Fundamental Engineering Maths 5

8/9/2019 Fundamental Engineering Maths 5

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PHAM

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Outline

1 Area problems

2 The fundamental theorems of calculus

3 Indefinite integrals and the net change theorem

4 The substitution rule

5 Integration by parts

6 Additional techniques of integration

7 Approximate integration

8 Improper integrals

Duong T. PHAM - EEIT2014 November 7, 2014 2 / 77


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Area problems

x

y

y = f(x)

a b

S?

x1 x2 x3 xi1 xi xi+1 xn1

S1 S2 S3 Si Sn

x1 x2 x3 xi1 xi xi+1 xn1

x

x1 x

2 x

3 x

i x

n

Rn =f(x1 ) x+f(x

2 ) x+ +f(xi ) x+ +f(xn ) x

S= limn

Rn



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Definite integrals

Def: Let f : [a, b]R.

Divide [a, b] byx0, x1, . . . , xn into n equal subintervals of width (b a)/n(a= x0


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Definite Integrals

Ex: Evaluate the Riemann sum for f(x) =x3

6xtaking the sample points to

be the right endpoints with a= 0, b= 3 and n= 6

Evaluate3

0(x3 6x) dx

Ans:

(a) For n= 6, the interval width isx= ban = 306 = 12 =0.5, and the rightendpoints are x1 = 0.5, x2 = 1, x3 = 1.5, x4 = 2, x5 = 2.5, x6 = 3.

The Riemann sum is

R6 =

6

i=1

f(xi) x= xf(0.5) +f(1) +f(1.5) +f(2) +f(2.5) +f(3)=

1

2(2.875 5 5.625 4 + 0.625 + 9 )= 3.9375



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Definite integrals

Ex: Evaluate3

0(x3 6x) dx

Ans: 0 3x1

3n

x2

23n

xn1

(n1)3n

With n subintervals, x= 3n

, and

x0 = 0, x1= 3

n, x2 = 2

3

n, . . . , xi=i

3

n, . . . , xn = 3.

30

(x3 6x) dx= limnn

i=1f(xi) x= limn

ni=1

f3i

n3

n

= limn

3

n

ni=1

3i

n

3 6

3i

n

= lim

n3

n

ni=1

27i3

n3 18i

n

= limn

81n4

ni=1

i3 54n2

ni=1

i

= limn

81n4

n(n+ 2)

2

2 54n2

n(n+ 1)2

= limn

81

4

1 +

1

n

2 27

1 +

1

n

=

81

4 27= 27

4



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The Midpoint Rule

Midpoint Rule:

ba

f(x) dxn

i=1

f(xi) x= x[f(x1) +f(x2) + +f(xn)]

where x= ban

and xi= xi1+xi

2 = midpoint of [xi1, xi].

Ex: Use Midpoint Rule with n= 5 to approximate 2

1

1

xdx

Ans: x= (2 1)/5 = 0.2. The endpoints of the subintervals arex0 = 1, x1 = 1.2, x2 = 1.4, x3 = 1.6, x4 = 1.8 and x5 = 2.

The midpoints are x1 = 1.1, x2 = 1.3, x3 = 1.5, x4 = 1.7, x5 = 1.9.

Applying the Midpoint Rule, 21

1

x dx 0.2[f(1.1) +f(1.3) +f(1.5) +f(1.7) +f(1.9)]

= 0.2 [ 1

1.1+

1

1.3+

1

1.5+

1

1.7+

1

1.9]

0.691908.



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Comparison Properties

Comparison properties:

1 Iff(x) 0 for a x b, thenb

a f(x) dx 0,

2 Iff(x)

g(x) for a

x

b, then ba

f(x) dx

b

a g(x) dx,

3 Ifm f(x) M fora x b, then

m(b a) ba

f(x) dx M(b a).



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Exercises

5.2:

14, 912 2122, 2728, 3540, 5254



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The fundamental theorems of calculus

The fundamental theorem of calculus, part I:Iffis continuous on [a, b], thenthe function gdefined by

g(x) =

xa

f(x) dx, a x b

is continuous on [a, b] and differentiable on (a, b) and g(x) =f(x).

Proof: Let x and x+h in (a, b) (Suppose h> 0) . Then

g(x+h) g(x) = x+ha

f(x) dx xa

f(x) dx=

x+hx

f(x) dx.

