F(x) + C$$

The symbol $\int$ is the integral sign, and $dx$ indicates that we are integrating with respect to $x$. But what is the $+ C$? It is called the constant of integration. Since the derivative of any constant is zero, there are infinitely many possible antiderivatives, all differing by a constant. For example, the derivative of $x^2$, $x^2+5$, and $x^2-100$ are all $2x$. So, the integral of $2x$ is $x^2+C$.

2. Integration as Area Under a Curve

The second, more visual interpretation is that integration represents the area under the graph of a function between two points. How can we find the area of a shape with a curved top?

The idea is to slice the area into an infinite number of infinitesimally thin rectangles. The area of each tiny rectangle is its height, $f(x)$, times its width, $dx$. The integral sign $\int$ is like a big 'S' for 'sum'. Integration is the process of summing up the areas of all these infinite, tiny rectangles.

The Fundamental Theorem of Calculus

The theorem that connects these two ideas is the central result of calculus. The Fundamental Theorem of Calculus states that the area under a curve can be found by evaluating its antiderivative at the endpoints. This result links the concept of slope (differentiation) to the concept of area (integration).

The Standard Rules of Integration

Before turning to the reference table below, it helps to collect the core rules of integration in one place. The first two are the linearity properties you have already used — integration distributes over addition, and constants pull straight out of the integral sign:

$$\int \big(f(x)+g(x)\big)\,dx = \int f(x)\,dx + \int g(x)\,dx$$

$$\int k\,f(x)\,dx = k\int f(x)\,dx$$

Two further rules handle integrals that do not reduce to the basic table directly.

Integration by Parts

When the integrand is a product of two functions, integration by parts replaces the original integral with a simpler one:

$$\int f(x)g(x)\,dx = f(x)\left(\int g(x)\,dx\right) - \int \left[\left(\int g(x)\,dx\right)\frac{d}{dx}\big(f(x)\big)\right]dx$$

In words: integrate one factor, multiply the result by the other factor, then subtract the integral of that running antiderivative times the derivative of the factor you left alone.

Integration by Substitution

When the integrand is a composite function multiplied by the derivative of its 'inner' function, substitution collapses it into a simpler integral in a new variable $u$:

$$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du,\qquad u=g(x)$$

This is the formal statement of the u-substitution procedure used throughout the worked examples in this chapter.

Table of Standard Integrals

This table is the chapter's central reference — memorize it, and every technique you meet later (substitution, integration by parts, definite-integral evaluation) becomes a matter of reducing an unfamiliar integral to one of these standard forms. Reading the table from left to right integrates; reading it from right to left differentiates (drop the arbitrary constant $C$ when differentiating).

Entries 12–13 and 18–21 will feel unfamiliar the first time you see them — they arise from inverse-trigonometric and 'irrational' substitutions you will practice in later units (e.g. trigonometric substitution). For now, treat the table as a lookup: whenever an integral matches one of these forms exactly, you can write down the answer immediately.

From a Family of Functions to a Single Number

While an indefinite integral gives us a whole family of functions (the antiderivative + C), a definite integral gives us a single numerical value. This value represents the net area under the graph of a function between two specific points, called the limits of integration.

The Fundamental Theorem of Calculus (Evaluation Part)

The link between the antiderivative and the area under a curve is the Fundamental Theorem of Calculus. It gives us a simple, powerful way to calculate definite integrals.

If $f(x)$ is a continuous function on the interval $[a, b]$ and $F(x)$ is any antiderivative of $f(x)$ (i.e., $F'(x)=f(x)$), then:

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

The notation $F(b) - F(a)$ is often written as $\left[F(x)\right]_{a}^{b}$.

Steps to Evaluate a Definite Integral:

1. Find the Antiderivative: First, find the indefinite integral of $f(x)$, which gives you $F(x)$. You can ignore the constant of integration, $C$, as it will cancel out.
2. Evaluate at the Limits: Calculate the value of the antiderivative at the upper limit ($F(b)$) and the lower limit ($F(a)$).
3. Subtract: Subtract the value at the lower limit from the value at the upper limit.

Example: Finding the Area

Find the area under the curve of $y = x^2$ from $x=1$ to $x=3$.

$$\text{Area} = \int_{1}^{3} x^2 \,dx$$

1. Antiderivative: The antiderivative of $x^2$ is $F(x) = \frac{x^3}{3}$.

2. Evaluate:
$F(3) = \frac{3^3}{3} = \frac{27}{3} = 9$
$F(1) = \frac{1^3}{3} = \frac{1}{3}$

3. Subtract:
$\int_{1}^{3} x^2 \,dx = F(3) - F(1) = 9 - \frac{1}{3} = \frac{26}{3}$

So, the area under the parabola from x=1 to x=3 is $26/3$.

What is 'Net' Area?

If a function dips below the x-axis, the definite integral will treat the area in that region as negative. The definite integral calculates the 'net' or 'signed' area: the sum of the areas above the axis minus the sum of the areas below the axis.

The same source material also works Examples 45–47, but these are near-identical restatements of Examples 38, 39, and 41 above — the same integrals, the same substitutions, and the same results — so they are not reproduced separately here.

Antidifferentiation

Finding an indefinite integral is the process of answering the question: 'What function, when differentiated, gives me the function I started with?' Since this process is the reverse of differentiation, the rules of integration are the reverse of the rules of differentiation.

The Reverse Power Rule

The most fundamental rule is the reverse of the power rule for derivatives.

$$\int x^n \,dx = \frac{x^{n+1}}{n+1} + C \quad (\text{for } n \neq -1)$$

In words: Add one to the power, then divide by the new power.

