The look and feel of Trevor Pythagoras Books will probably change greatly in the near future as I seek to make it fit in with the look of the rest of the site and make the navigation between the sections of the site as smooth as possible.

Also see

• Integration by Parts (0)
• Cartesian Equation of a Circle (0)
• Arithmetic Series and Progression (2)
• Confidence Intervals (8)
• Area of a Triangle (3)
• Proof by Contradiction (0)

∫2xe^3xdx

where 2x is one function and e^3x 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 e^3x 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

∫2xe^3xdx

To do this we first need to pick which function (2x or e^3x) 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 e^3x gives 3e^3x 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=e^3x

we then differentiate 2x to get 2

and integrate e^3x to e^3x/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

∫2xe^3xdx = 2xe^3x/3 – ∫e^3x2/3 dx

∫2xe^3xdx = 2xe^3x/3 – 2e^3x/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 e^3x dx = [2xe^3x/3]¹₀ – ∫¹₀ e^3x2/3 dx

=2e³ – 0 – 2e³/9 + 2/9 = (16e³ + 2) / 9

Also see

• Product Rule for Differentiation (0)
• Fundamental Theorem of Calculus (0)
• Implicit Differentiation (1)
• First Order Differential Equations (6)
• Common Intergrals (0)
• The new Trevor Pythagoras Book Store (0)

r² = (x-a)² + (y-b)²

Circle on Cartesian axis

However, we can expand these brackets out to get

r² = x² – 2ax + a² + y² – 2by + b²

but since a²+b²+r², -a and -b are all constant we can let

c = a²+b²-r²,
f = -a
g = -b

to get

x² + y² + 2gx + 2fy + c = 0
where the circle has a centre (-g,-f) and radius √(a²+b²-c²)

Also see

• Area And Circumference of a Circle : pi (38)
• The new Trevor Pythagoras Book Store (0)
• Integration by Parts (0)
• Arithmetic Series and Progression (2)
• Confidence Intervals (8)
• Area of a Triangle (3)

In the example i have written the series as
2 + 5 + 8 + 11 + 14 + 17 + …
instead of
2 , 5 , 8 , 11 , 14 , 17, …
because it is often interesting to look at sum of n items in a series. The sum S_n of the first n terms in a series can be found using
S_n = n(2a+(n-1)d)/2

Proof of sum of series
We can show that
S_n = n(2a+(n-1)d)/2
as follows

Write out the series in order and then in reverse order as follows
S_n = a + (a+d) + (a+2d) + …. + (a-(n-1)d) + (a+nd)
S_n = (a+(n-1)d) + (a+(n-2)d) + (a+(n-3)d) + …. + (a+d) + a
Now if we add each of the terms in the two sums we get
2S_n = (2a+(n-1)d) + (2a+(n-1)d) + (2a+(n-1)d) + … + (2a+(n-1)d) + (2a+(n-1)d)

where the 2a+(n-1)d repeatsn times. This is because we adding the rth term with the(n-r)th term hence we doing
(a+(r-1)d)+(a+(n-r)d)=2a+(n-1)d
so
2S_n=n(2a+(n-1)d)
therefore
S_n=n(2a+(n-1)d)/2

Also see

• How to use Surds – Add, Multiply and Rationalise (1)
• Geometric Series/Sum (0)
• The new Trevor Pythagoras Book Store (0)
• Fibonacci Sequence and PHI (3)
• Integration by Parts (0)
• Cartesian Equation of a Circle (0)

When taking a random sample it is much better to use a confidence interval for your results rather than just giving the mean, because it gives an idea of how reliable your mean is. This is because you can calculate an average value from a set of completely random results but this doesn’t mean that you can have any certainty that the next result will be similar to the mean (since we have stated that the results are completely random they are no more likely to be close to the mean than any other value).

