Introduction to Integration Shirleen Stibbe
www.shirleenstibbe.co.uk
OPEN UNIVERSITY
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M203 Pure Mathematics Summerschool
How to find the area under a graph Chop it into chunks – sum areas of rectangles
!
a 0
f(x) dx
Overestimate
Underestimate
Partition, P: Divides [a, b] into sub-intervals x0 = a < x1 < x2 < … < xn = b P = { [x0, x1], [x1, x2], … [xn-1, xn] } ith interval: [xi-1, xi], length: δxi = xi – xi-1 Standard partition for [0, 1] :
xi = i/n, δxi = 1/n
Pn = { [0, 1/n], [1/n, 2/n], …, [(n-1)/n, 1] }, Mesh P = ||P|| = max {δxi} mi
δxi
f(x)
On the ith interval: mi = inf f(x), Mi = sup f(x)
Mi
Areas: = miδxi
a
xi-1
xi
b
= Miδxi
n
Lower Riemann Sum : L(f, P) = ∑ mi δxi i=1 n
Upper Riemann Sum : U(f, P) = ∑ Mi δxi i=1
2
Add extra points: L(f, P) increases, U(f, P) decreases Suppose ||P|| → 0, and let
L = lim L(f, Pn ) n →∞
U = lim U(f, Pn ) n →∞
If L = U, then f is integrable on [a, b] and b
∫a f(x)dx = L = U General result: If f is monotonic (or continuous) on [a, b] then f is integrable on [a, b] To show f is integrable on [a, b]: Find ONE sequence {Pn} with ||Pn|| → 0, for which L = U To show f is not integrable on [a, b]: Find ONE sequence {Pn} with ||Pn|| → 0, for which L ≠ U 3
Stuff you'll need …
You must remember this n
n (n + 1) 1 + 2 + ... + n = ∑ i = 2 i=1 And by the way: n
∑i
n
=
i=1
n
∑ λi i=1
∑i
i= 0
n
= λ∑ i, i=1
n
∑ (i − 1) i=1
λ ≠i
n −1
=
∑i
i= 0
n −1
=
∑i i=1
⎛ n ⎞ n (n + 1) n2 + n − 2 ∑ i = ⎜⎜ ∑ i ⎟⎟ − 1 = 2 − 1 = 2 ⎝ i=1 ⎠ i=2 n
4
Example: Let f(x) =
x, 0 ≤ x < 1 0, x = 1
Find L(f, P) and U(f, P). Is f integrable on [0, 1]?
Standard partition for [0, 1] : Pn = { [0, 1/n], [1/n, 2/n], …, [(n-1)/n, 1] }, δxi = 1/n 1
Mi
Mn
mi
f increasing on [0, 1[, so mi = f((i-1)/n) = (i-1)/n, i ≤ n-1 Mi = f (i/n) = i/n, i ≤ n-1
mn i−1 n
i n
and mn = 0, Mn = 1 = n/n
1
n −1
n −1
i−1 1 1 n −2 L( f, Pn ) = ∑ mi δxi = ∑ = 2 ∑i n n i= 0 i=1 i=1 n 1 (n − 2)(n − 1) n2 − 3n + 2 = 2 = 2 n 2n2 n
n
i 1 1 U( f, Pn ) = ∑ Mi δxi = ∑ = 2 n i=1 i=1 n n
n
∑i i=1
1 n(n + 1) n2 + n = 2 = 2 n 2n2 L(f, P) =
1 3 2 − + 2 → 1/2 2 2n 2n
1 1 U(f, P) = + → 1/2 2 2n
as n → ∞
as n → ∞
L = U, so f is integrable on [0, 1] 1
∫0 f
= 1/2 5
Fundamental theorem of integration If f is integrable on [a, b], and F is a primitive for f on [a, b], i.e. F (x) = f(x), then b
∫a f(x) dx
= F(b) − F( a)
[ ∫f = F + c, c constant ]
Combination rules If f, g have primitives F, G respectively Sum rule:
∫ (f + g)
=F+G +c
Multiple rule:
∫ λf = λF + c
Scaling rule:
∫ f(λx)dx
= (1/ λ) F(λx) + c
Integrals you really really need to know
∫ xa
= ( 1 a +1 ) xa+1 , a ≠ -1
∫ x-1
= logex
∫ ex
= ex
∫ cos x
= sin x
Table of Standard Primitives HB p 95
∫ sin x = - cos x 6
Inequalities If f and g are integrable on [a, b], and x ∈ [a, b]:
1)
b
f(x) ≤ g(x) ⇒
2)
b
∫a f ≤ ∫a g b
m ≤ f(x) ≤ M ⇒ m(b − a) ≤ ∫ f ≤ M(b − a) a
Example: Prove that
Solution:
1
3 2
≤∫
For 0 ≤ x ≤ 1,
x2 (1 + x)
1
2
1 ≤ (1 + x)
⇒
dx ≤ 1
2
1 3
≤ 2
1 1 ≤ ≤1 1 2 (1 + x) 2
x2 x2 2 ≤ ≤ x ⇒ 2 (1 + x) 12 Then
Since
1 1 2 x dx ≤ ∫ 0 2
x2
∫0 (1 + x) 12
dx ≤
1
∫0
x2dx
1
1 3 ⎤ 1 ⎡ ∫0 x dx = ⎢⎣3 x ⎥⎦ 0 = 3 it follows that 1
2
1
3 2
Note:
1
0.2357
≤∫
≤
x2 (1 + x)
1
2
dx ≤
0.2533 ≤
1 3
0.3333
7
Differentiation: Product rule (fg)' = f'g + fg' ⇒ f'g = (fg)' – fg' Integration by parts: Reduction formula:
∫f'g
= ∫(fg)' - ∫fg' = fg - ∫fg'
Let I n =
!
b a
fn (x) dx, n " 0
Find a relationship between In and In-1, and use it to calculate individual values
1
x n Example: Let In = ∫ e x dx, n ≥ 0 0
Prove that In = e − nIn −1 and evaluate I0 and I2
Solution: where So Then
1
1
x n
In = ∫ e x dx = ∫ f'(x) g(x)dx 0
0
f'(x) = ex, f(x) = ex, g(x) = xn, g'(x) = nxn-1
In = e x xn
[
I0 =
1
∫0 e
]10 − n∫01 exxn−1dx
x
dx =
x 1 e 0
[ ]
= e − nIn −1
= e −1
I2 = e − 2I1 = e − 2(e − I0 ) = e − 2(e − (e − 1) ) = e − 2
(*)
(**) using (*) twice from (**) 8
Twiddles:
f, g positive functions, domain R f(n) ∼ g(n) as n → ∞ means
f(n) → 1 as n → ∞ g(n) n!
Stirling's formula:
∼ 2πn (n / e)n
2n λ 2 2 n Example: Find a real λ for which ⎛⎜ ⎞⎟ ∼ ⎝ n ⎠ n
Solution:
(
2n n
)
= = = ∼
=
(2n) ! n!(2n ! n) ! (2n) ! n!n! 2"2n (2n / e)2n 2"n (n / e)n 2"n (n / e)n 1
(2n)2n e n e n e2n
"n =
1 "n
2n
2
nn nn so # =
1 " 9