• Example 1: $\int x^3 \,dx = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C$
• Example 2 (Constant): $\int 5 \,dx = \int 5x^0 \,dx = 5 \frac{x^1}{1} + C = 5x + C$

Other Basic Rules

Just like with derivatives, we can integrate term by term and factor out constants.

• Constant Multiple: $\int kf(x) \,dx = k \int f(x) \,dx$
• Sum/Difference Rule: $\int (f(x) \pm g(x)) \,dx = \int f(x) \,dx \pm \int g(x) \,dx$

Table of Common Integrals

Example: Integrating a Polynomial

Find the indefinite integral of $f(x) = 3x^2 + 4x - 5$.

$$\int (3x^2 + 4x - 5) \,dx$$

We can integrate this term by term:

$$= 3 \int x^2 \,dx + 4 \int x \,dx - 5 \int 1 \,dx$$

$$= 3\left(\frac{x^3}{3}\right) + 4\left(\frac{x^2}{2}\right) - 5(x) + C$$

$$= x^3 + 2x^2 - 5x + C$$

(Note: We only need to add one constant of integration at the very end.)

Beyond the Basic Rules

While the basic integration rules are powerful, many integrals cannot be solved directly with them. We need more advanced techniques. The most common and important of these is u-substitution, which is essentially the chain rule in reverse.

The Idea Behind u-Substitution

U-substitution is used to transform a complicated integral into a simpler one that we know how to solve. It is designed to handle integrals of composite functions—where you have a function 'inside' another function.

Recall the chain rule for derivatives: $\frac{d}{dx}[F(g(x))] = F'(g(x))g'(x)$.
If we integrate both sides, we get: $\int F'(g(x))g'(x)\,dx = F(g(x)) + C$.
The expression on the left looks complicated. The goal of u-substitution is to take an integral that looks like $\int f(g(x))g'(x)\,dx$ and simplify it by letting $u=g(x)$.

Step-by-Step Procedure:

1. Choose 'u': Look for an 'inner function'. A good choice for $u$ is often a quantity that is raised to a power, inside a trig function, or in the denominator. Let $u = g(x)$.
2. Find 'du': Differentiate $u$ with respect to $x$ to find $\frac{du}{dx} = g'(x)$. Then, solve for $du$ to get $du = g'(x)dx$.
3. Substitute: Replace the inner function with $u$ and the remaining part of the integral with $du$. The entire integral should now be in terms of $u$, with no $x$'s left.
4. Integrate: Solve the new, simpler integral with respect to $u$.
5. Back-Substitute: Replace $u$ with the original expression in terms of $x$ to get your final answer.

Example: Using u-Substitution

Find $\int 2x \cos(x^2) \,dx$.

1. Choose u: The 'inner function' is $x^2$. Let $u = x^2$.

2. Find du: $\frac{du}{dx} = 2x \implies du = 2x\,dx$.

3. Substitute: Notice that our integral has both $x^2$ and $2x\,dx$. We can substitute these perfectly.
$\int \cos(x^2) (2x \,dx) = \int \cos(u) \,du$

4. Integrate: The integral is now much simpler.
$\int \cos(u) \,du = \sin(u) + C$

5. Back-Substitute: Replace $u$ with $x^2$.
The final answer is $\sin(x^2) + C$.

Integration as Summation of a Continuous Variable

Suppose a curve $y = f(x)$ traces out a region bounded below by the x-axis and on the sides by two vertical lines $x=a$ and $x=b$. We already know how to find the areas of triangles, rectangles, and circles — but how do we find the area under a curve whose height keeps changing? The method is to divide the region into many thin vertical strips, treat each strip as an ordinary rectangle, add up the rectangles, and then let the strips become infinitesimally thin.

Divide the interval $[a,b]$ into $n$ equal strips, each of width $\Delta x$, by marking the points

$$x_0 = a,\quad x_1 = a+\Delta x,\quad x_2 = a+2\Delta x,\quad x_3 = a+3\Delta x,\ \ldots,\ x_n = a+n\Delta x = b$$

This slices the area under the curve into $n$ thin rectangles, each of base $\Delta x$. The first rectangle has height $f(x_0)$, the second has height $f(x_1)$, and so on, with the last rectangle having height $f(x_{n-1})$.

Adding up the areas of all $n$ rectangles gives an approximate value for the area under the curve:

$$A_{approximate} = f(x_0)\Delta x + f(x_1)\Delta x + f(x_2)\Delta x + \cdots + f(x_{n-1})\Delta x$$

or, in compact summation notation,

$$A_{approximate} = \sum_{x=a}^{x=b} f(x_i)\,\Delta x$$

This approximation gets better as $n$ grows larger — the rectangles hug the curve more closely as their width $\Delta x$ shrinks. In the limit $\Delta x \to 0$ (equivalently $n \to \infty$), the jagged staircase of rectangles becomes indistinguishable from the smooth curve, and the sum gives the exact area:

$$A_{exact} = \lim_{\Delta x \to 0} \sum_{x=a}^{x=b} f(x_i)\,\Delta x$$

When $\Delta x$ becomes infinitesimally small it is given the special symbol $dx$, and this limiting process — the summation of a continuously varying quantity over infinitely many infinitesimal pieces — is called integration. The limit-of-a-sum notation is replaced by the integral sign:

$$A_{exact} = \int_{a}^{b} f(x)\,dx$$

It may seem strange that adding infinitely many pieces, each of vanishing size, can produce a finite, physically meaningful answer rather than zero or infinity — but this is exactly what happens, and it is the reason integration is such a powerful tool throughout physics.