Integration notes

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Intergration

Integration is the reverse process of differentiation and can be used to find the area under a curve.

Indefinite integration

Here we take a function, for example 2x and say that its integral is x^2; the answer is another function.

the 'c' is an arbitrary number that can take any value and is there for the constant of integration.

we call this process indefinite integration

Definite integration

In contrast, provides us with an exact numerical answer.

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Numerical integration

for questions asking for you to use the trapezium or Simpson's rules:

use values that are given to you in the form of a labeled graph or table.

if only some values are given use the table function on your calculator

substitute in values

where h is the value of the intervals between each given x coordinate.

and y0 is the y value at the first given x coordinate.

trapezium rule

simpson's rule

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sums and products

the sum and product formulas allow you to simplify a trig expression so that it can be integrated, etc. is given on formula sheet during the exam.

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Newton's Law of Cooling

Example- If a metal container of hot water at 70' cools to 40' in 5 min in a 20' room calculate its temperature after 10 min.

the integral of 1/(T-A(t)).dT=-k.dt

is ln| c (T-A(t)) | = -kt

natural log is removed from both sides of the equation to give, c(T-A(t)) =e^-kt

let 1/c = x , (T-A(t))=x.e^-kt

rearrange to find the T, T=x.e^-kt +A(t)

Substitute in the value for A(t) when A(t)=20'

find x by substituting into the equation the temp of the object, T, when the time, t, is zero.

use x=1/c solve for the value of the constant

c=1/50

find k by substituting into the equation the temp of the object at another given time, like T=40' when t=5min

change from exponential form to log form to solve for k

e^-5k=/5 so ln(2/5)=-5k then k=ln(0.4)/-5 therefore k=0.183

Then find the T after the asked for t (10 min) by substituting t=10 and solving for T

T=28'

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Differential equations

General solutions

when finding the general solution for Differential equations the equation is integrated and a constant is added

method-

separate values (for some equations the variables will already be separated and so you will be able to integrate straight away.)

integrate both sides

simplify

tidy so that y=function

Particular solutions

to then find a particular solution information of the equations boundary conditions is given, these consist of the value for x when y another value. the x and y values can be substituted into the equation so that you can solve for c the integration constant.

method-

separate values (for some equations the variables will already be separated and so you will be able to integrate straight away.)

integrate both sides

simplify

tidy so that y=function

sub into the equation the boundary conditions to solve for the integration constant

give a particular solution as the answer.

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definite integrals

Definite integrals are the solution of an integral which can be calculated between the limit of integration

method-

integrate equation

the integration constants will cancel and so can be left out when finding a definite integral.

integral is put into brackets with the limits shown outside (the limits are the x values for which the area between is being calculated)

when b and a are the limits of integration

firstly substitute b into the expression inside the brackets

then a and subtract.

note: area under the x-axis will give a negative result however area is always positive this is just indicating the negative position on the graph and should be taken into account when the area is being calculated.

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integrating quotients

If the quotient to be integrated has a numerator that is the derivative (or multiple of the derivative ) of the denominator then we use the given formula.