Introduction to Integration

In this section, we cover the following topics from the programme: Introduction, definite/indefinite integrals, properties of the integral, improper integrals, area under the curve of a function, area between two functions, volume of solid of revolution and arc length.

Previous knowledge required: Knowledge and properties of derivatives.

Table of Contents

Opening Problem

Say we want to find the area of the function $f(x)=x^2$ in the interval $[0,1]$ . We know that this area must be between 0 and 1 since the graph is inside a $1\times1$ square (whose area is 1). How can we increase this precision? We can take rectangles that are always slightly above the graph of our function or slightly below and add their areas to end up with an approximation.

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The desired area can be approximated as the sum of the areas of the lower (or upper) rectangles. The more rectangles we take, the more we refine our technique, the more we will get a better approximation.

If we take the 3 violet rectangles for the lower sum, the length of each rectangle is 0.25, the heights of the rectangle we can find using the function $f(x)=x^2$ . These heights are: $f(0.25)=\frac{1}{16},f(0.5)=\frac{1}{4},$ and $f(0.75)=\frac{9}{16}$ . The sum is therefore:

$\text{Lower Area}=0.25\cdot0.125+0.25\cdot0.25+0.25\cdot0.375=0.1875$

If we take the 4 red upper rectangles for the upper sum, the length of each is still 0.25, the heights are now: 0.125, 0.25, 0.375, and 1. The sum of areas of the upper rectangles is therefore:

$\text{Upper Area}=0.25\cdot0.125+0.25\cdot0.125+0.25\cdot0.375+0.25\cdot1=0.40625$

Both results here are rather different, this is because we are taking a very small number of rectangles. If we refine this technique, and take rectangles that are very small in length, both our results will start getting closer to one another. If we could take infinite of these rectangles (all infinitesimally small), the result would be equal.

The real answer, which we will give meaning to in a few minutes, is: $\int_0^1 x^2dx=\frac{1}{3}=0.3333...$ .

The Riemann Integral

With our opening problem understood, we can define what is meant by an integral.

Definition: Let $f(x)$ be an integrable function on $(a,b)$ . A function $F(x)$ is called a primitive function of $f(x)$ if $F'(x) = f(x)$ for any $x \in (a,b)$ .

The term antiderivative is sometimes used to also mean this. Since the derivative of a constant is 0, if $F(x)$ is a primitive function of $f(x)$ , then so is $F(x) + 1$ , $F(x) + 2$ , … and similarly with any constant addition.

Example: We know that if $F(x) = x^2$ then $F'(x) = f(x) = 2x$ . Therefore, we say that $x^2$ is a primitive of $2x$ . However, we could also have a different primitive $G(x) = x^2 + 1$ (or using any other constant) and the derivative of this is still $f(x) = 2x$ . This is why we refer to the collection of such functions (these are all the same up to a constant) as the indefinite integral.

Definition: The collection of all primitive functions $F(x)$ is called the indefinite integral of $f(x)$ .

Below is a list of common integrals we can easily understand from our knowledge of derivatives:

The $+C$ refers to all the different primitive functions. The symbol $dx$ serves to indicate with respect to which variable we are integrating. The symbol $\int$ marks the integral. The last two integrals should not make a big appearance, but we know them from knowing the derivatives of $\ln(x)$ and $e^x$ .

Example: To find the integral of $\int xdx$ , we think of $x$ as $x^1$ and use the formula $\frac{x^{n+1}}{n+1}$ . Here $n=1$ so:

$\int xdx=\int x^1dx=\frac{x^2}{2}+C$

Similarly, we have $\int 1dx=x+C$ because we can look say that $1=x^0$ so $n=0$ in this case. We can also understand that $\int 1dx$ because we know the primitive function. The derivative of $F(x)=x$ is $F'(x)=f(x)=1$ .

Let us see another example more in detail.

Example: Find the indefinite integral of the function $f(x) = x^4$ using the power rule for integration.

We defined the integral as the area under a curve $f(x)$ , but so far we have only found primitive functions. Our next theorem (the fundamental theorem of calculus) will link the primitive function and integral to the area calculation that we did above.

