Calculus and trapezium rule Flashcards
Sketch graph
area under graph
Split modulus up into pieces based on where the break occurs and integrate each part
Integral |x^2 - x| separate affected bounds and get rid of the modulus if needed and take negative of affected parts
Max/min complete the square
if using calculus, justify with second derivative
Inflection point
d^2y/dx^2 = 0
d^3y/dx^3 =/= 0
A =
0.5w(y1 + yn + 2(y2 +…+y(n-1)))
where w is the width of the strip
Overestimate when line curves upwards
concave/convex
Underestimate when line curves downwards
concave/convex
Trapezium rule, R =
= width/2 (y1 + yn + 2(y2 + y3 + y(n-1)))
= 1/2n(1+2(2^1/n..
width/2 = range of x values/2*number of trapeziums
The trapezium rule uses equal length intervals suggesting that the boundaries of the strips have to coincide with the dots in the diagram,
otherwise our trapeziums wouldn’t exactly match the function.
In general, it’ll be the lowest common multiple of the denominators
Overestimates and underestimates
concave, convex
concave curve overestimate
I(a) = area up to x=a DI/da = 0
when the area doesn’t change as ‘a’ changes therefore when y crosses x axis and the area stops increasing and is about to decrease
Integrating a transformed function:
change domain and look at what new equation is
integral (2,0) =
integral (2,1) + integral (1,0)
take constants outside of integral
integral with bounds is just a number/constant
trapezium rule = actual integral
when n is a multiple of the common denominator of points on the graph
di/DA(integral i da) = I
integral = summation of areas
Integral of f(-x) =
integral of f(x) between values due to area reflection
f(x) > 0
f(x)=f’(x)
therefore increasing function, concave curve oversestimate from trapezium rule
n steps of trap rule:
between 0 and 1
1/(2n)[f(0) + f(1) + 2f(1/n) + f(2/n)… f((n-1)/n))]
integral f(x^2 - 1)dx –>
change domain of f(x) to that of f(x^2-1), find equation of f(x) and change to f(x^2 - 1)
Point of inflection if
d^2y/dx^2 = 0
d^3y/dx^3 =/= 0
Cos(ax)
changes domain, solutions, graph
What is the maximum of an integral
The integral is at most the length of the range of integration times the largest value of the intergrand
Area is preserved under reflection
?
Integration questions
try acc do it
Integral of a polynomial =
For reals a0, a1, a2, …, ad and +ve integer d, Let the polynomial p(x) = a0 + xa1 + x^2a2 + x^3a3 + … + x^dad, Therefore the integral of p(x) dx = ∑(d)(r=0)ar∫x^r dx because integration is linear
Drawing trapezium rule
not equal heights, just equal width so must have common multiple of all points to be exact
Integral t, t-1 =
integral t, 0 – integral t-1, 0
i.e. different equations for the bounds seperated
When finding the max.min area under the curve
alongside your dy/dx = 0, check start and end points/bounds and max/mins of curve
Max/min of a function e.g. cubic
Differentiate set to 0, find max/min using second derivative
If quadratic complete the square
If a set domain/range look at end points of it
If cubic, look at end points to see if above or below max/min
