### Archive

Posts Tagged ‘calculus’

## 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

## Integration by Parts

July 19, 2009 1 comment

Integration by parts is a method that allows us to integrate the product of two function such as

∫2xe3xdx

where 2x is one function and e3x is another
To do this we use the formula

∫(u d/dx) dx = uv – ∫(v du/dx) dx

where u and v are both functions of x, 2x and e3x in the above example.

Proof of Integration by Parts Formula

This can be shown by considering the product rule for differentiation as shown below
the product rule states that
d(uv)/dx = u dv/dx + v du/dx

If we now integrate both side we get

uv = ∫( u dv/dx) dx + ∫(v du/dx) dx
since integration is the opposite of differentiation

now we can simply rearrange this to get

∫(u d/dx) dx = uv – ∫(v du/dx) dx

Simple Example (without limits)

To demonstrate this formula we shall integrate the example above

∫2xe3xdx

To do this we first need to pick which function (2x or e3x) to allocate to u and which to dv/dx. Picking the correct function is crucial to integration by parts since the method works by allowing us the differentiate the part of the function which is difficult to integrate. In order for this to work it is useful to use a function that will become simpler in some way. For example differentiating e3x gives 3e3x so making the product no easier to integrate, however integrating 2x gives 2 which does make it easier to integrate since we can take the 2 outside as a factor.

So we will let u=2x and dv/dx=e3x
we then differentiate 2x to get 2
and integrate e3x to e3x/3
(it can often be useful to put these in a table with columns u,v and the function on the first row and derivative on the second row)

These can then be put into the formula to find the integral
∫2xe3xdx = 2xe3x/3 – ∫e3x2/3 dx
∫2xe3xdx = 2xe3x/3 – 2e3x/9 + c

Example with Limits

usually, however integration is required to be carried out between limits. To perform integration by parts between limits is quite simple. You takes the value of uv as usual between limits and then set the limits of  ∫v du/dx dx to the limits of the original integral. To demonstrate this we will integrate the above example between 0 and 1

so

10 2x e3x dx = [2xe3x/3]10 – ∫10 e3x2/3 dx
=2e3 – 0 – 2e3/9 + 2/9 = (16e3 + 2) / 9

Categories: calculus, maths

## Implicit Differentiation

Implicit differentiation involves differentiating an equation that hasn’t been arranged such that all of one variable, eg y, is on one side and all of the other variable, eg x, is on the other side. For example differentiating the equation with respect to x (ie find dy/dx):

3x2 +2y3 + 6 = 3x2y

TO do this you need to remember that the derivative of y with respect to x is dy/dx, hence when you differentiate a function such as

y=3x

you get

dy/dx = 3

Where the right has gone to the derivative of 3x and the left has gone to the derivative of y, dy/dx.

If you want to differentiate more complex terms involving x you can use the chain rule, since y can be written as a function of x.
so if f(x) = g(y) then
f'(x) = dy/dx g'(y)

(or a simple way of doing it is treat any y’s like x’s and stick a dy/dx on the end).

So for example y2 differentiated becomes 2ydy/dx

All the other rules like the product rule and quotient rule still apply.
So to finish lets consider our original equation

3x2 +2y3 + 6 = 3x2y

This becomes

6x + 6y2dy/dx = 6xy +3x2dy/dx

Which we can re-arrange to get

dy/dx = (6xy-6x)/(6y2-3x2)=(2xy-2x)/(2y2-1x2)

By David Woodford

Categories: calculus, maths

## First Order Differential Equations

First order differential equations are equations which include a first derivative ie) dy/dx if the equation is in terms of x and y.

Solving these equations involves finding a expression for one of the variables in terms of the other without including a derivative.

To do this usually a general solution is found, one which includes a constant from integration, and then using a given set of conditions (usually initial conditions) the particular solution can be found.

This is done by “splitting the variables” and then integrating the resulting equation. To do this you treat the derivative like a fraction and then multiply through by the bottom of that one derivative is on one side (eg dx) and the derivative is on the other side (eg dy), you then want all the x’s to be multiplied by the dx and all the y’s to multiplied by the dy such that there are no terms not being multiplied by either. You can then integrate both sides, the one by dx and the other by dy – remembering to add a constant to one side.

Then simply re arranging will give the general solution. If you need the particular solution substitute the given values into the equation to find out the value of the constant and then substitute this value for the constant into the general solution.

If all of that was a bit hard to follow here is s a worked example.

Example

Find the general solution to
dy/dx = 3x + 4
and the particular solution in the case y = 4 when x = 0 .
So we can rearrange to get

dy = (3x+4)dx

And then by integrating both sides we get

∫dy = ∫3x+4 dx

y = (3x+4)2/6 + c

Which is the general solution.
We can use boundary conditions, ie) y = 4 when x = 0 to find a particular solution
4 = (0+4)2/6 +c
c = 4-16/6 = 8/6 = 4/3

so the particular solution is
y = (3x+4)2/6 +4/3

Categories: maths