So, g(x+h) g(x)h

= 1h

x+hx

f(x) dx. Since f is continuous on [x, x+h], by

the extreme value theorem, there are u, v [x, x+h] such that

f(u) = min{f(y) :y [x, x+h]} and f(v) = max{f(y) :y [x, x+h]}



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(continuing with the proof) We have m

f(y)

M forx

y

x+h. Hence,

f(u)h x+hx

f(x) dx f(v)h f(u) 1h

x+hx

f(x) dx f(v)

f(u) g(x+h) g(x)h

f(v) ()

When h 0, since u, v [x, x+h], we have u x, v x. Note that f iscontinuous on [a, b], thus f(u) f(x) and f(v) f(x). This together with()yields

limh0

g(x+h) g(x)h

=f(x).

This means that g(x) is differentiable (and then continuous) at x (a, b) andg(x) =f(x).The cases x=a and x=bcan be proved in the same manner, using one-sidedlimits.



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The fundamental theorem of calculus, part II: If f is continuous on [a, b],

then ba

f(x) dx=F(b) F(a)

where F is any antiderivative off.

Proof: Denote g(x) =xa f(x) dx. By part 1, g(x) =f(x). It means that g is

another antiderivative off. Thus g(x) =F(x) +C for some constant C. Then

F(b) F(a) = [g(b) C] [g(a) C] =g(b) g(a)

= ba

f(

x)

dx

a

a

f(

x)

dx

=

ba

f(x) dx



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Ex: Evaluate 3

1ex

dx

Ans: The function f(x) =ex is continuous everywhere and itsantiderivative is F(x) =ex. Thus

3

1exdx=e3 e1

Remark: People usually the notation

F(x)ba

=F(b) F(a)


Th f d l h f l l


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Ex: Find the area under the parabola y=x2 from 0 to 1

Ans:

y

0 x1

A

The area

A=

10

x2 dx= x3

3

10

=1

3



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I d fi it i t l


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Indefinite integrals

Notation: we denote

f(x) dx=F(x) to indicate that F =f.

Ex: We can write x2 dx= x3

3

+C as d

dxx

3

3

+C =x2

Ex:

sec2 x dx= ? tan x+C

because

d

dx(tan x+C) = sec

2

x


I d fi it I t l


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Indefinite Integrals

Indefinite integral table:

(i) cf(x) dx=c f(x) dx(ii)

[f(x) +g(x)] dx=

f(x) dx+

g(x) dx

(iii)

k dx=kx+c

(iv) xn dx= xn+1

n+1+c (n = 1)

(v)

1x

dx= ln |x| +c(vi)

ex dx=ex +c

(vii)

ax dx= e

x

ln a+c

(viii)

sin x dx= cos x+c(ix)

cos x dx= sin x+c

(x)

sec2 x dx= tan x+c (xi)

csc2 x dx= cot x+c(xii) 11+x2 dx= tan1 x+c (xiii) 11x2 dx= sin

1 x+c




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Ex: Evaluate 3

1(x3 6x) dx

Ans:

3

1

(x3 6x) dx= x44 3x2

31

=

81

4 27

1

4 3

= 6.75

Ex: Evaluate 9

1

2t2 +t2

t 1t2

dt

Ans: 91

2t2 +t2

t 1t2

dt=

91

2 +

t t2

dt

= 2t+2

3 t3/2

+t1

9

1

= 324

9


The net change theorem


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The net change theorem

Let y=F(x). Then F represents the rate of change ofy=F(x) w.r.t. x.

F(b) F(a)is the change in y when xchanges from a to b.

The net change theorem: The integral of a rate of change is the net change:

ba

F(x) dx=F(b) F(a)


Some applications


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Some applications

IfV(t)is the volume of water in a reservoir at time t, then its derivativeV(t)is the rate at which water flows into the reservoir at time t. So t2

t1

V(t) dt=V(t2) V(t1)

is the change in amount of water in the reservoir between t1 and t2.

If[C](t)is the concentration of the product of a chemical reaction at timet, then the rate of reaction is the derivative d[C]/dt. So

t2t1

d[C]

dt dt= [C](t2) [C](t1)is the change in the concentration of[C]from time t1 to time t2.


Some applications


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Some applications

If the mass of a rod measured from the left end to a point x is m(x), then

the linear density (x) =m(x). So ba

(x) dx=m(b) m(a)

is the mass of the segment of the rod that lies between x=a and x=b.

