Calculus and trapezium rule

Question 1

Sketch graph

Answer 1

area under graph

Question 2

Split modulus up into pieces based on where the break occurs and integrate each part

Answer 2

Integral |x^2 - x| separate affected bounds and get rid of the modulus if needed and take negative of affected parts

Question 3

Max/min complete the square

Answer 3

if using calculus, justify with second derivative

Question 4

Inflection point

Answer 4

d^2y/dx^2 = 0

d^3y/dx^3 =/= 0

Question 5

A =

Answer 5

0.5w(y1 + yn + 2(y2 +…+y(n-1)))

where w is the width of the strip

Question 6

Overestimate when line curves upwards

Answer 6

concave/convex

Question 7

Underestimate when line curves downwards

Answer 7

concave/convex

Question 8

Trapezium rule, R =

Answer 8

= 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

Question 9

The trapezium rule uses equal length intervals suggesting that the boundaries of the strips have to coincide with the dots in the diagram,

Answer 9

otherwise our trapeziums wouldn’t exactly match the function.
In general, it’ll be the lowest common multiple of the denominators

Question 10

Overestimates and underestimates

Answer 10

concave, convex

concave curve overestimate

Question 11

I(a) = area up to  x=a 
DI/da = 0

Answer 11

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

Question 12

Integrating a transformed function:

Answer 12

change domain and look at what new equation is

Question 13

integral (2,0) =

Answer 13

integral (2,1) + integral (1,0)

Question 14

take constants outside of integral

Answer 14

integral with bounds is just a number/constant

Question 15

trapezium rule = actual integral

Answer 15

when n is a multiple of the common denominator of points on the graph

Question 16

di/DA(integral i da) = I

Answer 16

integral = summation of areas

Question 17

Integral of f(-x) =

Answer 17

integral of f(x) between values due to area reflection

Question 18

f(x) > 0

f(x)=f’(x)

Answer 18

therefore increasing function, concave curve oversestimate from trapezium rule

Question 19

n steps of trap rule:

between 0 and 1

Answer 19

1/(2n)[f(0) + f(1) + 2f(1/n) + f(2/n)… f((n-1)/n))]

Question 20

integral f(x^2 - 1)dx –>

Answer 20

change domain of f(x) to that of f(x^2-1), find equation of f(x) and change to f(x^2 - 1)

Question 21

Point of inflection if

Answer 21

d^2y/dx^2 = 0

d^3y/dx^3 =/= 0

Question 22

Cos(ax)

Answer 22

changes domain, solutions, graph

Question 23

What is the maximum of an integral

Answer 23

The integral is at most the length of the range of integration times the largest value of the intergrand

Question 24

Area is preserved under reflection

Answer 24

?

Question 25

Integration questions

Answer 25

try acc do it

Question 26

Integral of a polynomial =

Answer 26

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

Question 27

Drawing trapezium rule

Answer 27

not equal heights, just equal width so must have common multiple of all points to be exact

Question 28

Integral t, t-1 =

Answer 28

integral t, 0 – integral t-1, 0

i.e. different equations for the bounds seperated

Question 29

When finding the max.min area under the curve

Answer 29

alongside your dy/dx = 0, check start and end points/bounds and max/mins of curve

Question 30

Max/min of a function e.g. cubic

Answer 30

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