Introduction to Integrals

From basic arithmetic we know that most operations have an inverse operation: subtraction is the inverse operation of addition, and vice versa. Multiplication is the inverse operation of division, and vice versa. In this topic, we will learn about the inverse operation of differentiation, as well as its application when it comes to measuring areas and accumulation over time.

Antiderivatives

At this point you're probably familiar with both the concept and the computational side of a derivative. As a quick recap, we can say that the derivative of a function $F$ evaluated at a specific point gives us the instantaneous rate of change of the function $F$ at that particular point. Then, the algebraic expression of this derivative as a function gives us information about the changing behavior of $F$ over its entire domain. Using Newton's notation, this derivative can be defined as:

$$F'(x)=\lim_{\Delta x \to 0}\frac{F(x + \Delta x)-F(x)}{\Delta x}$$Let us denote this derivative by $f$, so that

$$f(x) = F'(x)$$

We call function $F$ the antiderivative of $f$.

The antiderivative of a function can be found by taking the indefinite integral of said function. This is written as follows:

$$F(x) =\int f(x)\,dx$$The $\int$ symbol stands for integral and represents the integration operation. The $dx$ symbol tells us that we are integrating with respect to $x$, and it is read as differential of x.

As its name suggests, the antiderivative is the result of applying the inverse of the operation that produces derivatives. In other words, integration is the inverse operation of differentiation:

Let's say,

$$f(x) = 2x$$If we try to think of a function whose derivative is $2x$, we will probably think first of $x^2$, since

$$\frac{d}{dx}(x^2) = 2x$$

However, we know that

$$\frac{d}{dx}(x^2 + 3) = 2x$$and, also

$$\frac{d}{dx}(x^2 - 1) = 2x$$and in general,

$$\frac{d}{dx}(x^2 + C) = 2x$$for any constant value $C$, since the derivative of a sum is the sum of the derivatives and the derivative of a constant is zero.

Then, we have $$F(x) = x^2 + C$$

The existence of this arbitrary constant $C$ tells us that there are infinitely many antiderivatives of $f(x)$, which are also referred to as slope field, or family of curves.

In order to account for this constant, we can rewrite our integral expression as:

$$\int f(x)\,dx = F(x) + C$$This type of integral (yes, there are more) is also known as an indefinite integral. This term is often interchangeable with antiderivative.

For simple functions, integration can be as easy as thinking of a function whose derivative is the function being integrated. For example, for polynomials, we have:

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

However, as these integrands (the function being integrated) become more elaborate different techniques of integration were developed, such as trigonometric substitution or integration by parts. In another topic you will learn more about the computation of integrals. For now, let's move on to the next section!

Definite integral

Let's say you have the constant function $g(x)=2$

and, let's say we wanted to know the area under the graph of $g(x)$ from $x \in [0,3]$

Since this is a rectangle, we know we can determine its area by multiplying the length of its base times its height, like so

$$\begin{align*} A &= \Delta x \cdot g(x) = (3 - 0) \cdot g(3) = 3 \cdot 2\\ A &= 6 \end{align*}$$Now, let's say $g(x)=x$

We still want to know the area under the graph of $g(x)$. This time, it is the area of the triangle $g(x)$ makes with the $x$ axis, for the interval $x \in [0,6]$

Since it makes a right triangle, this is also pretty straightforward:

$$\begin{align*} A &= \frac{\Delta x \cdot g(x)}{2} = \frac{(6 - 0) \cdot g(6)}{2} = \frac{6 \cdot 6}{2}\\ A &= 18 \end{align*}$$However, if now we have some arbitrary function $f(x)$ with a graph like this:

Then, the problem of computing the area between $f(x)$ and the $x$-axis becomes much more difficult as we apparently cannot use our elementary formulas anymore. Or can we?

