Chapter 9 Theory of Differential and Integral Calculus• Section 1 Differentiability of a Function

Definition. Let y = f(x) be a function. The derivative of f is the function whose value at x is the limit

provided this limit exists.

If this limit exists for each x in an open interval I, then we say that f is differentiable on I.

E.g.1.2 (a) Show that f(x)=|tanx| is not differentiable at x=n*Pi

Proof of e.g.1.2

.at x abledifferentinot is f(x)

-1 xcos

1lim

h

sinhlim)(f Similarly,

1 11 xcos

1lim

h

sinhlim

h

|tanh|lim

h

|)tan(||)htan(|lim

h

)(f)h(flim)(f )a(

0h0h-

0h0h0h

0h

0h

0.at x abledifferentinot is xg(x) that Show )b( 3

2

h

1lim

h

)0(g)h0(glim ,fact In

0at x abledifferentinot is g(x)

h

1lim

h

0)h0(lim

h

)0(g)h0(glim

3

10h0h

3

10h

3

2

0h0h

Discussion: Ex.9.1, Q.1, 6Discussion: Ex.9.1, Q.1, 6

Solution to Ex.9.1Q.1

-1b i.e.

1b2(1)

xlim)bx2(lim

2 a

)1(f)1(f

2

1x1x

f(x).(x) fthat

and Rpoint xevery at abledifferenti is f that show

,1)x(glim and R xallfor xg(x),1f(x)

such that RR:gfunction fixed a is there(2)

and R,y x,allfor f(x)f(y), y)f(x (1)

:conditions following thesatisfies f If

R.on definedfunction valued-real a be f(x)Let 1.3 .g.e

0x

)x(f1)x(f

)h(glim)x(flim (x)g(h)flim h

)h(hg)x(flim

h

1)h(f)x(flim

h

)x(f)h(f)x(flim

h

)x(f)hx(flim(x)f

:1.3 e.g. of oofPr

0h0h0h

0h0h

0h0h

Discussion Ex.9.1, Q.9Discussion Ex.9.1, Q.9

Section 2 Mean Value Theorem

Rolle's Theorem. Let f be a function which is differentiable on the closed interval [a, b]. If f(a) = f(b) then there exists a point c in (a, b) such that f '(c) = 0.

Example of Rolle’s Theorem

Mean Value Theorem

Mean Value Theorem. Let f be a function which is differentiable on the closed interval [a, b]. Then there exists a point c in (a, b) such that

ab

)a(f)b(f)c('f

Proof of Mean Value Theorem

(Q.E.D.) a-b

f(a)-f(b)(c) f.e.i

0a-b

f(a)-f(b)-(c) f i.e.

0.(c)' such that b) ,a(c

Theorem, sRolle' by the and (b)(a)ab

)b(af)a(bfb

a-b

f(a)-f(b)-f(b)(b) and

ab

)b(af)a(bfa

a-b

f(a)-f(b)-f(a)(a) then

,xa-b

f(a)-f(b)-f(x)(x)function theConsider

Example 1 of Mean Value Theorem

Example 2 of Mean Value Theorem

Corollaries

(1) Let f be a differentiable function whose derivative is positive on the closed interval [a, b]. Then f is increasing on [a, b].

(2) Let f be a differentiable function whose derivative is negative on the closed interval [a, b]. Then f is decreasing on [a, b].

Proof of Corollary(1)

increasing is f(x) hence and )x(f)x(f

0)c('fxx

)f(x-)f(xsuch that

b)(a,c exists thereb,xxaany For

12

12

12

21

decreasing is f(x) hence and )x(f)x(f

0)c('fxx

)f(x-)f(xsuch that

b)(a,c exists thereb,xxaany For

12

12

12

21

Proof of corollary(2)

First Derivative Test. Suppose that c is a critical point of the function f and suppose that there is an interval (a, b) containing c. (1) If f '(x) > 0 for all x in (a, c) and f '(x) < 0 for all x in (c, b), then c is a local maximum of f. (2) If f '(x) < 0 for all x in (a, c) and f '(x) > 0 for all x in (c, b), then c is a local minimum of f.

