An integral helps to find out how much space is under a graph of something. Integrals undo derivatives. A derivative helps to find the steepness of a graph.

This is the symbol for integration: $\backslash int\_\{\backslash ,\}^\{\backslash ,\}$ It was first used by Gottfried Wilhelm Leibniz who used it as a stylized s (for summa, Latin for sum).

Integrals and derivatives are part of a branch of mathematics called calculus. The links between these two is very important, and is called the Fundamental Theorem of Calculus^[1].

Integration helps when trying to multiply units into a problem. For example, if a problem with rate ($distance\; \backslash over\; time$) needs an answer with just distance, one solution is to integrate with respect to time. This means multiplying in time (to cancel the time in $distance\; \backslash over\; time$). This is done by adding small slices of the rate graph together. The slices are close to zero in width, but adding them forever makes them add up to a whole. This is called a Riemann Sum.

Adding these slices together gives the equation that the first equation is the derivative of. Integrals are like a way to add many tiny things together by hand. It is like summation, which is adding $1+2+3+4....+n$. Integration is like we also have to add all the decimals and fractions in between as well.^[1]

Another time integration is helpful is when finding the volume of a solid. It can add two-dimensional (without width) slices of the solid together forever until there is a width. This means the object now has three dimensions: the original two and a width. This gives the volume of the three-dimensional object described.

Methods of Integration

Antiderivative

By the fundamental theorem of calculus, the integral is the antiderivative.

If we take the function $2x$, for example, and anti-differentiate it, we can say that an integral of $2x$ is $x^2$. We say an integral, not the integral, because the antiderivative of a function is not unique. For example, $x^2+17$ also differentiates to $2x$. Because of this, when taking the antiderivative a constant C must be added. This is called an indefinite integral. This is because when finding the derivative of a function, constants equal 0, as in the function

$f(x)\; =\; 5x^2\; +\; 9x\; +\; 15\backslash ,$.

$f\text{'}(x)\; =\; 10x\; +\; 9\; +\; 0\backslash ,$. Note the 0: we cannot find it if we only have the derivative, so the integral is

$\backslash int\; (10x\; +\; 9)\backslash ,\; dx\; =\; 5x^2\; +\; 9x\; +\; C$.

Simple Equations

A simple equation such as $y\; =\; x^2$ can be integrated with respect to x using the following technique. To integrate, you add 1 to the power x is raised to, and then divide x by the value of this new power. Therefore, integration of a normal equation follows this rule: $\backslash int\_\{\backslash ,\}^\{\backslash ,\}\; x^n\; dx\; =\; \backslash frac\{x^\{n+1\}\}\{n+1\}\; +\; C$

The $dx$ at the end is what shows that we are integrating with respect to x, that is, as x changes. This can be seen to be the inverse of differentiation. However, there is a constant, C, added when you integrate. This is called the constant of integration. This is required because differentiating an integer results in zero, therefore integrating zero (which can be put onto the end of any integrand) produces an integer, C. The value of this integer would be found by using given conditions.

Equations with more than one terms are simply integrated by integrating each individual term:

$\backslash int\_\{\backslash ,\}^\{\backslash ,\}\; x^2\; +\; 3x\; -\; 2\; dx\; =\; \backslash int\_\{\backslash ,\}^\{\backslash ,\}\; x^2\; dx\; +\; \backslash int\_\{\backslash ,\}^\{\backslash ,\}\; 3x\; dx\; -\; \backslash int\_\{\backslash ,\}^\{\backslash ,\}\; 2\; dx\; =\; \backslash frac\{x^3\}\{3\}\; +\; \backslash frac\{3x^2\}\{2\}\; -\; 2x\; +\; C$

Integration involving e and ln

There are certain rules for integrating using e and the natural logarithm. Most importantly, $e^x$ is the integral of itself (with the addition of a constant of integration): $\backslash int\_\{\backslash ,\}^\{\backslash ,\}e^\{x\}\; dx\; =\; e^\{x\}\; +\; C$

With e raised to a function of x ($f(x)$):

$\backslash int\_\{\backslash ,\}^\{\backslash ,\}\; e^\{f(x)\}\; dx\; =\; e^\{f(x)\}\backslash div\backslash int\_\{\backslash ,\}^\{\backslash ,\}f(x)\; dx$

