MATH 101 - Section 210

Instructor: Avner Segal (avners@math.ubc.ca)

February 9th 2017

Common course page:http://www.math.ubc.ca/~gerg/teaching/101-Winter2017/

Individual section page:http://www.math.ubc.ca/~avners/courses/math101-2017.html

Office hours: Thursdays 12:00-14:30 (LSK300)On Quiz Weeks: Tuesdays 12:00-13:00 (LSK300) - by appointment

only!Follow announcements on section web page in case of changes.

Quiz # 3 on Thursday(February 16th)

Material covers everything up to (and including) numericalintegration.

Don’t forget your UBC Card and a black pen!

Numerical Integration

Fix f (x) integrable on [a,b] and an integer n.

Trapezoidal Rule b−a2n (1,2,2,2, ...,2,2,1)

Tn = b−a2n

(f (a)+2f (a+∆x)+2f (a+2∆x)+·· ·+2f (a+ (n−1)∆x)+ f (b)

).

Simpson’s Rule b−a3n (1,4,2,4,2, ...,4,2,4,1)

Here n is even.

Sn = b−a3n

(f (a)+4f (a+∆x)+2f (a+2∆x)+4f (a+3∆x)+2f (a+4∆x)

+·· ·+4f (a+ (n−3)∆x)+2f (a+ (n−2)∆x)+4f (a+ (n−1)∆x)+ f (b))

Error BoundsDenote I =

∫ b

af (x)dx . Assume that |f ′(x)| ≤K for all a≤ x ≤ b,

|f ′′(x)| ≤M for all a≤ x ≤ b and that |f (4)(x)| ≤ L for all a≤ x ≤ b.

Method # points Error Bound

Left endpoint n |I −Ln| ≤ K2(b−a)2

n

Midpoint n |I −Mn| ≤ M24

(b−a)3

n2

Right endpoint n |I −Rn| ≤ K2(b−a)2

n

Trapezoidal rule n+1 |I −Tn| ≤ M12

(b−a)3

n2

Simpson’s rule n+1, n is even |I −Sn| ≤ L180

(b−a)5

n4

Question (Final 2012)

Let I =∫ 2

1

dxx.

(a) Write down Simpson’s rule approximation for I using 5 points,call it S4.

(b) Without computing I , find an upper bound for |I −S4|.

Solution

(a) We have f (x)= 1x , a= 1, b = 2 and n= 4. In particular, we have

S4 = 2−13 ·4

(f (1)+4f

(54

)+2f

(32

)+4f

(74

)+ f (2)

)

= 12

(1+ 16

5+ 43+ 16

7+ 12

).

Question (Final 2012)

Let I =∫ 2

1

dxx.

(a) Write down Simpson’s rule approximation for I using 5 points,call it S4.

(b) Without computing I , find an upper bound for |I −S4|.

Solution

(b) We first need to derive f (x)= 1x 4 times.

f ′(x)=− 1x2 , f ′′(x)= 2

x3 , f (3)(x)=− 6x4 , f (4)(x)= 24

x5 .

On the interval [1,2] the function∣∣∣f (4)(x)∣∣∣= 24

x5 is decreasingand hence it is bounded by L= f (1)= 24. Using the errorbound formula we get

|I −S4| ≤ 24180

· (2−1)5

44 = 11960

.

Important FactsRecall that

Ï |cos(x)|, |sin(x)| ≤ 1.Ï |a+b| ≤ |a|+ |b| (Called triangle inequality).Ï |a ·b| = |a| · |b|.

Question

Let I =∫ 6

4sin

(px)dx . Find n such that the estimate of I using the

midpoint rule will have an error of at most 13,000 .

SolutionÏ f ′(x)= 1

2p

x cos(p

x), f ′′(x)=− 14x3/2 cos(

px)− 1

4x sin(p

x).

Ï |f ′′(x)| ≤∣∣∣ 14x3/2 cos(

px)

∣∣∣+ ∣∣ 14x sin(

px)

∣∣ ≤∣∣∣ 14x3/2

∣∣∣ · |cos(px)|+ ∣∣ 14x

∣∣ · |sin(px)| ≤ 14x3/2 + 1

4x ≤ 14·43/2 + 1

16 = 110

Ï |I −Mn| ≤ 1/1024 · (6−4)3

n2 = 130·n2 .

Ï We want n such that 130·n2 ≤ 1

3,000 . This is the same asn2 ≥ 100 so n= 10 will do.