Let's focus on a smaller interval of our function, for $x \in [0,2]$

We can try to approximate the area under the curve with our rectangle formula, using the highest point of $f$ in this interval:

In this case, we can see that for our first approximation,

$$A_1=2\cdot15=30\quad ; \quad A < A_1$$Where $A$ is the actual area under the curve for $x \in [0,2]$

Now, let's try with the lowest point of $f$ in this interval:

Here we can see that for our second approximation,

$$A_2=2\cdot5=10\quad ; \quad A > A_2$$Then, we know that

$$10 < A < 30$$Now, if we further divide our interval of work for $x$ into smaller rectangular chunks, and sum the approximate areas from each rectangle, we will get a better overall approximation:

If we define a constant chunk width $\Delta x$ for $n$ chunks, we have

$$\begin{align*} A_{chunks} &= A_{chunk_1} + A_{chunk_2} + \cdots + A_{chunk_n} \quad \text{with} \quad A_{chunk_i} = f(x_i) \cdot \Delta x \\ A_{chunks} &= f(x_1) \cdot \Delta x + f(x_2) \cdot \Delta x + \cdots + f(x_n) \cdot \Delta x \\ A_{chunks} &= \sum_{i=1}^n f(x_i)\cdot \Delta x \end{align*}$$Then,$$A \approx A_{chunks} \implies A \approx \sum f(x_i)\cdot \Delta x$$For the plot above we have 8 chunks with equal width $\Delta x$, which means:

$$\begin{align*} \Delta x &=\frac{x_f -x_0}{n} = \frac{2 -0}{8} \\ \Delta x &= 0.25 \end{align*}$$

Thus,$$\begin{align*} \sum f(x_i)\cdot \Delta x &= [\,f(0.25) + f(0.5) + f(0.75) + f(1) + f(1.25) + f(1.5) + f(1.75) + f(2)\,] \cdot \Delta x \\ &= (7.48 + 9.80 + 11.82 + 13.42 + 14.49 + 14.98 + 14.85 + 14.10) \cdot 0.25 \\ \sum f(x_i)\cdot \Delta x &= 25.23 \end{align*}$$Then,

$$A \approx 25.23$$Another approximation would be to use the lowest point of $f$ for each chunk, or even the mid-point between the lowest and highest point:

However, if we want to improve our approximation there is only one real modification to be made: increase the number of chunks.

For $n=20$, we have

With $A \approx 24.6$

And for $n = 100$, we have

With $A \approx 24.3$

You probably know where we're going with this. We can determine the exact area under the curve by computing the limit of our sum when the number of chunks $n \to \infty$. Effectively, making it an infinite sum:

$$A = \lim_{n \to \infty} \sum_{i=1}^n f(x_i)\cdot \Delta x \quad \text{with }\, x \in [a, b]$$

This infinite sum is called Riemann Sum, named after the famous mathematician.

The integral symbol can also represent the infinite sum between two points $x=a$ and $x =b$, when we write those lower and upper bounds like this:

$$\int_a^b$$In addition, we can observe how the number of chunks increases as we decrease the width of each chunk. In other words, $n \to \infty$ as$\Delta x \to 0$

We can think of a differential as an infinitesimally small chunk. This way, we can think of $dx$ as an infinitesimally small version of $\Delta x$. Then, we can rewrite our previous expression as:

$$A = \int_a^b f(x) dx$$This expression has another name that is commonly used: definite integral. The definite integral of a function $f(x)$ is equal to the signed area between the curve given by $f$ and the $x$-axis, over a given interval; $x \in [a,b]$.

Notice the "signed" adjective there; it refers to the fact that areas can be either positive or negative. Areas can be negative when the function's curve is below the $x$-axis, since the value of the integrand would be negative for the interval of integration. For example, let's take $f(x) = -3$ and an interval $x \in [0,2]$

In this case, $A = 2 \cdot (-3) = -6$

There are many ways to compute a definite integral. Computers almost always use numeric methods like our rectangle method above to approximate the result; they sum a large number of chunks. However, this will always be an approximation. In order to get the exact number, we need to determine the limit of this sum.

For example, by solving the corresponding definite integral for our example, the exact area $A$ between $x = 0$ and $x = 2$ can be found to be:

$$A = 24.17$$Which means our approximation for $n=100$ was off only by about 0.5% (and if we were to use $n=10\ 000$, the approximation would only be about 0.004% off!)

A few useful conventions arise when working with definite integrals:

$$\begin{align*} 1. \ &\int_a^a f(x) dx = 0 \\ \\ 2. \ &\int_a^c f(x) dx = \int_a^b f(x) dx + \int_b^c f(x) dx \quad ; \ a < b<c \\ \\ 3. \ &\int_b^a f(x) dx = - \int_a^b f(x) dx \\ \\ \end{align*}$$These can be remembered as: $(1)$ The area under a curve $f(x)$ from and to the same point is equal to zero. $(2)$ The area over a given interval can be expressed as the sum of the areas over adjacent subintervals. $(3)$ Swapping integration bounds yields the same area with opposite sign.

