-defint.pdf

The Differential

Definition 1. If the function f is defined by the equation y = f(x), thenthe differential of y, denoted by dy, is given by

dy = f ′(x)∆x

where x is in the domain of f ′ and ∆x is an arbitrary increment of x.

Definition 2. If the function f is defined by the equation y = f(x), thenthe differential of x, denoted by dx, is given by

dx = ∆x

where x is in the domain of f ′ and ∆x is an arbitrary increment of x.

Chapter 1: Definite Integral and Integration 1

Example 1. Given y = 4x2 − 3x + 1, we determine ∆y, dy and ∆y − dyfor x = 2, ∆x = 0.1, 0.01, 0.001.

x ∆x ∆y dy ∆y − dy2 0.1 1.34 1.3 0.042 0.01 0.1304 0.13 0.00042 0.001 0.013004 0.013 0.000004

We observe that as ∆x gets closer to zero, the difference between ∆yand dy gets smaller. Hence, dy is an approximation of ∆y when ∆x issmall.

For a fixed value of x, say x0,

dy = f ′(x0)dx = f ′(x0)∆x.

Also,f(x0 + ∆x) ≈ f(x0) + dy.


Properties of Differentials

In the following, u and v are functions, and k and n are constants.

Derivative Rule Differential Rule

dk

dx= 0 dk = 0

d(ku)

dx= k

du

dxd(ku) = kdu

d(u+ v)

dx=du

dx+dv

dxd(u+ v) = du+ dv

d(uv)

dx= u

dv

dx+ v

du

dxd(uv) = udv + vdu

d(u/v)

dx=vdudx − u

dvdx

v2d(u/v) =

vdu− udvv2

d(un)

dx= nun−1du

dxd(un) = nun−1du


Example 2. Use differentials to approximate the value of√

26.2.

We observe that 26.2 is close to 25, and we already know that thesquare root of 25 is 5. So, we begin with the equation y =

√x and we

estimate the change in y when x is increased from 25 to 26.2.

dy =dx

2√x

=1.2

2√x

= 0.12

Hence,

f(x0 + ∆x) ≈ f(x0) + dy

f(26.2) ≈ f(25) + 0.12

= 5.12

The estimated value of√

26.2 is 5.12.


Example 3. Use differentials to approximate the volume of a spherical shellwhose inner radius is 4 in. and whose thickness is 1

16 in.

The volume of a sphere is given by V =4

3πr3 which gives us

dV = 4πr2dr

Since r = 4 nd dr = 116, we have

dV = 4π(4)2 1

16= 4π

The approximate volume of the spherical shell is 4π in.3


Example 4. A closed container in the form of a cube having a volume of1000 in.3 is to be made by using six equal squares of material costing 20cents per square inch. How accurately must the side of each square bemeasured so that the total cost of material will be correct to within $3.00?

Let C dollars be the total cost of material.

C = 0.20(6x2) = 1.2x2 and dC = 2.4xdx

For the volume of the cube to be 1000 in.3, x = 10. Hence, the cost ofthe material will be exactly $120 if the length of a side of the squares is 10in. We wish to find |∆x| so that |∆C| ≤ 3.

We have ∆C = 2.4x∆x and |∆C| = 24|∆x| when x = 10. Since wewant 24|∆x| ≤ 3 it follows that |∆x| ≤ 0.125.

The side of each square should be measured to within 0.125 in so thatthe total cost of material will be correct to within $3.00.


Exercises.

Find dy.

1. y = (2x− 3)−4

2. y = (sinx+ cosx)2

3. y = (x10 +√

sin 2x)2

Solve each of the following.

1. Use differentials to approximate the value of 3√

26.91.

2. Assuming that the equator is a circle whose radius is

approximately 4000 miles, how much longer than the equator


would a concentric coplanar circle be if each point on it were

2 feet above the equator? Use differentials.

3. A tank has the shape of a cylinder with hemispherical ends.

If the cylindrical part is 100 cm lone and has a radius of 10

cm, about how much paint is required to coat the outside of

a tank to a thickness of 1 millimeter?

4. Einstein’s Special Theory of Relativity says that mass m is

related to velocity v by the formula

m =m0√

1− v2/c2.

Here m0 is the rest mass and c is the velocity of light. Use


differentials to determine the percent increase in mass of an

object when its velocity is increased from 0.9c to 0.92c.


Antidifferentiation

Definition 3. A function F is called an antiderivative of the

function f on an interval I if F ′(x) = f(x) for every value x

in I.

Theorem 1. If f and g are two functions defined on an

interval I, such that

f ′(x) = g′(x) for all x ∈ I

then there is constant K such that

f(x) = g(x) +K for all x ∈ I.


Theorem 2. If F is a particular antiderivative of f on an

interval I, then every antiderivative of f on I is given by

F (x) + C (1)

where C is an arbitrary constant, and all antiderivatives of

fon I can be obtained from (1) by assigning particular values

to C.