Question (Final 2008, enhanced)

Let∫ 1

0cos(x2)dx . Find the smallest positive integer n you can such

that the Simpson’s Rule approximation Sn for I has error less thenor equal to 0.001.

SolutionFirst, we derive f (x)= cos(x2):f ′(x)=−2x sin(x2),f ′′(x)=−2sin(x2)−4x2 cos(x2)f (3)(x)=−12x cos(x2)+8x3 sin(x2),f (4)(x)= (16x4−12)cos(x2)+48x2 sin(x2).Hence|f (4)(x)≤ |(16x4−12)cos(x2)|+ |48x2 sin(x2)| ≤|16x4−12|+48|x2| ≤ 4+48= 52 .In the original problem, it was given that 60 is a bound.

Question (Final 2008)

Let∫ 1

0cos(x2)dx . It can be shown that the 4th derivative of

cos(x2) has absolute value at most 60 on the interval [0.1]. Usingthis bound, find the smallest positive integer n you can such thatthe Simpson’s Rule approximation Sn for I has error less then orequal to 0.001.

SolutionIt is known that the error in the n-th is bounded by

En = 60180

(1−0)5

n4 = 13 ·n4 .

We want to find the smallest value of n so that 13·n4 ≤ 0.001, or

otherwise the smallest n so that n4 ≥ 1,0003 . One checks that

54 = 635> 10003 and 44 = 256< 1000

3 so n= 5.

Question (Final 2009, rephrased)

Give the Simpson’s rule approximation S4 to∫ 2

0sin(ex)dx .

Variant: Give an upper bound for the error in this approximation.

SolutionÏ f (x)= sin(ex), a= 0, b = 2, n= 4.

Ï S4 = 2−03·4

(sin(e0)+4sin

(e

12

)+2sin(e1)+4sin

(e

32

)+ sin(e2)

)= 1

6

(sin(e)+4sin(

pe)+2sin(e)+4sin

(pe3

)+ sin(e2)

)≈ 0.37 .

Ï f ′(x)= ex cos(ex), f ′′(x)= ex cos(ex)−e2x sin(ex),f ′′′(x)= ex cos(ex)−3e2x sin(ex)−e3x cos(ex),f (4)(x)= ex cos(ex)−7e2x cos(ex)−6e3x cos(ex)+e4x sin(ex).

Ï |f (4)(x)| ≤ |ex |+7|e2x |+6|e3x |+|e4x | ≤ e2+7e4+6e6+e8 ≤ 6,000.

Ï |I −S4| ≤ 6,000180 · (2−0)5

44 ≈ 4.1.

Ï More careful analysis will show that |f (4)(x)| ≤ 1300 whichimproves the bound to 0.89. Which is still larger than theapproximation. Actually, I ≈ 0.55.

Improper IntegralsWe wish to be able to talk about integrals of the form∫ 1

0

dxx

,∫ ∞

0

dx1+x2 ,

∫ 1

−1

dxx2

An integral having either an infinite limit of integration or anunbounded integrand is called an improper integral.

The idea: To approximate it by proper definite integrals.

Example

Determine∫ ∞

1

dxx2 ?

Ï For R > 1 we have∫ R

1

dxx2 =

[−1

x

]R

1=− 1

R+1.

Ï Taking the limit R →∞ yields∫ ∞

1

dxx2 = lim

R→∞

∫ R

1

dxx2 = 1.

We say that this improper integral converges.

Example

Determine∫ 1

0

dxpx?

Ï For T > 0 we have∫ 1

T

dxpx= [

2p

x]1T = 2

p1−2

pT .

Ï Taking the limit T → 0+ yields∫ 1

0

dxpx= lim

T→0+

∫ 1

T

dxpx= 2.

We say that this improper integral converges.

Example

Determine∫ 1

0

dxx?

Ï For T > 0 we have∫ 1

T

dxx

= [logx ]1T = log1− logT .

Ï Taking the limit T → 0+ yields∫ 1

0

dxx

= limT→0+

∫ 1

T

dxx

=∞.

We say that this improper integral diverges.

Improper Integral with Infinite Domain of IntegrationÏ If the integral

∫ R

af (x)dx exists for all R > a, then∫ ∞

af (x)dx = lim

R→∞

∫ R

af (x)dx

when the limit exists (and is finite).

