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D.T.
PHAM
-VGU
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
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PHAM
-VGU
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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PHAM
-VGU
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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PHAM
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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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D.T.
PHAM
-VGU
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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PHAM
-VGU
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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-VGU
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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PHAM
-VGU
Exercises
5.2:
14, 912 2122, 2728, 3540, 5254
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PHAM
-VGU
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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PHAM
-VGU
The fundamental theorems of calculus
(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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PHAM
-VGU
The fundamental theorems of calculus
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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PHAM
-VGU
The fundamental theorems of calculus
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)
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Th f d l h f l l
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PHAM
-VGU
The fundamental theorems of calculus
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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PHAM
-VGU
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
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I d fi it I t l
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PHAM
-VGU
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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Indefinite Integrals
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PHAM
-VGU
Indefinite Integrals
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
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The net change theorem
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PHAM
-VGU
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)
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Some applications
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PHAM
-VGU
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.
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Some applications
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PHAM
-VGU
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.)
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Exercises
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PHAM
-VGU
Exercises
5.4: 110, 2130, 48, 4952
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The substitution rule
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PHAM
-VGU
The substitution rule
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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The substitution rule
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PHAM
-VGU
The substitution rule
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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The substitution rule
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PHAM
-VGU
The substitution rule
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.
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The substitution rule for definite integrals
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PHAM
-VGU
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
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-VGU
5.5: 115, 5160, 7374, 77, 86.
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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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Integration by parts
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PHAM
-VGU
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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Integration by parts
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PHAM
-VGU
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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Integration by parts
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PHAM
-VGU
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
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PHAM
-VGU
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
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Exercises
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PHAM
-VGU
7.1: 120, 3338, 4445,
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sinm x cos2k+1 x dx
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PHAM
-VGU
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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PHAM
-VGU
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
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tanm x sec2k x dx
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PHAM
-VGU
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)
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tan2k+1 x secn x dx
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PHAM
-VGU
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)
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Exercises
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T.PHAM
-VGU
7.2: 130
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Trigonometric substitution
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D.
T.PHAM
-VGUEx: Evaluate
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
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Trigonometric substitution
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D.
T.PHA
M-VGU
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
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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
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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
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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
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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
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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))
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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
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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
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Exercises
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D.T
.PHA
M-VGU
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).
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Improper integrals of type I
E E l t
0xd
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D.T
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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
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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
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Improper integrals of type II
Improper integrals of type II:
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D.T
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M-VGU
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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D.T
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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.
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Exercises
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D.T
.PHA
M-VGU
7.8: 12, 3, 530, 55, 5759, 75
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