Definition 4. Antidifferentiation is the process of finding the

set of all antiderivatives of a given function. The symbol∫

denotes the operation of antidifferentiation, and we write∫f(x)dx = F (x) + C

where F ′(x) = f(x) and d(F (x)) = f(x)dx. The expression

F (x) + C is the general antiderivative of f .


Theorem 3. ∫dx = x+ C

Theorem 4. ∫af(x)dx = a

∫f(x)dx

where a is a constant.

Theorem 5.∫[f(x) + g(x)]dx =

∫f(x)dx+

∫g(x)dx


Theorem 6. If f1(x), f2(x), . . . , fn(x) are defined on the same

interval,∫[c1f1(x) + c2f2(x) + · · ·+ cnfn(x)]dx =

c1∫f1(x)dx+ c2

∫f2(x)dx+ · · ·+ cn

∫fn(x)dx

where c1, c2, . . . , cn are constants.

Theorem 7. If n is a rational number,∫xndx =

xn+1

n+ 1+ C n 6= −1.


Example 5.

1.

∫dx

x3=

∫x−3dx =

x−2

−2+ C = −

1

2x2+ C

2.

∫ √x7dx =

∫x7/2dx =

x9/2

9/2+ C =

2x9/2

9+ C

3.

∫36x5dx = 36

∫x5dx = 6x6 + C

4.

∫−

2dx√x

= −2

∫x−1/2dx = −4

√x+ C


5.

∫(4x2 − 5)dx =

4x3

3− 5x+ C

6.

∫(x− 4)(2x+ 3)dx =

∫(2x2 − 5x− 13)dx

=2x3

3−

5x2

2− 13x+ C


Theorem 8. ∫sinxdx = − cosx+ C

Theorem 9. ∫cosxdx = sinx+ C

Theorem 10. ∫sec2 xdx = tanx+ C

Theorem 11. ∫csc2 xdx = − cotx+ C


Theorem 12. ∫secx tanxdx = secx+ C

Theorem 13. ∫cscx cotxdx = − cscx+ C


Example 6.

1.

∫(3 secx tanx− 5 csc2 x) dx =

3

∫secx tanx dx− 5

∫csc2 x dx = 3 secx+ 5 cotx+ C

2.

∫2 cotx− 3 sin2 x

sinxdx = 2

∫cscx cotx dx− 3

∫sinx dx

= −2 cscx+ 3 cosx+ C


Some Techniques of Antidifferentiation

Theorem 14. The Chain Rule for Antidifferentiation.

Let g be a differentiable function, and let the range of g be

an interval I. Suppose that f is a function defined on I and

that F is an antiderivative of f on I. Then∫f(g(x))[g′(x)]dx = F (g(x)) + C

Proof. It is known that F ′(g(x)) = f(g(x)). Hence,

Dx[F (g(x))] = F ′(g(x))[g′(x)] = f(g(x))g′(x).


Therefore, it follows that∫f(g(x))[g′(x)]dx = F (g(x)) + C.

Theorem 15. If g is a differentiable function and n is a

rational number,∫[g(x)]n[g′(x)]dx =

[g(x)]n+1

n+ 1+ C n 6= −1


Example 7. Evaluate the following indefinite integrals.

1.

∫x(2x2 + 1)6dx 7.

∫r2 sec2 r3dr

2.

∫3x√

4− x2dx 8.

∫ √1

t− 1

dt

t2

3.

∫s√

3s2 + 1ds 9.

∫sin3 θ cos θdθ

4.

∫t√t+ 3

dt 10.

∫x(x2 + 1)

√4− 2x2 − x4dx

5.

∫(x3 + 3)1/4x5dx 11.

∫ (t+

1

t

)3/2(t2 − 1

t2

)dt

6.

∫t cos 4t2dt 12.

∫secx tanx cos(secx)dx


Exercises. (TC7, pages 334-335)

1.

∫ √1− 4y dy

2.

∫x

3√x2 − 9 dx

3.

∫x2(x3 − 1)10 dx

4.

∫y3

(1− 2y4)5dy

5.

∫x√x+ 2 dx

6.

∫2r

(1− r)7dr

7.

∫ √3− 2xx2 dx

8.

∫cos 4θ dθ

9.

∫6x2 sinx3 dx

10.

∫y csc 3y2 cot 3y2 dy


Area

Definition 5.

n∑i=m

F (i) = F (m) + F (m+ 1) + · · ·+ F (n− 1) + F (n)

where m and n are integers, and m ≤ n.

