### Archive

Posts Tagged ‘integration’

## Fundamental Theorem of Calculus

This theorem forms much of the basis of calculus and the uses of differentiation and integration. It basically states that differentiation and integration are opposites so if you differentiate and integral you’ll get the function you started with. This can be stated as follows:

if $F(x) = \int_a(x)^b(x) \! f(t) \, dx$ then $\frac{dF}{dx} = f(a(x))\frac{da}{dx} - f(b(x))\frac{db}{dx}$

or in the more simple case

if $F(x) = \int_0^x \! f(t) \, dx$ then $\frac{dF}{dx} = f(x)- f(0)$

It is this idea that allows us to know, for example,
$\int \! \frac{1}{1+x^2} \, dx = tan^-1(x) + c$
from the knowledge that
\$frac{d(tan^-1(x))}{dx} = \frac{1}{1+x^2}$

This makes much of integration easier as it is often much easier to work out the derivative a function than work out the integral of one so we can look for functions which when differentiated give us the function that we want to integrate and then know that the integral is that function plus a constant.

Categories: calculus, maths

## Integrating Fractions – using the natrual logarithm – Example tan(x)

From result found be differentiating the natural logarithm,
$\frac{d}{dx} (ln(f(x))) = \frac{f'(x)}{f(x)}$
for some function f(x),

and the fundamental theorem of calculus we cay say that

$\int \! \frac{f'(x)}{f(x)} \, dx = ln|f(x)| + c$ where c is the integration constant

### Simple Example

The most basic example of this is the integration of 1/x,

$\int \! \frac{1}{x} \, dx = ln|x| + c$

### More complex example: Integration of tan(x)

A slightly more complicated example of this is the integration of tan(x). To do this we must remember that $tan(x) = \frac{sin(x)}{cos(x)}$ and notice that $\frac{d}{dx}(cos(x)) = -sin(x)$. This means that -tan(x) is of the form $\frac{f'(x)}{f(x)}$ as required. Using this we can get

$\int \! tan(x) \, dx = \int \! \frac{sin(x)}{cos(x)} \, dx = lan|cos(x)| + c$

### Trick for using this identity

Sometimes we get integrals that are almost in this form but not exactly, eg) $\int \! \frac{x}{5 + x^2} \, dx$, however to solve these we can often factorise a constant so that it is in the required form. In this example we can take out a 2 so we get $\frac{1}{2} \int \! \frac{2x}{ 5 + x^2} \, dx = ln|5 + x^2| + c$

Categories: calculus