Ï If the integral∫ b

rf (x)dx exists for all r < b, then∫ b

−∞f (x)dx = lim

r→−∞

∫ b

rf (x)dx

when the limit exists (and is finite).

Ï If the integral∫ R

rf (x)dx exists for all R > r , then∫ ∞

−∞f (x)dx = lim

r→−∞

∫ c

rf (x)dx + lim

R→∞

∫ R

cf (x)dx

when both limits exist (and are finite). Any c can be used.In this cases we say that the improper integral is convergent,otherwise it is divergent.

Improper Integral with Unbounded IntegrandÏ If the integral

∫ b

tf (x)dx exists for all a< t < b, then∫ b

af (x)dx = lim

t→a+

∫ b

tf (x)dx

when the limit exists (and is finite).

Ï If the integral∫ T

af (x)dx exists for all a<T < b, then∫ b

af (x)dx = lim

T→b−

∫ R

af (x)dx

when the limit exists (and is finite).

Ï Let a< c < b. If the integrals∫ T

af (x)dx and

∫ b

tf (x)dx exist

for all a<T < c and c < t < b, then∫ b

af (x)dx = lim

T→c−

∫ T

af (x)dx + lim

t→c+

∫ b

tf (x)dx

when both limits exist (and are finite).In this cases we say that the improper integral is convergent,otherwise it is divergent.

WarningConsider the integral

∫ ∞

−∞xdx . Since x is an odd function, for any

R one has∫ R

−Rxdx = 0.

It might be tempting to say that∫ ∞

∞xdx = 0 but always remember

that the definition required that we look at∫ 0

rxdx and

∫ R

0xdx

separately.

Since both∫ 0

−∞xdx and

∫ ∞

0xdx diverge then so does

∫ ∞

−∞xdx .

Basic Example (I)

Consider the improper integral∫ 1

0

dxxp for some p. For which values

of p will it converge and for which it will diverge?

SolutionÏ We first recall that∫ dx

xp ={ 1

(1−p)xp−1 +C , p 6= 1

log |x |+C , p = 1

Ï For p = 1, we have∫ 1

T

dxx

= [log |x |∣∣1T = log(1)− log(T )=− log(T ).

Ï Taking the limit T → 0+ we get

limT→0+

dxx

=− limT→0+ log(T )=∞,

so the improper integral is divergent.

Example Continued

Ï For p 6= 1 we have∫ 1

T

dxxp =

[1

(1−p)xp−1

∣∣∣∣1T

= 11−p

− 1(1−p)T p−1 .

Ï Taking the limit T → 0+ we get

limT→0+

∫ 1

T

dxxp = lim

T→0+

(1

1−p− 1(1−p)T p−1

)=

{∞, p > 1

11−p , p < 1

.

Ï Conclusion, the improper integral∫ 1

0

dxxp will converge if and

only if p < 1.Ï Note that for p ≤ 0 this is, in fact, a proper integral.

Basic Example (II)

Consider the improper integral∫ ∞

1

dxxp for some p. For which

values of p will it converge and for which it will diverge?

SolutionÏ Recall, again, that

∫ dxxp =

{ 1(1−p)xp−1 +C , p 6= 1

log |x |+C , p = 1

Ï For p = 1, we have∫ R

1

dxx

= [log |x |∣∣R1 = log(R)− log(1)= log(R).

Ï Taking the limit R →∞ we get

limR→∞

dxx

= limR→∞

log(R)=∞,

so the improper integral is divergent.

Example Continued

Ï For p 6= 1 we have∫ R

1

dxxp =

[1

(1−p)xp−1

∣∣∣∣R1= 1(1−p)Rp−1 − 1

1−p.

Ï Taking the limit R →∞ we get

limR→∞

∫ 1

R

dxxp = lim

R→∞

(1

(1−p)Rp−1 − 11−p

)=

{∞, p < 1

1p−1 , p > 1

.

Ï Conclusion, the improper integral∫ ∞

1

dxxp will converge if and

only if p > 1.

Question (Final 2010)

Evaluate∫ −1

−∞e2xdx . Simplify your answer as much as possible.

Solution

Ï∫ −1

re2xdx =

[e2x

2

∣∣∣∣−1

r= e−2−er

2.

Ï Taking the limit r →−∞ yields

∫ −1

−∞e2xdx = lim

r→−∞e−2−er

2= e−2

2.