Corollary 3If f ’(x)=0 for all x in an interval I, then f(x) is a constant function in I.

I.in function constant a is f

)x(f)f(x

0)c(fxx

)x(f)f(x

such that I)x,(xc Theorem, ValueMean by

I,in x any x For

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12

12

12

21

21

Corollary 4If f and g are differentiable functions on I and f ’(x)=g’(x) for all x in I, then f(x)=g(x) +c for some constant c.

I.on c g(x)f(x) thus

and Rc , ch(x) 2,corollary By

0.(x)g-(x)f(x)h then g(x),-f(x)h(x) Let

:oofPr

Example: 0.any xfor x1e that Show x

.x1e hence and x1e .e.i

1x

1e i.e. ,1ee

x

1e

.0-x

f(0)-f(x)(c) fsuch that )0 ,x(c 0, xWhen

x1e .e.i

x1e i.e. ,1eex

1e

.0-x

f(0)-f(x)(c) fsuch that )x,0(c 0, xWhen

.e(x) f then ,1ef(x) Let

xx

x0c

x

x

x0cx

xx

Section 3 Convex Functions

Definition 3.1

• A real-valued function f(x) defined on an interval I is said to be convex on I iff

• for any two points x1, x2 in I and any two positive numbers p and q with p+q=1,

• f(px1+qx2) pf(x1) + qf(x2).

I.on convex is f then 0,(x) f If

3.1 Theorem

Example

ab2

ba.e.i

abln)2

ba(ln- .e.i

abln)b2

1a

2

1(ln- .e.i

bln2

1aln

2

1)b

2

1a

2

1(ln- .e.i

)b(f2

1)a(f

2

1)b

2

1a

2

1(f

.Ron convex is f ,0x

1(x)f

0. xallfor ,xln)x(f Let

2

Theorem 3.2

)x(fxf then

,1such numbers negative-non are ..., , , and

Ion are x ..., ,x ,x I,on convex is f(x)function a If

i

n

1iii

n

1ii

n

1iin21

n21

Proof of Theorem 3.2

proved. is 3.2 Theorem induction, of principle By the 1.knfor truealso is it

xfxf...xfxf

xfxf...xfxf

xxf...xfxf

)1x... (Since xf...xfxf

x...xxf

xx...xxf

xx...xxfxf then

,1 that such ,,..., , and I x,x,..., x,consider x and

k,nfor trueisit that assume Now

2.nfor trueisit convexity, of definition By the

2.non induction by theorem theprove togoing now are We

1k1kkk2211

1k1kk

1kk

1kk

k1kk2211

1k1kk

1kk

1kk

k1kk2211

kk21kk2211

kk2211

1k1kk

1kk

1kk

k1kk2211

1k1kkk2211

1k

1iii

1k

1ii1kk211kk21

Example n21

nn21

n21

a... a an

1...a a a

,a..., ,a ,a numbers real positiveany for that ovePr

G.M.A.M.

a...aaaan

1.e.i

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1ln

alnalnalnn

1a

n

1ln -

then,a xand n

1let and 3.2 Theorem By the

0.for xconvex is f hence and 0x

1(x)fthen

0,for x-lnx f(x) Let

nn21

n

1i

n

1

i

n

1ii

n

1i

n

1

i

n

1ii

n

1i

n

1

i

n

1i

n

1

i

n

1ii

n

1ii

iii

2

§4 Definite Integral is the Limit of a Riemann Sum

∫ ba dx)x(f

n

1

n

2

n

i

n

n

n

y=f(x)Find the sum of the areas of the rectangles in terms of n and f.

dx)x(f1

0

nnn

f...

nn3

f

nn2

f

nn1

f

nlim

y=f(x)

Ai

222n

n

n

1

0

222n

1

0

n

n...

n

2

n

1lim

n

n...