Fundamental theorem of calculus

In its first part, the Fundamental Theorem of Calculus brings together both differentiation and integration as part of the fundamental operations in Calculus. From this, a second part that brings together both antiderivatives and definite integrals is derived.

$$\text{Part one:} \quad F(x) = \int_a^x f(t)dt \implies F'(x) = f(x)$$

$$\text{Part two:} \quad \int_a^b f(x)dx = F(b) - F(a) \quad \text{if} \quad F'(x) = f(x)$$This gives us a very straightforward way to compute definite integrals. Instead of finding the limit of the Riemann Sum, we just need to find the antiderivative of $f(x)$ using an indefinite integral, evaluate this antiderivative at both $x =a$ and $x = b$, and subtract the value corresponding to the lower bound from the one corresponding to the upper bound.

For example, let's say

$$f(x) = 2x$$And we wanted to find the area for $x \in [1,3]$

We can see the highlighted area can be computed as the area of a rectangle plus the area of the triangle sitting on top of it:

$$A = A_{rectangle} + A_{triangle} = (3 -1) \cdot 2 + \frac{(3-1) \cdot (6-2)}{2} = 4 + 4 = 8$$

However, this area can also be computed with the following definite integral:

$$A=\int_1^3 2x dx$$We find the antiderivative of $2x$,

$$\begin{align*} F(x) &= \int 2x dx = 2 \cdot \int x dx \\ &= 2\cdot \frac{x^{(1+1)}}{1+1} +C\\ &= 2 \cdot \frac{x^2}{2} +C\\ F(x) &=x^2 + C \end{align*}$$We evaluate it at both $x=1$ and $x=3$:

$$\begin{align*} &F(3) = (x^2 + C) |_{x=3} \\ &F(3) = 3^2 + C = 9+ C \\ \\ &F(1) = (x^2 + C) |_{x=1} \\ &F(1) = 1^2 + C = 1+ C \end{align*}$$Finally,$$\begin{align*} A&=\int_1^3 2x dx = F(3) - F(1) \\ &=(9+C)-(1+C)\\ A&=8 \end{align*}$$We have obtained the same value for the area as before!

As a corollary of the Fundamental Theorem of Calculus, we can say that for a function $f(t)$, the function $F(x)$ describes the accumulated behavior of $f(t)$ up to an arbitrarily chosen value $t=x$, since it is equal to the area under the curve $f(t)$ that has been growing since an initial $t=a$, plus the corresponding initial value at that point, $F(a)$.

$$\int_a^x f(t)dt = F(x) - F(a) \implies F(x) = \int_a^x f(t)dt +F(a)$$This concept is very useful in physics for determining the state of elements that accumulate energy over time, such as the voltage in a capacitor at a specific point in time for a given function of its current. Because of this practicality, integrals are used in control systems to keep track of the deviation from a set point over time, in order to automatically reduce it and eventually eliminate it.

Linearity of integration

Just like differentiation, integration is a linear operation. This implies the following:

$$\begin{align*} 1. \quad & \int k \cdot f(x) dx = k \cdot \int f(x) dx \\ \\ 2. \quad & \int ( f(x) + g(x) ) dx = \int f(x) dx + \int g(x)dx \end{align*}$$With $k$ being constant.

Since we can define definite integrals in terms of indefinite integrals using the Fundamental Theorem of Calculus, it follows that definite integrals are also linear operators.

Conclusion

In this topic, we've learned that:

• Integration is the inverse operation to differentiation, and antiderivatives and derivatives are the results of those respective operations.
• Antiderivatives are also known as indefinite integrals since they are not bounded to any given interval. Because of this, an antiderivative is a function and not a number.
• Definite integrals are integrals bounded in a given interval. The result of computing a definite integral is a number that represents the algebraic sum of the area below a given curve $f(x)$ and the $x$-axis.
• Definite integrals can be approximated using numerical methods. Computers often solve integrals using numerical methods.
• Definite integrals can be determined using the second part of the Fundamental Theorem of Calculus, by evaluating the antiderivative of the function at each bound, and subtracting the value corresponding to the lower bound from the value corresponding to the upper bound.
• Definite integrals are also useful to represent the accumulated behavior of a time-dependent function over time.
• Integration is a linear operation, in both indefinite and definite forms.