P1 Chapter 16 CIE Centre A-level Pure Maths © Adam Gibson

Integration Differentiation Integration Sum Difference Gradient Area Calculus

Integrals Definite IntegralIndefinite Integral Noun (Action): integration Verb: integrate Noun: integral Derivatives Read it as “integral [of] f [of] x dx [between a and b]”

Integration as the reverse of differentiation The first, and easiest, way to understand integration: What function g(x) has f(x) as its gradient at all points x in the domain? The indefinite integral helps us find the answer. Example: We remember the rule that Think….

Integration as the reverse of differentiation – contd. We note that we can change to (I won’t give a formal proof here, but you can verify it). ! Warning … What if ? So we construct the rule: where k is an “arbitrary constant”. This is the indefinite integral.

Implications Does the formula apply for all rational numbers n? No! At n = -1 the formula “explodes”. We will consider this in detail later. What does “arbitrary constant k” actually mean? For this point, think about moving the graph of a function up and down by adding a number. Does the gradient change? So there are an infinite number of functions with the same gradient formula. We can only calculate k if we are given the coordinates of at least one point on the graph. If you don’t have specific information, always write “+k”.

TASKS 1.Find the integral of 2.Find 3.Find 4.Try Q15 and Q16 p. 240 given that g(1)=0

area Integration as area How do we make it more accurate? …

area – continued. Integration as area – continued. Just like finding the gradient, this is a limit problem. The exact area is given by We take the sum of an infinite number of trapezia, each with infinitely narrow width. Thus, if the function is f(x), we get: Remember – dx is an infinitely small ∆x Section 16.2 and Section 16.3 give a more detailed derivation

area – continued. Integration as area – continued. Let’s find the area under the curve between x=0 and x=2 “Limits of integration”

area – continued. Integration as area – continued. We write this quantity as the definite integral: Higher limit at the top Lower limit at the bottom You should perform the integration, but not include the arbitrary constant k (can you see why?). Then you should put the answer in square brackets with the limits marked on the right.

area – continued. Integration as area – continued. Now for the last step. Evaluate the expression inside the brackets at x=upper limit and x=lower limit, and subtract the latter from the former. What are the units for the answer? To be clear: the definite integral is the difference between the indefinite integral evaluated at the upper and lower limits … here we could write it as I(2) – I(0). The constant “k” is not needed as it’s the same in both cases.

Your turn … Answer: Be careful! The area is negative?! Is it a mistake …?

No mistake! The area is just a sum of function values. It can be any real number, positive, negative or zero. So be careful when interpreting the result! You have just calculated the blue area.

Infinite and improper integrals Let’s consider difficult cases like: We don’t have a formula for this (n+1=0), so we’ll ignore it for now (actually ) Improper integrals are integrals over an interval including a value of x where the function is undefined. Infinite integrals are those involving at least one limit at ±∞ This has a finite value This also has a finite value Calculate please A B C

Infinite and improper integrals continued In order to better understand the difference between A and B, let’s take a close look at the graph: The area under between 1 and ∞ is finite but unbounded.

Infinite and improper integrals continued The problem of infinite integrals is of course similar to the problem for infinite sums: but… we say it “diverges” Example Evaluate:

Notes on Integration Integration by substitution – ONLY use this for examples where the “inner” function is linear. Other examples are too advanced for P1. Remember the “n+1 over n+1” rule and use it correctly! Indefinite integrals – you MUST use +k. ! n = -4. a = 10, so divide by (-3x10)

Notes on Integration Areas found by definite integration are positive if.. … the function values are positive (and vice versa) Areas of regions bounded by two functions can be calculated … by subtracting one function from the other and then integrating. Result will be positive if … you subtract the “lower” function from the higher one. Is the integral (area) positive for: a=8 b=10 a=-10 b=-8 a=1 b=3 a=1 b=7 ?