n

2

n

1

n

1lim

)n

)01(n0(f...)

n

)01(20(f)

n

010(f

n

01limxdx

x.f(x) Consider:oofPr

.n

n...

n

2

n

1limxdx that Show

:Example

Group DiscussionExpress each of the following integrals as a limit of sum of areas:

1

0

x

1

0

1

0

2

1

0

dxe .4

sinxdx .3

dxx .2

xdx .1

2222n n

n...

n

3

n

2

n

1lim

3

2

3

2

3

2

3

2

n n

n...

n

3

n

2

n

1lim

n

1in n

isin

n

1lim

n

1i

n

i

ne

n

1lim

Group DiscussionExpress each of the limits as a definite integral :

1

0

x-

1

0

1

0

1

0

3

dxe

cosxdx

dxx1

1

dxx

4

3

4

3

4

3

4

3

n n

n...

n

3

n

2

n

1lim .1

nn

1...

3n

1

2n

1

1n

1lim .2n

n

1in n

icos

n

1lim .3

n

1i

n

i

ne

n

1lim .4

4

1

2ln

1sin

1e1

Example 4.1(a)

4

π=]x[tan=

x+1

dx=

nn

+1

1+...+

n3

+1

1+

n2

+1

1+

n1

+1

1

n

1lim=

n+n

1+...+

3+n

1+

2+n

1+

1+n

1n lim

:Solution

10

110 2

22

22

22

22∞→n

22222222∞→n

∫

.nn

1...

3n

1

2n

1

1n

1nlim Evaluate )a(

22222222n

Read example 4.1(b)

Classwork Ex.9.4 Q.3

Read example 4.1(b)

Classwork Ex.9.4 Q.3

Area bounded by the curve, x-axis, x=a and x=b

a bn

aba

n

)ab(2a

n

)ab(3a

n

)ab)(1n(a

n

abaf

n

)ab(2af

n

)ab)(1n(af

)b(f

n

abrectangles of width

n

1in

b

a

n

abiaf

n

a-blim

dxf(x)

b and abetween curve under the area the

Homework Ex.9.4Homework Ex.9.4

Example 4.1(a)

4

π=]x[tan=

x+1

dx=

nn

+1

1+...+

n3

+1

1+

n2

+1

1+

n1

+1

1

n

1lim=

n+n

1+...+

3+n

1+

2+n

1+

1+n

1n lim

:Solution

.n+n

1+...+

3+n

1+

2+n

1+

1+n

1nlim Evaluate )a(

10

110 2

22

22

22

22∞→n

22222222∞→n

22222222∞→n

∫

n

1kn

b

a n

)ab(kaf

n

ablimdx)x(f

Section 5 Properties of Definite Integrals

.0dxf(x) then b], ,a[x 0,f(x)

and b] [a,on integrable is f(x) If 5.1 Theoremb

a

0n

)ab(kaf

n

ablimdx)x(f

0n

)ab(kaf

n

ablim Hence

0n

)ab(kaf

0n

)ab(kaf

:oofPr

n

1kn

b

a

n

1kn

n

1k

When does the equality fail?When does the equality fail?

Theorem 5.4Theorem 5.4

Discussion:Ex.9.5, Q.1,2Discussion:Ex.9.5, Q.1,2

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

g(x)dx f(x)dx

0g(x)dx f(x)dx

0dx g(x)-f(x)

,0h(x)dx 5.1, TheoremBy

0.h(x) then g(x),-f(x)h(x) Let

:oofPr

.g(x)dx f(x)dx

thenb], [a, xallfor g(x),f(x) if

and b] [a,on integrable are g(x) and f(x) If 5.2 CorollaryWhen does the equality fail?When does the equality fail?

