Useful Background
By Chris Patterson, Head of Maths, Kainos International School
Some useful ideas in the background while doing integration problems.
Vocabulary
Base: a factor in a multiplication. For {z\ =\ ax}^p\times\ {by}^q, x and y are bases.
Power: the number of times the base is a factor in a multiplication calculation. For {z\ =\ ax}^p\times\ {by}^q, p and q are powers of their respective bases. Instead of ‘power’ we can say ‘exponent’, ‘index’, or ‘logarithm’.
Coefficient: a constant multiplying a variable term. For {z\ =\ ax}^p\times\ {by}^q, a and b are coefficients of their respective bases.
Factoring and indices (or powers):
60\ =\ 15\ \times\ 4\
60\ =\ 5\ \ \times\ 3\times\ 2\ \times\ 2
60\ =\ \ 2^2\times\ 3^1\times\ 5^1\ \
The power counts the number of times the base is used as a factor.
If a factor x is used once, its power is 1. If a factor is not used, its power is 0. For example, 11 is not a factor of 60. Therefore, I could write 60\ =\ \ 2^2\times\ 3^1\times\ 5^1\ \times\ {11}^0.
Fractional, negative and zero powers
x^{1/2} and \sqrt x both represent the same concept:
\sqrt x \times\sqrt x\ =\ {\sqrt x}^2=\ x
x^{1/2} \times x^{1/2}\ \ =\ {\ x}^{1/2\ +\ 1/2}=\ x^1\ =\ x
x^0=\ 1.
Consider the sequence 2, 4, 8, 16, … . The next number in the sequence is \ 16\ \times\ 2\ =\ 32.\ But, consider a different perspective. What if I want the next number on the left? …, 2, 4, 8, 16. The next number in the sequence is \ 2\ \div\ 2\ =\ 1.\ Writing my doubling sequence with powers, I have 1,\ 2,\ 4,\ 8,\ 16,\ 32,...\ =\ 2^0,\ 2^1,\ 2^2,\ 2^3,\ 2^4,\ 2^5,\ \ \ .... I could use different bases, the basic idea would be the same: 1,\ 3,\ 9,\ 27,...\ \ =\ 3^0,\ 3^1,\ 3^2,\ \ \ ....
Negative powers represent factors in the denominator.
Consider the sequence 2, 4, 8, 16. What if I want the next three numbers on the left?
2\ \div\ 2\ =\ 1\
1\ \div\ 2\ =\ \frac{1}{2}\
\frac{1}{2}\ \div\ 2\ =\ \frac{1}{4}
Quite naturally the sequence changes from "2^1,\ 2^2,\ 2^3,\ 2^4" to "2^{-2},2^{-1},\ 2^0,\ \ {2^1,2}^2,\ 2^3,\ 2^4"
Undoing an operation:
Maths operations often appear in pairs of opposites. Adding undoes subtraction, and vice versa. Multiplication undoes division, and vice versa.
We begin in calculus by supposing
we know a function
we want find a formula for the gradient of the tangent.
For example, f(x)\ =\ y\ =\ {5x}^3. What is the gradient of the tangent? \frac{dy}{dx\ }\ =\ {15x}^2.
However, in real life we more often know the rate and would like to have a formula for the function.
I am driving along the highway. Sometimes my speed is 110km/h; sometimes it is 80km/h; sometimes I might be stopped. I know \frac{dy}{dx\ }\ at every point on my journey. I want to know the distance I have travelled.
I know my cash flow. Sometimes it is $200/day;\ sometimes it is -$1000/day. I want to know my bank balance.
I know \frac{dy}{dx\ }\ =\ {15x}^2. Can I find the ‘antiderivative’?
To differentiate a power of x, we i) multiply the coefficient of the variable by the given power and ii) subtract one from the power. For {5x}^3\ the coefficient is 5 and the power is 3. The derivative is \ {15x}^2.
The opposite procedures are to i) add 1 to the power and ii) divide the coefficient by the raised power. The power is 2; adding 1 gives power 3. The coefficient is 15. Divide 15 by 3 =\ 5. The antiderivative is {5x}^3.
Derivative of a constant
There is more than one antiderivative. For any constant C, we know \frac{dC}{dx\ }\ =\ 0.\ When we antidifferentiate 0, we get back C. Since \ y\ =\ \ {5x}^3\ =\ {5x}^3\ +\ 0\ we can differentiate to get {15x}^2+\ 0.\ \ When we antidifferentiate we get {5x}^3\ +C,\ which could be {5x}^3\ +7, {5x}^3-\ \sqrt2 and so on.
The derivative of {\ ax}^p using the product rule:
Suppose u\ =f(x)\ =\ 5 and g(x)\ =\ x^3.
u\prime\ =\frac{du}{dx}\ \ =\ 0 and v\prime\ =\frac{dx^3}{dx}\ \ =\ {3x}^2
\frac{d{5x}^3}{dx}\ =\ u\prime v\ +\ v\prime u\ =\ 0v\ +\ 5\frac{dx^3}{dx} . In simple English, “constants pull theough” the differentiation operator.
To be continued…
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