# Integrals

Delving into the world of infinitesimal calculus, we find two primary types of integrals: definite and indefinite. Let's take a closer look at these intriguing mathematical constructs.

## The Indefinite Integral

In a nutshell, the **indefinite integral** is essentially the inverse operation of differentiation. $$ \int f(x) \ dx $$

So, if we have a function f(x), the indefinite integral of that function, written as ∫f(x)dx, represents a collection of antiderivative functions F(x).

What's fascinating about these functions is that when you take their derivative, they magically transform back into f(x).

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

The dx here acts as the variable of integration - x in this case, and C is our constant of integration.

Now, when we take the derivative of our antiderivative function, F(x)+C, it matches the original "integrand" function, f(x).

$$ \frac{d \ F(x)+C}{dx} = f(x) $$

Interestingly, the result is not merely a single function but an infinite family of functions. Why is that? The derivative of the constant always equals zero, no matter the value of the constant.

Consider a **practical example**: the integral of f(x)=2x

$$ \int 2x \ dx $$

The output is a family of functions **F(x)=x ^{2}+C**

$$ \int 2x \ dx = x^2 + C $$

This is because any function of the form F(x)=x^{2}+C, when differentiated, mirrors the integrand function f(x)=2x.

$$ \frac{d \ x^2+C}{dx} = 2x $$

Here, C is any arbitrary constant.

Remember, its value is irrelevant to the process because the derivative of a constant always equals zero.

For instance, whether you differentiate x^{2}+1 or x^{2}+2, the result is still 2x

$$ \frac{d \ x^2+1}{dx} = 2x $$

$$ \frac{d \ x^2+2}{dx} = 2x $$

$$ \vdots $$

**Now, finding the antiderivative may seem like a daunting task, but don't fret**. There are various techniques at your disposal, such as substitution integration, integration by parts, or integration of rational functions through polynomial division, among others.

## The Definite Integral

The **definite integral**, unlike its indefinite counterpart, isn't a function but a number. It's the numerical representation of the area under the function's graph within a specific interval (a,b). $$ \int_a^b f(x) \ dx $$

When working with definite integrals, you're required to denote the limits of the integration interval (a,b).

For instance, the definite integral of the function f(x) from the lower limit "a" to the upper limit "b" would be denoted as:

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

This figure represents **the area between the curve f(x) and the x-axis within the interval (a,b)**.

Take, for example, the integral of the function 2x within the interval a=1 and b=3

$$ \int_1^3 2x \ dx $$

To solve this, you'd first need to find the **antiderivative function F(x)**, by solving the indefinite integral

$$ F(x) = \int 2x \ dx = x^2 + C $$

You can leave out the constant C for now

$$ F(x) = x^2 $$

Once you've got the antiderivative function F(x), apply **the fundamental theorem of calculus**.

This ingenious theorem links the two forms of integrals and provides an efficient method to calculate definite integrals.

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

In this case, the limits are a=1 and b=3

$$ \int_3^5 2x \ dx = F(3) - F(1) $$

We already established that the antiderivative function is **F(x)=x**^{2}.

Therefore, F(1)=1^{2} and F(3)=3^{2}

$$ \int_3^5 2x \ dx = 3^2 - 1^2 $$

$$ \int_3^5 2x \ dx = 9 - 1 $$

$$ \int_3^5 2x \ dx = 8 $$

This tells us that the area under the graph of the function f(x)=2x within the interval (1,3) equates to 8.

Now, in this case, the graph of the function is a straightforward linear graph. The area it spans comprises a square and a right triangle. If you need further reassurance, you can visually confirm that the area beneath the graph equals 8 in the interval (1,3).