MST124 BOOK C UNIT 8

- for finding an approximate area under a curve, divide the interval [a,b] into a number, say n, of subintervals of equal width
- the width of each subinterval is then (b-a)/n
- for each subinterval, you calculate the product:
f(left endpoint of subinterval) x (b-a/n)
- add up all these products 
- the larger the number n of subintervals, the closer your answer will be to the area between the graph of f and the x-axis, from x = a to x = b 

- if the graph lies on or below the x-axis throughout an interval [a,b], the left endpoints of each subinterval is -ve or 0
- so each product will be the -ve of the area between the line segment that approximated the curve and the x-axis 
- when you add all the products, you'll obtain an approximate value for the -ve of the area between the curve and the x-axis, from x = a to x = b 

- simply remove the minus sign to obtain the approximate value for the area that you want

- consider any region on a graph that lies either entirely above or entirely below the x-axis
- the signed areas of the region is its area with a plus or minus sign according to whether it lies above or below the x-axis

- if you have a collection of regions on a graph, each of which lies entirely above or below the x-axis, then the total signed area of the collection is the sum of the signed areas of the individual regions 

- if a and b are equal, then the signed area is 0

- if f is a continuous function whose domain includes the interval [b,a], the signed area between the graph of f and the x-axis from x = a to x = b is defined to be the -ve of the signed area between the graph of f and the x-axis from x = b to x = a 

- if f is a continuous function and a and b are numbers in it domain, then the signed area between the graph of f and the x-axis from x = b to x = b is called the definite integral of f from a to b (see page 119 for how it is denoted)

- the numbers a and b are called the lower and upper limits of integration, respectively 

Standard properties of definite integrals
(page 120 of book C)

Algebraic definition of a definite integral
Suppose that f is a continuous function and a and b are numbers in its domain
- then the definite integral of f from x = a to x = b is given by the equation in the blue box on page 124 
- w = (b-a)/n

Definite integrals
Suppose that f is a continuous function, and a and b are numbers in its domain
- the signed area between the graph of f and the x-axis from x = a to x = b is called the definite integral of f from a to b
- denoted by the formula in blue box on page 126 

A(x) = signed area between the graph of f and the x-axis from s to x
- A = signed-area-so-far-function 
(bottom of page 127 book C)

- if a and b are any two numbers in the domain of f, then the signed area between the graph of f and the x-axis, from x = a to x = b, is given by
A(b) - A(a)
(Figure 18 page 128 book C)

- for any constant function f(x) = k, and any signed-area-so-far function A for f, the value of A(x) changes at the rate of k square units for every unit by which x changes 
- the rate of change of A(x) is k 

Theorem
Suppose that f is a continuous function, and s is any number in its domain. Let A be the signed-area-so-far function for f with starting point s. Then A is an antiderivative of f. 

Fundamental theorem of calculus (in words)
Suppose that f is a continuous function whose domain contains the numbers a and b, and that F is an antiderivative of f. Then the signed area between the graph of f and the x-axis from x = a to x = b is equal to the change in the value of F as x changes from x = a to x = b 

- every continuous function has an antiderivative 

- for any function F, the expression
F(b) - F(a)
can be denoted by
[F(x)]^b_a 
For example
[sin]^pi/2₀ = sin(pi/2) - sin(0) = 1 - 0 = 1

Fundamental theorem of calculus (square bracket notation)
[F(x)]^b_a 

Constant multiple rule and sum rule for the square bracket notation
[kF(x)]^b_a = k[F(x)]^b_a, where k is a constant
[F(x) + G(x)]^b_a = [F(x)]^b_a + [G(x)]^b_a 

Constant multiple rule and sum rule for definite integrals
blue box on page 140 book C

- for the velocity-time graph of any object, and for any two points in time t = a and t = b in the time period covered by the graph, the following hold:
- the change in displacement of the object from t = a to t = b is the total signed area between the graph and the time axis from t = a to t = b
- the total distance traveled by the object from t = a to t = b is the total area between the graph and the time axis from t = a to t = b (this applies when a ≤ b) 

page 151 of book c for notation for indefinite integrals
- must always have an arbitrary constant (c) on the other side if this notation is used 

Constant multiple rule and sum rule for indefinite integrals
Blue box, page 153, Book C 

Integration by substitution
1 - recognise that the integrand is of the form
f(something) x the derivative of the something
where f is a function you can integrate
2 - set the something equal to u, and find du/dx
3 - write the integral in the form
(elongated s) f(u) du,
by using the fact that (elongated s) f(u) du/dx dx = (elongated s) f(u) du
4 - do the integration
5 - sub back for u in terms of x 

EXAMPLE 15 UNIT 8 BOOK C
- you can multiply by a number inside the integral, as long as you divide by the same number outside it too 

- you can use integration by substitution to find any integral of the form
(Elongated s) the derivative of the something / something dx
or
(Elongated s) the derivative of something / somethingⁿ dx, where n is a constant 

Indefinite integral of a function of a linear expression
If f is a function with antiderivative F, and a and b are constants with a not equaling 0, then
(elongated s) f(ax + b)dx = (1/a)F(ax + b) + c 

Indefinite integral of a function of a multiple of the variable
If f is a function with antiderivative F, and a is a non-zero constant, then
(elongated s)f(ax)dx = (1/a)F(ax) + c 

DISPLACEMENT IS AN ANTIDERIVATIVE OF VELOCITY 

Integration by parts formula (lagrange notation)
(elongated s)f(x)g(x)dx = f(x)G(x) - (elongated s)f'(x)G(x)dx
G is an antiderivative of g 

Integration by parts formula (informal)
Integral of product = (first) x (antiderivative of second) - integral of((derivative of first) x (antiderivative of second)) 

- if you're trying to use integration by parts, but you find that it leads to an integral that's more complicated than the original integral, then try swapping the two expressions in the original integrand 

Integrands of the form xⁿg(x) 
- n is a +ve integer and g(x) is an expression that you can integrate
- use integration by parts n times 
- integrate by parts the whole thing, then integrate by parts the second part of the expression (with the elongated s) 

Integrands of the form x^rlnx
- r is a constant
- you cant apply the integration by parts formula with lnx as the second expression
- apply the integration by parts formula with lnx as the first expression 

Integrands of the form e^axsin(bx) or e^axcos(bx)
- you can integrate these expressions, where a and b are non-zero numbers, by integrating by parts twice
- you obtain an equation that expresses the original integral in terms of itself, and you can rearrange this equation to find the original integral
- is easier to take exp as the g(x) term 

Integration by parts formula for definite integrals
Blue box, page 186, Book C 

Integration by parts formula (Lagrange notation, alternative version) and Integration by parts formula (Leibniz notation)
Blue boxes on page 188 of book C 

Three trig identities
sin²(theta) = 1/2(1 - cos(2*theta))
cos²(theta) = 1/2(1 + cos(2*theta))
sin(2*theta) = 2sin(theta)cos(theta) 

- when using integration by parts to find an area, use the interval endpoints as the upper and lower limits 

Choosing a method for finding an integral
Is it a standard integral? - consult the table in the Handbook
Can you rearrange the integral to express it as a sum of constant multiples of simpler integrals?
Is the integrand the form f(ax) or f(ax + b)? - rule for integrating a function of a linear expression, or substitution
f(something) x the derivative of the something? - set something to u
f(x)g(x) - integration by parts
If the integrand contains trig functions, try using trig identities to rewrite it in a form that's easier to integrate
ALWAYS ADD + c