Calculating a confidence interval (for a normal distribution)

When calculating a confidence interval you must first decide on the percentage certainty that you are going to use for the interval (a common value to use is a 95% confidence interval)

You then use the reverse tables for the normal distribution to work the value of the standardised normally distributed variable to use.
Note: You must use the value half way between the certainty level and 100%, ie if you want a 95% confidence interval use 97.5% since you only want 2.5% on either side of the distribution.

You can now calculate the interval. To do this you need the standard deviation and mean of the sample. However to correct the standard deviation for the entire sample of possible tests divide by the square root of the number of items in your sample

ie) if you sample of n items is Y and the entire sample of possible results is X
sd(X) = sd(Y)/√n

Now the confidence interval is

X – z sd(X), X + z sd(X)

Where z is the value obtained from the inverse tables.

Also see

• The new Trevor Pythagoras Book Store (0)
• Integration by Parts (0)
• Cartesian Equation of a Circle (0)
• Arithmetic Series and Progression (2)
• Area of a Triangle (3)
• Proof by Contradiction (0)

A = ½ wh

triangle of width w and height h

Proof

In order to work out the area of a triangle we can draw it within a rectangle that touches all three corners and has a base of equal width as shown below.

triangle inside rectangle

From this diagram we can now consider the are of the triangle on the left and the one on the right seperatley and then add them together to get the area of the whole triangle. We can see that each of these triangles cuts the rectangle they are in in half so they half the area of that rectangle,

ie) the left hand triangle has the area ha/2 and the right hand one has the area hb/2 so the total are of the triangle is

A = ha/2 + hb/2

Now we can take out the h/2 as a common factor to get

A =(a+b) h/2

but a+b = w since that was how the triangle was constructed hence we get the formula for the area

A = wh/2

Also see

• Proof by Contradiction (0)
• Rational and Irrational Numbers (5)
• Derive Quadratic Formula (16)
• Trigonometry Identities (11)
• Ellipse’s: equation, foci, eccentricity (5)
• Proof of Pythagorases Theroem (0)

For example consider the proof √2 is irrational that follows

Assume root 2 is rational, ie that it can be written as r/s where s≠0 and r and s are both integers. We can choose r and s such that they have no common factors, since any common factors can be cancelled out.

then 2=r²/s²
so r² = 2s² —(1)
hence 2 is a factor of r² and since 2 is prime 2 must also be a factor of r so we can write r = 2k where k is an integer.

From (1) we can now write
4k²=2s²
so s = 2k²
so 2 is also a factor of s
But we assumed r and s had no common factors, Contradiction therefore root 2 must be irrational.

So to conclude, here we have taken the statement root 2 is irrational, assumed it to be false, shown this leads to a contradiction and therefore concluded that root 2 must be irrational.

Also see

• Area of a Triangle (3)
• Proof there are infinitley many prime numbers (0)
• Rational and Irrational Numbers (5)
• Derive Quadratic Formula (16)
• Trigonometry Identities (11)
• Ellipse’s: equation, foci, eccentricity (5)

Prime numbers are natural numbers (1,2,3 etc) whose only factors are one and themselves. The first few primes are:

2,3,5,7,11,13……

note that 1 isn’t a prime.

Primes are important because every number can be written uniquely as a product of primes. This idea can also be used to prove that there infinitely many primes. The proof which follows is an example of proof by contradiction.

Proof

Assume that there are a finite number of primes.

Then all these primes can be written in a list

P0, P1, P2 … Pn

Then we could multiply all these numbers together and add one to get a new number Q

Q = P0P1P2…Pn + 1

Suppose that Q+1 had a prime factor, say p. The there exist natural numbers a and b such that Q=ap and Q+1 = bp since p divides both Q and Q+1, and a and b are distinct since Q and Q+1 are distinct. But this means
Q+1-Q = p(b-a) which implies divides 1 which in turn implies p = 1, since there are no other natural numbers which divide 1. But p is a prime so p doesn’t equal 1 which is a contradiction and therefore our assumption p divides Q+1 is wrong.