Corollary 5.5Corollary 5.5

Discussion: Ex.9.5, Q.3Discussion: Ex.9.5, Q.3

Corollary 5.3

dx)x(f dx)x(f

thenb], [a,on integrable is f(x) Ifb

a

b

a

b

a

b

a

b

a

b

a

b

a

dx|)x(f|dx)x(f.e.i

dx|)x(f|dx)x(fdx|)x(f|

|)x(f|)x(f|)x(f|

:oofPr

|x|K iff -Kx K|x|K iff -Kx K

a)f(b),-(b f(x)dx a)f(a)-(b

then there,increasingstrictly (a)

is f(x) if and b] [a,on continuous is f(x) If

5.6 Corollary

b

a

dxf(b)f(x)dxdxf(a)

f(b)f(x)f(a)

bxa b), (a,any x For:oofPr

b

a

b

a

b

a

f(b)

f(a)

a b

a)f(a)-(b f(x)dxa)f(b)-(b

then there,decreasingstrictly (b)b

a

a)-f(b)(bf(x)dxa)-f(a)(b

dx 1 f(b)f(x)dxdx 1 f(a)

b

a

b

a

b

a

b

a

)x(fy

Example 5.1 Prove the following:integer. positive a isn where,

6x1

dx

2

1 )a( 2

1

0 n2

6

xsindxx-1

1

2

1

dxx1

1dx

x-1

1dx 1

2

1

012

1

0 2n

2

1

0 2

2

1

0 2n

2

1

0

,xx0 have we

,2

1x0 i.e. ],

2

1 ,0[any x For

:oofPr

22n

22n

22n

x1

1

x-1

11

x1x-11

Where do they come from?

6]

2

x[sin

xx4

dx

2

1 1

011

0 32

0x4xx44

,0x x1], ,0[any x For

:oofPr

232

32

6xx4

dx

2

1 (b)

1

0 32

232 x4

1

xx4

1

2

1

1

0 2

1

0 32

1

0 x4

dx

xx4

dxdx

2

1

32 xx4

1y

2

1y

2x4

1y

Homework:Ex.9.5,3-7Homework:Ex.9.5,3-7

Example 5.2

0.nfor trueis it

e1dteI

:Proof

.K and Jinteger somefor eKJI

n,integer negative-nonany for that induction,by Prove, (a)

integer. negative-nonany isn where,dtteILet

1

0

t0

nnnnn

1

0

ntn

.... 2, 1, 0,nfor trueisit induction, of principle By the

1.knfor truealso is it

eKJeK)1k(1J)1k( 1n1nnn

)eKJ)(1k(e1)I(k-e nnk

1

0

kt1

0

1

ot1kt1k1

0

1kt1k dtt)1k(eetedtdtteIthen

k,nfor trueisit that assume Now

1n allfor ,n

edtetI

1n

1 that Show

(b) 5.2 Example1

0

nn

1

0

ntn dtteI

How to get n+1? How to get n?

1n and 1t0for etettet

:Proof1nnntn

1

0

1n1

0

n1

0

nt1

0

n etdtetdttedtt

n

edtetI

1n

1.e.i

n

etdtetI

1n

t.e.i

1

0

nn

1

0

n1

0

nn

1

0

n

n

ppKqJ

1n

q .e.i

qn

p

q

pKJ

1n

1 .e.i

,n

eeKJ

1n

1 (a),By

.n

eI

1n

1 have we(b),by

thenq, and p numbers natural somefor q

pe that contrary, on the Assume,

:Proof

number.

irrationalan bemust e that show ion,contradictby and (b) and (a) Using)c(

nn

nn

nn

n

.irrational bemust e

integer.an is pKqJ and

1 and 0between lyinginteger no is theresinceion contradict a is which

,11p

ppKqJ

1n

q 0 have we1,pn Setting

nn

nn

What kinds of nos are they?

How to get contradiction?How to get contradiction?