If the rate of growth of a population is dn/dt, then

t2

t1

dn

dt dt=n(t2) n(t1)

is the net change in population during the time period from t1 to t2. (Thepopulation increases when births happen and decreases when deaths occur.The net change takes into account both births and deaths.)


Exercises


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Exercises

5.4: 110, 2130, 48, 4952


The substitution rule


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The substitution rule: Ifu=g(x)is a differentiable function whose range is aninterval I and f is continuous on I, then

f(g(x)) g(x) dx=

f(u)du

Proof: Suppose that F is an antiderivative off, i.e., F=f. Then the chain rule

ddx

[F(g(x))] =F(g(x)) g(x).

This implies

F(g(x)) g(x) dx=F(g(x)) +C.

Using u=g(x), we obtain F(g(x)) g(x) dx=F(u) +C=

F(u) du=

f(u) du




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Ex: Find x3 cos(x4 + 2) dxAns: We use a change of variables u=x4 + 2. Then du= 4x3 dx. Thusx3 dx= du4 . Using the change of variables, we obtain

x

3

cos(x

4

+ 2) dx=

cos u

du

4

=1

4

cos u du

=1

4

sin u+C

=1

4sin(x4 + 2) +C.




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Ex: Find 2x+ 1 dxAns: We use a change of variables u= 2x+ 1. Then du= 2 dx. Thusdx= du2 . Using the change of variables, we obtain

2

x+ 1

dx=udu

2

=1

2

u du

=1

2

2

3

u32 +C

=1

3(2x+ 1)

32 +C.


The substitution rule for definite integrals


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The substitution rule for definite integrals

The substitution rule for definite integrals: Ifg is continuous on [a, b] andf is continuous on the range ofu=g(x), then

ba

f(g(x)) g(x) dx=

g(b)g(a)

f(u) du

Proof: Let Fbe an antiderivative off. Then

d

dxF(g(x)) =F(g(x)) g(x) =f(g(x)) g(x).

Fundamental Theorem for Calculus (part 2) gives

b

a

f(g(x)) g(x) dx=F(g(b)) F(g(a)).

Since F=f, using Fund. Theo. for Cal. (II) again, we have g(b)g(a)

f(u) du= F(u)|g(b)g(a) =F(g(b)) F(g(a)).

Comparing the two above equalities yielding the conclusion.Duong T. PHAM - EEIT2014 November 7, 2014 30 / 77


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Integrals of symmetric functions


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g y

Integrals of symmetric functions: Let fbe a continuous funtion on[a, a].1 Iff is even (f(

x) =f(x)), then aaf(x) dx= 2 a0 f(x) dx

2 Iff is odd (f(x) = f(x)), then aaf(x) dx= 0.Proof:

1 (f is even). We have aaf(x) dx= 0

a

f(x) dx+ a0 f(x) dx.Denote u(x) = x. Then du= dx and u(a) =a, u(0) = 0.Noting that f(u) =f(u) and using substitution rule, we obtain

0

a

f(x) dx= 0

a

f(u)(du) = 0

a

f(u)du= 0

a

f(u)du

=

a0

f(u)du=

a0

f(x)dx.

Hence,

a

af(x)dx= 2

a

0 f(x)dx


Exercises


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5.5: 115, 5160, 7374, 77, 86.


Integration by parts


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g y p

The product gives

ddx[f(x)g(x)] =f(x) g(x) +f(x)g(x).

Then

[f(x) g(x) +f(x)g(x)] dx=f(x)g(x)

Integration by parts: f(x)g(x) dx=f(x)g(x)

f(x)g(x) dx

Remark: Another form of the integration by parts is u dv=uv

v du




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g y p

Ex: Evaluate xsin x dxAns: Let u(x) =x and v(x) = sin x. Then u(x) = 1 and v(x) = cos x.Hence,

xsin x dx=u(x)v(x) v(x)u(x) dx

=x( cos x)

( cos x) dx

=

xcos x+ cos x dx= xcos x+ sin x+C.




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Ex: Evaluate

ln x dx

Ans: Denote u= ln x and dv=dx. Then du= dx

x and v=x.

Integration by parts gives ln x dx=xln x

x

dx

x

=xln x

dx

=xln x x+C




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Ex: Evaluate

t2etdt

Ans: Denote u=t2 and dv=et

dt. Then du= 2t dt and v=et

.Integration by parts gives

t2etdt=t2et

2tetdt=t2et 2

tetdt.