Therefore Q is either a new prime or there is a prime which isn’t on the list which is a factor of Q. So the list doesn’t contain all the primes because it doesn’t contain this new prime. This is a contradiction because we assumed that the list contained all the primes, therefore there cant be a finite number of prime so there must be an infinite number of them, hence they cannot be written in a list.

By David Woodford

Also see

• Proof by Contradiction (0)
• Area of a Triangle (3)
• Rational and Irrational Numbers (5)
• Log Laws: Adding and Subtracting Logs (0)
• Derive Quadratic Formula (16)
• Trigonometry Identities (11)

It involves rewriting letting asinx + bcosx = rsin(x+y) (or you could use cos(x+y)) where y is acute and then finding values for r and y, then with only one trig function to deal with the equation can be solved more easily.

For example

consider 2sinx + 3cosx = 3

Let 2sinx + 3cosx = rsin(x+y)

Now expand the sin(x+y) to get

2sinx + 3cosx = rsinx cosy + rcosx siny

Since y is constant and therefore cosy and sin y are constant we can compare the coefficients to get

2 = rcosy —–(1)
3 = rsiny ——(2)

We can solve these to find values for r and y.
To find y consider (2)/(1) to get

3/2 = tany
since sin/cos = tan and the r’s cancel
so y = 56.3 °

To find r consider (1)2+(2)2 to get
22+32 = r2
since sin2+cos2 = 1
so r =√13

So we can write

2sinx + 3 cosx = √13 cos(x+56.3) = 3

so x = cos^-1(3/√13) -56.3

so x = cos-1(3/√13) -56.3

since cos-1(3/√13) = 33.7 for solutions between 0° and 90°

x = ±33.7 -56.3 + 180n where n is an integer

In General

asinx + bcosx = √(a2+b2) sin(x+tan-1(b/a))

If you have any questions, comments or corrections please leave them as a comment below

By David Woodford

Also see

• Trigonometry Identities (11)
• Proof of Cosine Rule (4)
• Understand the Sine and Cosine Rules (11)
• Area of a Triangle (3)
• Proof by Contradiction (0)
• Rational and Irrational Numbers (5)

A number x is rational if it can be written in the form a/b where a and b are integers and b≠0 and a number which cant be written in this form is irrational.

Most of the time we deal with rational numbers, for example all the integers are rational as they can be written in the form of themselves over 1, all fractions are obviously rational and also are terminating or repeating decimals. When an irrational number is written out it will appear as an infinite non repeating decimal. Examples of irrational numbers are π (pi) and √2, in fact many square roots are irrational numbers.

Numbers can usually be proved to be irrational by contradiction, by assuming that they are rational and writing them in the form p/q where p and q are co-prime. Then a contradiction is usually produced by showing that they both have common factor other than 1. For example when proving √2 is irrational you can show that p and q must both have a common factor of 2.

There infinitely many rational and irrational numbers, however whilst it is possible to use a method to count all the rational numbers, if you had an infinite amount of time, this cannot be done for the irrational numbers.

Showing a repeating decimal is rational

A repeating decimal can be shown to be rational using the following method:

Consider the repeating decimal x,

Multiply x by 10 ⁿ where n is the number of repeating digits
Now subtract x from x10 ⁿ to get (10ⁿ-1)x
This number will be a finite decimal as all the repeating terms after the decimal point should cancel.
divide through by (10ⁿ-1)x to get an expression for x in the form of a fraction.

eg let x = 1/7 = 0.142857142857142857………
then 1000000x = 142857.142857142857………
so 999999x = 142857.0
x=142857/999999
so 1/7 = 142857/999999 therefore 1/7 is rational

if you have any questions, comment or corrections please leave them as a comment below

By David Woodford

Also see

• Area of a Triangle (3)
• Proof by Contradiction (0)
• Derive Quadratic Formula (16)
• Trigonometry Identities (11)
• Ellipse’s: equation, foci, eccentricity (5)
• Proof of Pythagorases Theroem (0)