Theorem (Newton-Leibniz Formula): Let $f(x)$ be an integrable function with primitive $F(x)$ . Then, the area under the curve of $f(x)$ from $a$ to $b$ is given by:

$\int_a^b f(x)dx = F(b) - F(a)$

The expression $\int_a^bf(x)dx$ is called the definite integral of $f$ , it calculates the net area between the function $f(x)$ and the $x$ -axis over the specific interval $[a,b]$ . When no interval is assigned, i.e.: when we take $\int f(x)dx$ , we refer to it as the indefinite integral.

A small disclaimer here: If your function is negative anywhere in the interval $(a,b)$ , then that area will be counted as negative. We can think of the integral as how much “more” area is above the $x$ -axis. We will get used to this after a few examples.

Example: In our opening problem, we had:

$Area=\int_0^1 x^2dx=\frac{x^3}{3}\Bigg|_0^1=\frac{1^3}{3}-\frac{0^3}{3}=\frac{1}{3}$

Here we are using the fact that the primitive function of $f(x)=x^2$ is $F(x)=\frac{x^3}{3}+C$ .

What happened with $+C$ ? Since it is constant, it is cancelled when we calculate $F(b)-F(a)$ because it appears twice (once with each sign). Let us see it more in detail:

$\int_0^1f(x)dx=\frac{x^3}{3}+C\Big|_0^1=\underbrace{\left(\frac{1^3}{3}+C\right)}_{F(b)}-\underbrace{\left(\frac{0^3}{3}+C\right)}_{F(a)}=\frac{1}{3}+C-0-C=\frac{1}{3}$

The $C$ cancelling is a common occurrence when we have a definite integral, so we can ignore it entirely as above.

We have used the word “integrable” a few times. Within the S7 programme, what makes a function “integrable” is not defined or discussed, the word is added for correctness. In simple terms, a function is integrable on some interval when we can calculate the net area over the given interval. This is similar to how not every function is differentiable everywhere. The function $\frac{1}{x}$ is not integrable on $[0,1]$ because the area is infinite (we have discussed already that it has a vertical asymptote at $x=0$ ).

Properties of Integral

The properties we can expect from the integral are relatively straightforward.

Claim: If $f(x)$ and $g(x)$ are integrable and $k,k_1,$ and $k_2$ are constants, then:

$\int \big(f(x) \pm g(x)\big)dx = \int f(x)dx \pm \int g(x) dx$

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

$\int \big(k_1 \cdot f(x) \pm k_2\cdot g(x)\big)dx = k_1 \int f(x)dx \pm k_2 \int g(x)dx$

These will become a second nature with a bit of practice. Either way, it should not surprise us to see these properties, we know that integration and differentiation are connected, and the derivative works in the same way!

Example: To calculate the integral of $10x^4+3x-1$ we separate them into more workable integrals:

$\int(10x^4+3x-1)dx=\int10x^4dx+\int3xdx-\int1dx=10\int x^4dx+3\int xdx-1\int dx$

$=10\cdot\frac{x^5}{5}+3\cdot\frac{x^2}{2}-1\cdot x+C=2x^5+1.5x^2-x+C$

For each of the integrals above, we used the formula $\int x^n dx=\frac{x^{n+1}}{n+1}$ .

Claim: Let $f(x)$ and $g(x)$ be two integrable functions. Assume that $f(x) \leq g(x)$ for all $x \in [a,b]$ . Then:

$\int_a^b f(x)dx\geq\int_a^b g(x)dx$

Think about this property as follows: if a function is greater than another function on some interval, it makes sense that it will have a bigger net area between its graph and the $x$ -axis.

You will have noticed by now that a symbol $dx$ is recurring in every integral. The symbol $dx$ indicates that the integral is being taken with respect to the variable $x$ . You will not see functions with many variables, so this seems less necessary, but we must include it nonetheless. Moreover, if you recall that we defined the integral as an area, $dx$ represents the infinitesimally small lengths of the rectangles we draw in our Darboux sums. This is more background knowledge, so do not worry if it seems difficult.

When we can find C

The indefinite integral is a set of functions, all differing by a constant. We could be given a point on the graph of $F(x)$ and use that to find the specific function that is a primitive and whose graph goes via that point.

Example: Find the primitive function $F(x)$ of the polynomial $f(x) = 3x^2 - 4x + 5$ given that $F(2)=7$ .