Denote u=t and dv=etdt. Then du=dt and v=et. Applyingintegration by parts again yields

tetdt=tet etdt=tet et +C.Hence,

t2etdt=t2et 2(tet et +C) =t2et 2tet + 2et +C1




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Integration by parts:

b

a

u dv=uvba

b

a

v du

Ex: Evaluate

/20

ex sin x dx

Ans: Denote u=ex and dv= sin x. Then du=ex dx and v=

cos x. We have /2

0

ex sin x dx= ex cos x/2

0 /2

0

ex( cos x)dx= 1 + /2

0

ex cos x dx

Denote u=ex and dv= cos x. Then du=ex dx and v= sin x. Thus

/2

0

ex cos x dx=ex sin x/20

/2

0

ex sin x dx=e/2

/2

0

ex sin x dx

Hence,/2

0 ex sin x dx= 1 +e/2 /2

0 ex sin x dx. This implies

/2

0

ex sin x dx=1 +e/2

2


Exercises


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7.1: 120, 3338, 4445,


sinm x cos2k+1 x dx


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Ex: Evaluate

sin2 x cos3 x dx

Ans: Noting thatcos2 x= 1

sin2 x, sin2 xcos3 xdx=

sin2 xcos2 x cos xdx=

sin2 x(1 sin2 x) cos xdx.

Denote t= sin x. Then dt=cos xdx and sin2 xcos3 xdx=

t2(1 t2) dt=

(t2 t4)dt= t3

3 + t

5

5 +C

=sin3 x

3 +

sin5 x

5 +C

sinmxcos2k+1x dx=

sinmxcos2kxcos xdx=

sinmx(1 sin2x)k cos xdx

=

tm(1 t2)k dt (by substitution t= sin x)



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sin2m x cos2n x dx


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Ex: Evaluate sin2 x dxAns: Applying the following identity

sin2 x=1 cos2x

2

to obtain sin2x dx=

1

2

(1 cos2x)dx= 1

2

xsin2x

2 +C

sin2mxcos2nx dx=

1 cos2x

2

m 1 + cos 2x2

ndx


tanm x sec2k x dx


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Ex: Evaluate tan3 x sec4 x dxAns: Note that 1 + tan2 x=sec2x and dtan x

dx = sec2 x. We have

tan3 x sec4 x dx=

tan3 x(1 + tan2 x)sec2 x dx=

t3(1 +t2)dt

= t4

4 + t

6

6 +C=tan

4 x4

+tan6 x

6 +C

tanmx sec2kx dx= tanmx sec2k2xsec2xdx= tanmx(1+ tan2x)k1 sec2xdx=

tm(1 +t2)k1 dt (denote t= tan x)


tan2k+1 x secn x dx


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Ex: Evaluate tan3 x sec3 x dx

Ans: Denote u= sec x. Then du= sec x tan x dx. tan3x sec3x dx=

tan2xsec2x secxtanxdx=

(1 sec2x)sec2x secxtanx dx

= (1 u2)u2 du= u3

3 u5

5 +C=

sec3 x

3 sec5 x

5 +C

tan2k+1x secnx dx=

tan2kx secn1xsecx tan xdx

=

(1 sec2x)k secn1x secxtanxdx

=

(1 t2)ktn1 dt (denote t= sec x)


Exercises


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7.2: 130


Trigonometric substitution


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9 x2x2

dx

Ans: Denote x= 3 sin . Then dx= 3 cos d. We then have

9

x2

x2 dx=

9 9sin2

9sin2 3cos d=

3cos 9sin2 3cos d

=

cot2 d=

(csc2 1) d

= cot

+C


Trigonometric substitution


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Expression Substitution Identity

a2 x2 x=a sin , 2 2 1 sin2 = cos2

a2 +x2 x=a tan , 2 < < 2 1+tan2 = sec2

x2 a2 x=a sec , 0 deg(Q), then divide P byQ to obtain

P(x)

Q(x)=S(x) +

R(x)

Q(x), deg(R)< deg(Q)

Ex:

x3+xx1 dx=

x2+x+2+

2

x1

dx= x3

3 +x2

2 + 2x+ 2 ln |x 1| +C

Step 2: Factorize Q(x):

2.1: IfQ(x) = (a1x+b1) (akx+bk) has no reapeated factor, then write

R(x)Q(x)

