Integral

Integrating a function is the reverse of differentiating it.

Antiderivatives

We start with an easy problem. Let's take the function f(x) = 2x. Now let's find a function F that equals f after differentiating. Written formally, we are looking for a function F such that F′ = f.

From the basic formulae for differentiation we know that this is the function F(x) = x². The derivative of x² is exactly 2x. The question is whether this is the only function that satisfies the condition.

The answer is that it is not – if we add any real constant to x², we get 2x again after differentiating, because the derivative of a constant is zero. Any function of the form x² + c, where c is a real constant, can be taken as the function F we are looking for.

We can generalise this problem. Take a real function f and find a function F such that F′(x) = f(x) for all x in the domain of f, or in some interval within the domain. This function need not be unique, but on the other hand, such a function need not exist at all.

We call this function F an antiderivative of f (a function whose derivative is f; at A level it is simply called “an integral of f”). We write:

$$F(x)=\int f(x)\text{ d}x$$

The function f is called the integrand, and the process of finding an antiderivative is called integration. The expression dx shows which variable we are integrating with respect to. If we write an example like this

$$\int 4x^3\text{ d}x,$$

we want to find an antiderivative of the function 4x³, integrating with respect to the variable x. The result can be the function x⁴, which equals 4x³ after differentiating. Of course, we can add any real constant to this function.

It can be proved that an antiderivative either does not exist, or is always of the form F(x) + c, where c is some real constant – and then there are infinitely many of them, because there are infinitely many real constants. So if we find one function F(x), we can generate all the antiderivatives. At the same time, no other function, one we would not generate in this way, is an antiderivative of the function we are looking for.

A few more words about the expression dx. It looks very mysterious, but it really just shows which variable we are integrating with respect to. For example, if we have the integral

$$\int a\cdot b+c\text{ d}b,$$

it means that we are integrating with respect to the variable b, and the other letters a and c behave like constants. That is, they behave just as if they were numbers.

Integrating elementary functions

We can easily derive formulae for the antiderivatives of some elementary functions. Throughout this chapter, let's assume that c ∈ ℝ.

We start with the function f(x) = 0, the zero function. Which function is equal to zero after differentiating? Any function that contains only a constant:

$$\int 0\text{ d}x=c$$

Next, the function f(x) = 1. We know that x′ = 1, so the integral of one is

$$\int 1\text{ d}x=x+c$$

Next comes the function f(x) = x. Which function must we differentiate to get x? If we differentiate x², we get 2x. That is almost where we wanted to be. We just need to divide by two. So if we differentiate (x²)/2, we get exactly x.

$$\int x\text{ d}x=\frac{x^2}{2}+c$$

If we generalise this to the function xⁿ, where n ∈ ℝ, we get the formula:

$$\int x^n\text{ d}x=\frac{x^{n+1}}{n+1}+c$$

Notice that this formula does not hold for n = −1, because we would be dividing by zero, and that is simply not done – it isn't nice. So if n = −1, we get the function f(x) = x⁻¹, which is the function 1/x. Which function must we differentiate to get 1/x? It is the logarithm.

$$\int \frac{1}{x}\text{ d}x=\ln x+c$$

The problem is that the logarithm is defined only for positive real numbers, so this formula does not hold for all x in the domain of f. It does hold, however, on the interval (0, ∞). For the interval (−∞, 0) we simply flip the sign of the argument of the logarithm: ln(−x). Putting these together, we can write

$$\int \frac{1}{x}\text{ d}x=\ln |x|+c$$

This formula holds for all x in the domain of f. You can find further integrals, for example for trigonometric functions, in the following table:

$$\begin{array}{rcl}\int a^x\text{ d}x& =& \frac{a^x}{\ln a}+c\text{for }a>0,a\ne 1\\ \int \sin x\text{ d}x& =& -\cos x+c\\ \int \cos x\text{ d}x& =& \sin x+c\end{array}$$

The integral of a sum

Working out the derivative of a function is usually fairly easy and – as they say – even a trained monkey can do it, because you usually just follow the formulae. With integrals the situation is different. The methods for working out integrals are considerably more complicated, which makes finding an integral much harder than working out a derivative. This is mainly because we do have a handy formula for the integral of a sum, but not for the integral of a product.

Let's try to work out this integral:

$$\int x+1\text{ d}x$$

How could we work it out? We need a function that gives x after differentiating, and then another function that gives 1. From the formulae we already know that these are the functions (x²)/2 and x. If we add these functions, we get f(x) = (x²)/2 + x. Differentiating this function gives x + 1. So we have the result:

$$\int x+1\text{ d}x=\frac{x^2}{2}+x+c$$

While working it out, we used one property – the integral of a sum is equal to the sum of the integrals. That is,

$$\int f(x)+g(x)\text{ d}x=\int f(x)\text{ d}x+\int g(x)\text{ d}x$$

This also holds for subtraction:

$$\int f(x)-g(x)\text{ d}x=\int f(x)\text{ d}x-\int g(x)\text{ d}x$$

That is why we could split up the previous example like this:

$$\int x+1\text{ d}x=\int x\text{ d}x+\int 1\text{ d}x$$

Taking out a constant

A similar rule holds for taking a constant out in front of an integral. If we have an integral of the form:

$$\int a\cdot f(x)\text{ d}x,$$

where a is some real constant, we can take this constant out in front of the integral:

$$\int a\cdot f(x)\text{ d}x=a\cdot \int f(x)\text{ d}x$$

Example

Work out the integral:

$$\int x^2+6x+42\text{ d}x$$

In the first step we split it into three integrals:

$$\int x^2+6x+42\text{ d}x=\int x^2\text{ d}x+\int 6x\text{ d}x+\int 42\text{ d}x$$

We will deal with the integrals one at a time. The first integral can be done directly using the formula for powers xⁿ:

$$\int x^2\text{ d}x=\frac{x^3}{3}+c$$

In the second integral we first take out the factor 6:

$$\int 6x\text{ d}x=6\cdot \int x\text{ d}x$$

The integral of x is again a standard integral, and it is equal to (x²)/2. So we get:

$$6\cdot \int x\text{ d}x=6\cdot \left(\frac{x^2}{2}\right)=3x^2+c$$

In the last integral we also first take out 42:

$$\int 42\text{ d}x=42\cdot \int 1\text{ d}x$$

And that is again a standard integral:

$$42\cdot \int 1\text{ d}x=42x+c$$

We add these partial integrals together and we have the result:

$$\int x^2+6x+42\text{ d}x=\frac{x^3}{3}+3x^2+42x+c$$

One more note – in each partial integral we got a constant c, so when adding we should write +3c in the result instead of +c. However, c is an arbitrary real constant, so 3c is again equal to some real number. We could replace c in the partial integrals with c₁, c₂, c₃ and then say that c = c₁ + c₂ + c₃.