Theorem Value teIntermedia

.L)f( such that

b] [a, M,Lmany for then b], [a,any xfor

Mf(x)m and b] [a,on function continuous a is f(x) If

M

m

L

Section 6 Theorem 6.1Mean Value Theorem of Integral

).(f)ab(dx)xf(such that

b] [a,point a exists e then therb], [a,on continuous is f(x) Ifb

a

A

A

)(f

)x(fy

a b

)ab(Mdx)x(fa)-m(b

Mdxdx)x(fmdx

M.f(x)m

such that M and m then b], [a,on continuous is f If

:oofPr

b

a

b

a

b

a

b

a

a-b

dx)x(f)f(

such that b],a[ Theorem, Value teIntermedia Byb

a

Ma-b

dx)x(fm

b

a

= L

)f()ab(dx)x(fb

a

M

m

L

Differentiation of Integrals

Bdxx1

0

2 Adxx 2 3

1

3

x1

0

3

C

3

x 3

B? andA between iprelationsh theis What

10]A[B

b

af(x)dx? integral for theresult your generalizeyou Can

b

a

b

adx)x(fdx)x(f

f(x)(x)F where),a(F)b(F

Theorem s'LeibnizNewton

( )

it? generalizeyou Can

discover?you do What (c)

.F(x)dx

d Find (b)

F(x). Evaluate )a(

.dtte=F(x) Let

discussion groupfor oblemPr

∫ x0

t

∫ xa ).x(f=dt)t(f

dx

d :Answer

Theorem 6.2 Fundamental Theorem of Calculus

f(x), (x) F' and b] [a,point xevery at abledifferenti is

f(t)dt F(x)function then theb],[a,on continuous is f(x) Ifx

a

h

dt)t(f lim

h

dt)t(f dt)t(flimdt)t(f

dx

d

:oofPr

hx

x

0h

hx

a

x

a

0h

x

a

)f()ab(dx)x(fb

a

Since f(x) is continuous.

f(x)

)lim(f x0h

)(flim

h) x,x(,h

)(f)xhx(lim

x0h

xx

0h

x

a).x(fdt)t(f

dx

d i.e.

Newton-Leibniz Formula

f(x)(x)F i.e. f(x) offunction primitive

a is F(x) where),a(F)b(Fdx)x(fb

a

0f(x)-f(x)(x)F-f(x)(x)g then

),x(Fdt)t(f g(x)function a Considerx

a

)a(F)b(Fdt)t(f b

a

.c)a(Ff(t)dt F(a)- .e.i

get we, a xPuttinga

a

c)x(Fdt)t(f i.e.

c.constant somefor cg(x) i.e.x

a

)a(F)x(Fdt)t(f x

a

get web, xPutting

Newton-Leibniz’s contribution in Calculus

x

a)x(fdt)t(f

dx

d

)f()ab(dx)x(fb

a

)a(F)b(Fdx)x(fb

a

Integral of

Theorem Value Mean

Calculus of

Theorem lFundamenta

Formula

LeibnizNewton

3

1

)n

12)(

n

11(

6

1lim

)1n2)(1n(6

n

n

1lim

kn

1lim

n

k

n

01limdxx

n

3n

n

1k

23n

n

1k

2

n

1

0

2

correct? is

method Which

?dxx1

0

2 3

1

3

x1

0

3

em?between th

difference theis What

What will happen if we don’t know Newton-Leibniz’s Theorem?

Questions for discussion

)x(g

)x(h

x

x

t

x

0

t

1x2

0

t

1x

0

t

x

1

t

dtf(t) dx

d .6

dtedx

d .5

dtedx

d .4

dtedx

d .3

dtedx

d .2

dtedx

d .1

:following theEvaluate

22

22

2

2

2

Application of Fundamental Theorem of Calculus

.dx)x(xfdxf(t)dt

that show function, oddan is f(x) If 1 Example

n

n

n

n-

x

n-

dxf(t)dtdx

dxdt)t(fxdxf(t)dt

:oofPr

n

n-

x

n-

n

n

x

n

n

n-

x

n-

dx)x(xf

dx)x(xf-0n0n

n

n-

n

n-

dx)x(xfdt)t(fndt)t(fn n

n-

n

n

n

n

Example 2

b]. [a, xallfor 0f(x) then ,0f(t)dt if that deduce Hence

b]. [a,on function increasingan is F that Show

b]. [a,for x dt)t(f F(x) Define

b]. [a,on function continuous negative-non a be f Let

b

a

x

a

b]. [a, xallfor 0(x)Ff(x)

b] [a, xallfor 0F(x)