= A1a1x+b1)

+ + Akakx+bk)

Ex: Evaluate

x2 + 2x 12x3 + 3x2 2xdx


Integration of rational fucntions


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Ex:

x2 + 2x 12x3 + 3x2 2xdx. We have2x

3 + 3x2 2x=x(2x 1)(x+ 2). Then

x2 + 2x 12x3 + 3x2 2x =

A1

x +

A2

2x 1+ A3

x+ 2 x

= x2 + 2x 1 =A1(2x 1)(x+ 2) +A2x(x+ 2) +A3x(2x 1) x x2 + 2x 1 = (2A1+A2+ 2A3)x2 + (3A1+ 2A2 A3)x 2A1 x

=

2A1+A2+ 2A3 = 1

3A1+ 2A2 A3 = 22A1 = 1

=

A1 = 12A2 =

15

A3 = 110Hence

x

2

+ 2x 12x3 + 3x2 2xdx=

12x

+ 12(2x 1) 110(x+ 2)

dx

=1

2ln |x| +1

4ln |2x 1| 1

10ln |x+ 2| +C



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Integration of rational functions


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D.

T.PHA

M-VGUHence

4x

(x

1)2(x+ 1)

dx= 1

x

1

+ 2

(x

1)2

1x+ 1 dx

=A ln |x 1| Bx 1+Cln |x+ 1| +K,

where K is a constant.



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Exercises


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7.4 16, odd numbers from 738, 3940


Why we need approximate integration


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Many integrals can not be computed exactly, e.g., 10

ex2

dx

11

1 +x3 dx

Many integrals arising from science and real life do not have a closedintegrands.

= Approximate integration



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Error of Midpoint rule

Error bound for Midpoint rule: Suppose that |f (x)| K for a x b


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Error bound for Midpoint rule: Suppose that|f (x)| K for a x b.Then

En

M K(b

a)3

24n2

Ex: Approximate

21

1

xdxby Midpoint method with n= 5.

x1 2x1 x2 x3 x4

n= 5= x= 215 = 0.2 and x0 = 1,x1 = 1.2, x2 = 1.4, x3 = 1.6, x4 = 1.8, x5 = 2

The midpoints: x1 = 1.1, x2 = 1.3, x

3 = 1.5,

x4 = 1.7, x5 = 1.9

Mn =5

i=1

f(xi )x= 0.2 ( 1

1.1+

1

1.3+

1

1.5+

1

1.7+

1

1.9) = 0.691907885715935

Meanwhile, 21 1xdx= ln 2=EnM= 0.001239294844010


Error of Midpoint rule


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We applied the Midpoint rule for different divisions:

n Mn EnM

5 0.691907885715935 0.001239294844010

10 0.692835360409960 3.118201499850981e 0420 0.693069098225587 7.808233435824263e 0540 0.693127651979310 1.952858063514196e 0580 0.693142297914324 4.882645621484549e 06

200 0.693146399314218 7.812457272216022e

07

1000 0.693147149309952 3.124999337078549e 08



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Error of Trapezoidal rule

Error bound for Trapezoidal rule: Suppose that |f (x)| K for a x b


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Error bound for Trapezoidal rule: Suppose that|f (x)| K for a x b.Then

E

n

T :=

b

a f(x)dx Tn , EnM K(b

a)3

12n2

Ex: Approximate

21

1

xdxby Trapezoidal method with n= 5.

x1 2x1 x

2 x

3 x

4

n= 5= x= 215 = 0.2 and x0 = 1,x1 = 1.2, x2 = 1.4, x3 = 1.6, x4 = 1.8, x5 = 2

Tn = x

2 (f(1) + 2f(1.2) + 2f(1.4) + 2f(1.6) +2f(1.8) +f(2))

Tn =0.2

2

1

1+ 2

1

1.2+ 2

1

1.4+ 2

1

1.6+ 2

1

1.8+f(2)

= 0.695634920634921

Meanwhile, 21 1xdx= ln 2=EnT = 0.002487740074976


Error of Trapezoidal rule


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We applied the Trapezoidal rule for different divisions:

n Tn EnT

5 0.695634920634921 0.002487740074976

10 0.693771403175428 0.00062422261548320 0.693303381792694 0.000156201232749

40 0.693186240009141 0.000039059449195

80 0.693156945994225 0.000009765434280

200 0.693148743055062 0.000001562495117

1000 0.693147243059937 0.000000062499992


Simpson Rule

y f ( )


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y

x

y = f(x)

a bx1 x2 x3 x4 x5x0 = = x6

P0P1

P2

P3

P4

P5

P6

Simpson rule: Let n be even.

b

a

f(x)dx Sn =x3

(f(x0) + 4f(x1) + 2f(x2) + 4f(x3) +. . .

