Antiderivative

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Antidifferentiation (or indefinite integration) is a part of mathematics. It is the opposite of differentiation. It is integrating with no limits. The answer is an equation.

It is written as $\int x\ dx$

with the integral sign that has no limits $\int$
the equation you are integrating $x$
and the $d x$ which means "with respect to $x$ ", which does not mean anything with simple integration.

1 Simple integration
- 1.1 Examples
2 Integrating a bracket ("chain rule")
- 2.1 Examples
3 See also

[change] Simple integration

To integrate $a x n$

add 1 to the power $n$ , so $a x n$ is now $a x n + 1$
divide all this by the new power, so it is now $\frac{ax^{n+1}}{n+1}$
and a constant $c$ should be added, so it is now $\frac{ax^{n+1}}{n+1} + c$

This can be shown as:

$\int ax^n\ dx = \frac{ax^{n+1}}{n+1} + c$

When there are many $x$ terms, integrate each part on its own:

$\int 2x^6 - 5x^4\ dx = \frac{2x^7}{7} - \frac{5x^5}{5} + c = \frac{2}{7}x^7 - x^5 + c$

(This only works if the parts are being added or taken away.)

[change] Examples

$\int 3x^4\ dx = \frac{3x^5}{5} + c$

$\int x + x^2 + x^3 + x^4\ dx = \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^5}{5} + c$

$\int \frac{1}{x + 4}\ dx = \ln (x + 4) \times 1 + c = \ln (x + 4) + c$

Changing fractions and roots into powers makes it easier:

$\int \frac{1}{x^3}\ dx = \int x^{-3}\ dx = \frac{x^{-2}}{-2} + c = -\frac{1}{2x^2} + c$

$\int \sqrt{x^3}\ dx = \int x^{\frac{3}{2}}\ dx = \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + c = \frac{2}{5}x^{\frac{5}{2}} + c = \frac{2}{5}\sqrt{x^5} + c$

[change] Integrating a bracket ("chain rule")

If you want to integrate a bracket like $(2 x + 4) 3$ , we need to do it a different way. It is called the chain rule. It is like simple integration. It only works if the $x$ in the bracket has a power of 1 (it is linear) like $x$ or $5 x$ (not $x 5$ or $x - 7$ ).

To do $\int (2x+4)^3\ dx$

add 1 to the power $3$ , so that it is now $(2 x + 4) 4$
divide all this by the new power to get $\frac{(2x+4)^4}{4}$
divide all this by the derivative of the bracket $\left (\frac{d(2x+4)}{dx} = 2 \right )$ to get $\frac{(2x+4)^4}{4 \times 2} = \frac{1}{8}(2x+4)^4$
and add a constant $c$ to give $\frac{1}{8}(2x+4)^4 + c$

[change] Examples

$\int (x+1)^5\ dx = \frac{(x+1)^6}{6 \times 1} + c = \frac{1}{6}(x+1)^6 + c \left ( \because \frac{d(x+1)}{dx} = 1 \right )$

$\int \frac{1}{(7x+12)^9}\ dx = \int (7x+12)^{-9}\ dx = \frac{(7x+12)^{-8}}{-8 \times 7} + c = -\frac{1}{54}(7x+12)^{-8} + c = -\frac{1}{54(7x+12)^8} + c \left ( \because \frac{d(7x+12)}{dx} = 7 \right )$