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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 \(f(x)\).

Tutorial

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. \(\int \frac{1}{x} dx \)

Answers Without Working

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

Answers With Working

  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: \[\int \frac{1}{x} dx = ln(x) + c \]