0F(b)F(x)F(a)0

b]. [a,on increasing is )x(F

0)x(f)x(F

:Solution

Example 3

f(x). Find

.R xallfor f(t)dt3t xf(x)

satisfyingfunction a be f(x) Letx

0

23

))x(f1(x3

)x(fx33x(x)f

equation, theatingDifferenti

:Solution

2

22

1e)x(f

e)u(f1

u)u(f1ln

u)x(f1ln

3

3

x

u

3

3u0

u0

3u

0|x)x(f1d

))x(f1(

1

2x3))x(f1(

(x)f

dxx3dx))x(f1(

(x)f u

0

2u

0

Example 4

1..g(0) and 0f(0) C.

R;for x -f(x)(x)g B.

R;for x g(x)(x)f A.

:conditions following the

satisfying Ron defined functions abledifferenti be g and f Let

Rfor xcosx g(x) andsinx f(x) that show otherwise, or

,]xcos)x(g[sinx]-[f(x)h(x) atingdifferentiBy 22

R.for xcosx g(x) andsinx f(x)

0cosx-g(x) and 0xsin)x(f

0]xcos)x(g[sinx]-[f(x)

0h(x)

0

1]-[10

]0cos)0(g[sin0]-[f(0)h(0)c

get we0, xPutting

c.constant somefor c)x(h

0

)xsin)x(f)(xcos)x(g(2

)xcos)x(g)(xsin)x(f(2

)xsin)x(g)(xcos)x(g(2

)xcos)x(f)(xsin)x(f(2)x(h

:Solution

22

2

22

Example 5

..dx)x(ff(x)dx ab

thenf(a),b0 if that (a) from Deduce (b)

a]. [0,u allfor 0F(x) that Prove (a)

).u(ufdx)x(f dx)x(f F(x)Let

0.f(0) and 0a where

a], [0,on function increasingstrictly and abledifferenti a is f

a

0

b

0

1-

f(u)

0

1-u

0

a]. [0,u allfor 0F(x) that Prove (a)

).u(ufdx)x(f dx)x(f F(x)Let f(u)

0

1-u

0

a

f(a)

u

f(u)

b

u

0dx)x(f

f(u)

0

1- dx)x(f

0

)u(f)u(fu)u(fuf(u)

)u(f)u(fu)u(f))u(f(f)u(f)u(F

:oofPr1

a] [0,u allfor 0F(x)

0

)0(f0dx)x(f dx)x(f F(0)

get we0, xPuttingf(0)

0

1-0

0

c.constant somefor cF(x)

a

0

b

0

1- .dx)x(ff(x)dx ab

thenf(a),b0 if that (a) from Deduce (b)

f(u)

0

1-u

0).u(ufdx)x(f dx)x(f F(x)Let

0

)u(f)u(fu)u(fuf(u)

)u(f)u(fu)u(f))u(f(f)u(f)u(F

:oofPr1

a] [0,u allfor 0F(x)

0

)0(f0dx)x(f dx)x(f F(0)

get we0, xPuttingf(0)

0

1-0

0

c.constant somefor cF(x)

a

f(a)

u

f(u)

b

a

0dx)x(f

b

0

1- dx)x(f

0

0u

∫∫∫∫∫

∫∫

)f(u0 00

1-u0

au

)f(u0 000

1-u0

00

f(u)0

-1u0

000

00

.b)ua(+bu>dx)x(f +dx)x(f +dx)x(f

.bu=)u(fu=dx)x(f +dx)x(f

then,b=)f(usuch that u=u Let

).u(uf=dx)x(f +dx)x(f (a), By

ab=