+2f(xn2) + 4f(xn1) +f(xn))


Simpson rule

Error bound for Simpson rule: Suppose thatf(4)(x) K for a x b


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Error bound for Simpson rule: Suppose that f (x) K for a x b.Then

EnS

K(b

a)5

180n4

Ex: Approximate

21

1

xdxby Simpson rule with n= 10.

x1 2

n= 10= x= 2

110 = 0.1 and

x0 = 1, x1 = 1.1, x2 = 1.2, x3 = 1.3,

x4 = 1.4, x5 = 1.5, x6 = 1.6, x7 = 1.7,x8 = 1.8, x9 = 1.9, x10 = 2

S10 =0.1

3

1

1+

4

1.1+

2

1.2+

4

1.3+

2

1.4+

4

1.5+

2

1.6+

4

1.7+

2

1.8+

4

1.9+

1

2

= 0.693150230688930 = E10S = 0.000003050128985


Simpson rule


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We applied the Simpson rule for different divisions:

n Sn EnS

10 0.693150230688930 0.000003050128985

20 0.693147374665116 0.00000019410517140 0.693147192747956 0.000000012188011

80 0.693147181322587 0.000000000762642

200 0.693147180579475 0.000000000019530

1000 0.693147180559975 0.000000000000030


Exercises


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7.7: 712, 21, 22



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Improper integrals of type I


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Ex: Determine the convergence of 1

1

xdx

Ans: By definition, we have

1

1

xdx = lim

t t

1

1

xdx = lim

tln

|x

| t

1

= limt

(ln |t| ln1)= lim

tln |t|

=

.

Hence, the

11x

dx is divergent (not convergent).


Improper integrals of type I

E E l t

0xd


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Ex: Evaluate

xexdx

Ans: By definition, we have

0

xexdx= limt

0

t

xexdx.

Denote u=x and dv=exdx. Then du=dx and v=ex. Integration byparts gives

0

t

xexdx=xex0t

0

t

exdx= tet ex0t

= tet 1 et

Hence,

0

xexdx= limt

(tet 1 et) = limt

(tet) 1 limt

et

= 1


Improper integrals of type II

1


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Ex: Determine the convergence of

1

1

xpdx

Ans: When p= 1, the integral is divergent (see previous example). when p= 1,by definition, we have

1

1

xpdx= lim

t t

1

xpdx = limt

xp+1

p+ 1

t

1

= limt

t1p

1 p 11 p

= 1

1 p limt t1p 1

1 p

=

11p ifp>1 ifp1divergent () ifp 1


Improper integrals of type II

Improper integrals of type II:


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1 Let f : [a, b) R be continuous and f be discontinuous at b. Then b

a

f(x)dx := limtb

t

a

f(x)dx

provided that the limit exists.

2 Let f : (a, b]R be continuous and f be discontinuous at a. Then

ba

f(x)dx := limta+

bt

f(x)dx

provided that the limit exists.

3 Iff has a discontinuity at c, where a


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Ex: Evaluate

21

x

2

dx

Ans: We first note that 1x2 is discontinuous at 2. By definition, we have

5

2

1x

2

dx= limt

2+

5

t

1x

2

dx

= limt2+

2

x 25t

= limt2+

2

3 2t 2

= 2

3.

Hence,

52

1x 2 dx= 2

3



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Comparison test for improper integrals

C i Th L f d b i f i i h 0 f ( )


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Comparison Theorem: Let f and gbe continuous functions with 0 f(x)g(x) for x

a.

Ifa

g(x) dx is convergent thena

f(x) dx is convergent.

Ifa

f(x) dx is divergent thena

g(x) dx is divergent.

Ex: Determine whether the integral

1

dx

x2 ex is convergent?

Ans: When x 1, ex >1 and hence,1

x2 ex 1

x2.

Since

1

1

x2dx is convergent (? ), by Comparison Theorem,

1

dx

x2ex is

convergent.


Exercises


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7.8: 12, 3, 530, 55, 5759, 75


Fundamental Engineering Maths 5

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