#### The Internet belongs to everyone. Let’s keep it that way.

Protect Net Neutrality

### Present Remotely

Send the link below via email or IM

• Invited audience members will follow you as you navigate and present
• People invited to a presentation do not need a Prezi account
• This link expires 10 minutes after you close the presentation

Do you really want to delete this prezi?

Neither you, nor the coeditors you shared it with will be able to recover it again.

# Integration and Differentiation

A short presentation on the need for integration and differentiation
by

## Laura Chapman

on 24 May 2013

Report abuse

#### Transcript of Integration and Differentiation

Integration by Parts Integration Integration is often introduced as the reverse process to differentiation, and has wide applications, for example in finding areas under curves and volumes of solids.  It can also be used to find the relationship between two variables. For example, we may know the velocity of an object at a particular time, but we need integration to discover the position of the object. What is it
and why do
we need it? Differentiation Differentiation is primarily used for calculating rates of change, and more specifically to find the gradient at a specific point on a graph. In Mechanics, the rate of change of displacement (with respect to time) is the velocity and the rate of change of velocity (with respect to time) is the acceleration. When illustrating a function on a graph the rate of change is represented by the gradient. Chain Rule Let's use this equations as an example:

y = (3x − 2)4

If we let u = 3x − 2 to give us y = u4.

We can now differentiate this to give us:

dy/du = 4u^3

Integration and Differentiation Methods include:
Integration by parts
Partial Fractions
Substitution Methods include:
Product rule
Quotient rule

ʃ ln x dx

we need to label each part to complete the calculation:

uv - ʃ vdu Integration by Parts From this we can multiply out the brackets:

x ln x - ʃ dx

And we now have our answer:

x ln x - x Integration by Parts We can let:

u = lnx
du = 1 / x
dv = dx
v = x

Therefore substituted into the equation:

ln(x) - ʃ x (1/x) dx Chain Rule The only problem is that we want dy/dx, not dy/du, and this is where we use the chain rule. The chain rule states that:

dy/dx = (dy/du) * (du/dx)

As we already know u and du/dx, we can substitute them into the equation as follows:

du/dx = (4u^3) * 3 = 12 (3x - 2)^3