The natural logarithm, ln, is useful when integrating equations with $1/x$. These cannot be integrated using the formula above (add one to the power, divide by the power), because adding one to the power produces 0, and a division by 0 is not possible. Instead, the integral of $1/x$ is $\backslash ln\; x$: $\backslash int\_\{\backslash ,\}^\{\backslash ,\}\backslash frac\{1\}\{x\}\; dx\; =\; \backslash ln\; x\; +\; C$

In a more general form: $\backslash int\_\{\backslash ,\}^\{\backslash ,\}\backslash frac\{f\text{'}(x)\}\{f(x)\}\; dx\; =\; \backslash ln\; \{|f(x)|\}\; +\; C$

The two vertical bars indicated a absolute value; the sign (positive or negative) of $f(x)$ is ignored. This is because there is no value for the natural logarithm of negative numbers.

Properties

Sum of functions

The integral of a sum of functions is the sum of each function's integral. that is,

$\backslash int\backslash limits\_\{a\}^\{b\}\; [f(x)\; +\; g(x)]\backslash ,\; dx\; =\; \backslash int\backslash limits\_\{a\}^\{b\}\; f(x)\backslash ,\; dx\; +\; \backslash int\backslash limits\_\{a\}^\{b\}\; g(x)\backslash ,\; dx$.

Proof is straightforward: The definition of an integral is a limit of sums. Thus

$\backslash int\backslash limits\_\{a\}^\{b\}\; [f(x)\; +\; g(x)]\backslash ,\; dx\; =\; \backslash lim\_\{n\; \backslash to\; \backslash infty\}\; \backslash sum\_\{i=1\}^n\; \backslash left(f(x\_i^*)\; +\; g(x\_i^*)\backslash right)$

$=\; \backslash lim\_\{n\; \backslash to\; \backslash infty\}\; \backslash sum\_\{i=1\}^n\; f(x\_i^*)\; +\; \backslash sum\_\{i=1\}^n\; g(x\_i^*)$

$=\; \backslash lim\_\{n\; \backslash to\; \backslash infty\}\; \backslash sum\_\{i=1\}^n\; f(x\_i^*)\; +\; \backslash lim\_\{n\; \backslash to\; \backslash infty\}\; \backslash sum\_\{i=1\}^n\; g(x\_i^*)$

$=\; \backslash int\backslash limits\_\{a\}^\{b\}\; f(x)\backslash ,\; dx\; +\; \backslash int\backslash limits\_\{a\}^\{b\}\; g(x)\backslash ,\; dx$

Note that both integrals have the same limits.

Constants in integration

When a constant is in an integral with a function, the constant can be taken out. Further, when a constant c is not accompanied by a function, its value is c * x. That is,

$\backslash int\backslash limits\_\{a\}^\{b\}\; cf(x)\backslash ,\; dx\; =\; c\; \backslash int\backslash limits\_\{a\}^\{b\}\; f(x)\backslash ,\; dx$ and

This can only be done with a constant.

$\backslash int\backslash limits\_\{a\}^\{b\}\; c\backslash ,\; dx\; =\; c(b-a)$

Proof is again by the definition of an integral.

Other

If a, b and c are in order (i.e. after each other on the x-axis), the integral of f(x) from point a to point b plus the integral of f(x) from point b to c equals the integral from point a to c. That is,

$\backslash int\backslash limits\_\{a\}^\{b\}\; f(x)\backslash ,\; dx\; +\; \backslash int\backslash limits\_\{b\}^\{c\}\; f(x)\backslash ,\; dx\; =\; \backslash int\backslash limits\_\{a\}^\{c\}\; f(x)\backslash ,\; dx$ if they are in order. (This also holds when a, b, c are not in order if we define $\backslash int\backslash limits\_\{a\}^\{b\}\; f(x)\; \backslash ,dx=\; -\backslash int\backslash limits\_\{b\}^\{a\}\; f(x)\backslash ,\; dx$.)

$\backslash int\backslash limits\_\{a\}^\{a\}\; f(x)\backslash ,\; dx\; =\; 0$. This follows the fundamental theorem of calculus (FTC): F(a)-F(a)=0

$\backslash int\backslash limits\_\{a\}^\{b\}\; f(x)\backslash ,\; dx\; =\; -\backslash int\backslash limits\_\{b\}^\{a\}\; f(x)\backslash ,\; dx$ Again, following the FTC: $F(b)-F(a)\; =\; -[F(a)-F(b)]$

References