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# What's an Integral? What's Integration?

Given a function $$f(x)$$, we'll often need to find a function whose derivative is equal to $$f(x)$$.

The function, which has derivative $$f(x)$$, is known as the integral of $$f(x)$$ and we'll frequently refer to it using the capital letter $$F(x)$$.

In other words: the integral of $$f(x)$$ is a function $$F(x)$$ such that $$F'(x) = f(x)$$.

For instance, given the function $$f(x) = 2x$$, the function $$F(x) = x^2$$ is an integral of $$f(x)=2x$$ since $$F'(x) = 2x$$.

In fact any function $$F(x) = x^2 + c$$, where $$c\in \mathbb{R}$$ will have derivative $$F'(x) = 2x$$.

This generic function $$F(x) = x^2 + c$$ is known as the antiderivative (sometimes called the primitive) of $$f(x)$$.

## Definition - Integral & Antiderivative

Given a continuous function $$f(x)$$, we define its antiderivative $$F(x)$$ as the function whose derivative equals $$f(x)$$.
That's: $F'(x) = f(x)$ Note: the antiderivative is usually referred to using the capitalized letter of the letter used for the initial function; $$F(x)$$ for $$f(x)$$, $$G(x)$$ for $$g(x)$$, $$H(x)$$ for $$h(x)$$, ... .

The function $$F(x)$$ is also known as the indefinite integral, loosely said "integral", of $$f(x)$$ and we write: $F(x) = \int f(x) dx$ Note: this notation will make more sense once we have studied definite integrals, for now it should be accepted and known.

For instance, if $$f(x) = 3x^2$$ we define its antiderivative as the function: $F(x) = x^3 + c$ Since $$F'(x)=3x^2$$.

We can also write this as the indefinite integral: $\int 3x^2 dx = x^3 + c$ Where $$c$$ is known as the constant of integration and should not be forgotten.

No matter the value of $$c \in \mathbb{R}$$, when we differentiate $$x^3 + c$$: $$c$$ will disappear. This shows us that there is an infinite number of functions that have derivative $$3x^2$$, all of which can be written $$F(x) = x^3 + c$$.

## Exercise

Try finding an expression for each of the following integrals:

1. $$\int 3x^2 dx$$

2. $$\int cos(x) dx$$

3. $$\int 6x^2 dx$$

4. $$\int \frac{1}{2\sqrt{x}}dx$$

5. $$\int \frac{1}{x} dx$$

1. $$\int 3x^2 dx = x^3 + c$$

2. $$\int cos(x) dx = sin(x) + c$$

3. $$\int 6x^2 dx = 2x^3 + c$$

4. $$\int \frac{1}{2\sqrt{x}} dx = \sqrt{x} + c$$

5. $$\int \frac{1}{x} dx = ln(x) + c$$

1. To find $$\int 3x^2 dx$$ we use the fact that: $\begin{pmatrix} a.x^n \end{pmatrix}' = n\times a.x^{n-1}$ so the function that, when differentiated, will equal to $$3x^2$$ must have an $$x^3$$.
And, since $$\begin{pmatrix} x^3 \end{pmatrix}' = 3x^2$$, we can state: $\int 3x^2 dx = x^3 + c$ Where $$c$$ could be any real number, which would disapear when differentiated.
2. To find $$\int cos(x) dx$$ we think in the opposite way as we would for differentiation.
Since we know that: $\begin{pmatrix} sin(x) \end{pmatrix}' = cos(x)$ Thinking in the opposite way leads to: $\int cos(x) dx = sin(x) + c$ Again, $$c$$ could be any real number, which would disapear when differentiated.
3. To find $$\int 6x^2 dx$$ we, like in question 1), use the fact that: $\begin{pmatrix} a.x^n \end{pmatrix}' = n\times a.x^{n-1}$ so the function that, when differentiated, will equal to $$3x^2$$ must have an $$x^3$$.
And, since $$\begin{pmatrix} x^3 \end{pmatrix}' = 3x^2$$ and $$6x^2 = 2 \times 3x^2$$, we can state: $\int 6x^2 dx = 2x^3 + c$ Again, $$c$$ could be any real number, which would disapear when differentiated.
4. To find $$\int \frac{1}{2\sqrt{x}} dx$$ we use the fact that $$\begin{pmatrix} \sqrt{x} \end{pmatrix}' = \frac{1}{2\sqrt{x}}$$ and thinking in the opposite "direction" leads to: $\int \frac{1}{2\sqrt{x}} dx = \sqrt{x} + c$ Again, $$c$$ could be any real number, which would disapear when differentiated.
5. To find $$\int \frac{1}{x} dx$$, we use that fact that: $\begin{pmatrix} ln(x) \end{pmatrix}' = \frac{1}{x}$ So, thinking in the opposite "direction", we find: $\int \frac{1}{x} dx = ln(x) + c$