QuestionFind a constant C such that

∫ ∞

−∞Cdx1+x2 = 1.

Solution

Ï Recall that∫ dx

1+x2 = arctan(x)+C .

Ï∫ R

0

Cdx1+x2 =C arctan(R)−C arctan(0)=C arctan(R) −→

R→∞Cπ2

.

Ï Similarly, limr→−∞

∫ 0

r

Cdx1+x2 = Cπ

2.

Ï It follows that∫ ∞

−∞Cdx1+x2 =Cπ.

Ï We need to take C = 1π

.

Question (Final 2009)

For what values of p does∫ ∞

e

dxx (logx)p

converge?

Ï For R > e we make the substitution u(x)= log(x),∫ R

e

dxx (logx)p

=∫ log(R)

1

duup

Since limR→∞ log(R)=∞ the left hand side converge if and

only if∫ ∞

1

duup converges, which happens if and only if p > 1.

Question (Math 103 Final, 2013)Evaluate the integral if it exists, otherwise show that it doesn’t:

I =∫ 2

0

dx1−x2 .

SolutionsÏ The function 1

1−x2 is discontinuous at x = 1.Ï Recall that

11−x2 = 1

2

(1

1−x+ 11+x

).

Since 11+x is continuous on [0,2] it is integrable there so we

start by checking whether∫ 2

0

dx1−x

is convergent or divergent.

Ï In order for∫ 2

0

dx1−x

to be convergent, both∫ 1

0

dx1−x

and∫ 2

1

dx1−x

needs to be convergent.

Solutions ContinuedWe consider

∫ 1

0

dx1−x

and make the substitution u = 1−x . We have

∫ 1

0

dx1−x

=∫ 1

0

duu

which is divergent. Hence, so does I =∫ 2

0

dx1−x2 .

Convergence Tests for ImproperIntegrals

Sometimes, we only want to check whether an integral isconvergent or not.

Comparison Test

Let a be a real number. Let f and g be functions that are definedand continuous for all x ≥ a and assume that g(x)≥ for all x ≥ 0.

1. If |f (x)| ≤ g(x) for all x ≥ a and if∫ ∞

ag(x)dx converges then∫ ∞

af (x)dx also converges.

2. If f (x)≥ g(x) for all x ≥ a and if∫ ∞

ag(x)dx diverges then so

does∫ ∞

af (x)dx .

Similarly for a bounded domain with unbounded integrand.

Example

Determine whether the improper integral∫ ∞

1

dxx2+1

is convergentor not.

SolutionÏ We note that for any x we have x2+1> x2 and hence

1x2+1 < 1

x2 .

Ï Recall that∫ ∞

1

dxx2+1

is convergent.

Ï By the comparison test,∫ ∞

1

dxx2+1

is convergent.

Example

Determine whether the improper integral∫ ∞

2

dxx2−1

is convergentor not.

SolutionÏ It is not true that 1

x2−1 < 1x2 so the simple minded comparison

will not help here.Ï Warning: The fact that 1

x2−1 > 1x2 doesn’t men that the

integral is divergent. More generally, being smaller than adivergent integral doesn’t mean anything and being biggerthan a convergent integral doesn’t mean anything (a dogwhich is tamer than a wild dog can still bite and a dog whichis wilder than a tamed dog doesn’t have to bite).

Ï Heuristically, 1x2−1 should behave similarly to 1

x2 so the integralshould converge.

Example

Determine whether the improper integral∫ ∞

2

dxx2−1

is convergentor not.

SolutionÏ We still have x2−1> x2−2x +1= (x −1)2. Indeed

x2−1> x2−2x +1⇔ 2x > 2⇔ x > 1.

Ï By the comparison test, if we’ll show that∫ ∞

2

dx(x −1)2

is

convergent it will follow that so does∫ ∞

2

dxx2−1

.

Ï Indeed, ∫ R

2

dx(x −1)2

=[− 1

x −1

]R

2= 1− 1

R −1−→

R→∞1

and hence∫ ∞

2

dx(x −1)2

is convergent.

ExampleDetermine whether the improper integral

∫ ∞

1

1+ sinxx2 dx is

convergent or not.

SolutionÏ Note that

∣∣∣1+sinxx2

∣∣∣≤ 2x2 for x ≥ 1.

Ï∫ ∞

1

dxx2 converges.

Ï By the comparison test, so is∫ ∞

1

1+ sinxx2 dx .