Theorem 16.

n∑i=1

c = cn, where c is any constant


Theorem 17.

n∑i=1

c · F (i) = c

n∑i=1

F (i), where c is any constant

Theorem 18.

n∑i=1

[F (i) +G(i)] =n∑i=1

F (i) +

n∑i=1

G(i)


Theorem 19. If n is a positive integer, then

n∑i=1

i =n(n+ 1)

2

n∑i=1

i2 =n(n+ 1)(2n+ 1)

6

n∑i=1

i3 =n2(n+ 1)2

4

n∑i=1

i4 =n(n+ 1)(2n+ 1)(3n2 + 3n− 1)

30


Definition 6. Suppose that the function f is continuous on the

closed interval [a, b], with f(x) ≥ 0 for all x ∈ [a, b], and that

R is the region bounded by the curve y = f(x), the x axis,

and the lines x = a and x = b. Divide the interval [a, b] into

n subintervals, each of length ∆x = (b − a)/n, and denote

the ith subinterval by [xi−1, xi]. Then if f(ci) is the absolute

minimum function value on the ith subinterval, the measure of

the area of region R is given by

A = limn→+∞

n∑i=1

f(ci)∆x

This equation means that for any ε > 0 there is a number

N > 0 such that if n is a positive integer and


if n > N then

∣∣∣∣∣n∑i=1

f(ci)∆x−A

∣∣∣∣∣ < ε.

Example 8. Find the area of the region bounded by the curve

y = x2, the x axis, and the line x = 3 by taking inscribed

rectangles.


We can also define the measure of the area of the region R

by

A = limn→+∞

n∑i=1

f(di)∆x

where f(di) is the absolute maximum value of f in [xi−1, xi].

In this case, the area is computed by taking circumscribed

rectangles.

Example 9. Find the area of the region bounded by the curve

y = x2, the x axis, and the line x = 3 by taking circumscribed

rectangles.


The Definite Integral and Its Properties

Definition 7. Let f be a function whose domain includes the

closed interval [a, b]. Then f is said to be integrable on [a, b]

if there is a number L satisfying the condition that, for any

ε > 0, there exists a δ > 0 such that for every partition ∆ for

which ||∆|| < δ, and for any wi in the closed interval [xi−1, xi],

i = 1, . . . , n, then

∣∣∣∣∣n∑i=1

f(wi)∆ix−A

∣∣∣∣∣ < ε


For such a situation we write

lim||∆||→0

n∑i=1

f(wi)∆ix = L

Definition 8. If f is a function defined on the closed interval

[a, b], then the definite integral of f from a to b, denoted by∫ baf(x)dx, is given by

∫ b

a

f(x)dx = lim||∆||→0

n∑i=1

f(wi)∆ix

if the limit exists.


Theorem 20. If a function is continuous on the closed

interval [a, b], then it is integrable on [a, b].

Definition 9. Let the function f be continuous on [a, b] and

f(x) ≥ 0 for all x in [a, b]. Let R be the region bounded by

the curve y = f(x), the x axis, and the lines x = a and x = b.

Then the measure A of the area of region R is given by

A = lim||∆||→0

n∑i=1

f(wi)∆ix

=

∫ b

a

f(x)dx


Theorem 21. If a > b and∫ baf(x)dx exists, then

∫ b

a

f(x)dx = −∫ a

b

f(x)dx

Theorem 22. If f(a) exists, then∫ a

a

f(x)dx = 0.

Theorem 23. If ∆ is any partition of the closed interval [a, b],

then

lim||∆||→0

n∑i=1

∆ix = b− a


Theorem 24. If f is defined on the closed interval [a, b], and

if

lim||∆||→0

n∑i=1

f(wi)∆ix

exists, where ∆ is any partition of [a, b], then if k is any

constant,

lim||∆||→0

n∑i=1

kf(wi)∆ix = k lim||∆||→0

n∑i=1

f(wi)∆ix


Theorem 25. If f is defined on the closed interval [a, b], and

if

lim||∆||→0

n∑i=1

f(wi)∆ix

exists, where ∆ is any partition of [a, b], then if k is any

constant, then ∫ b

a

kf(x)dx = k

∫ b

a

f(x)dx

Theorem 26. If f and g are integrable on the closed interval

[a, b], then f + g is integrable and∫ b

a

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

∫ b

a

f(x)dx+

∫ b

a

g(x)dx


Theorem 27. If f is integrable on the closed intervals [a, b],

[a, c], and [c, b],∫ b

a

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

∫ c

a

f(x)dx+

∫ b

c

f(x)dx

Theorem 28. If f is integrable on a closed intervals

containing a, b and c, then∫ b

a

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

∫ c

a

f(x)dx+

∫ b

c

f(x)dx

regardless of the order of a, b and c.


The Mean-Value Theorem for Integrals

Theorem 29. It the functions f and g are integrable on the

closed interval [a, b], and if f(x) ≥ g(x) for all x in [a, b],

then ∫ b

a

f(x)dx ≥∫ b

a

g(x)dx.


Theorem 30. Suppose that the function f is continuous on

the closed interval [a, b]. If m and M are, respectively, the

absolute minimum and absolute maximum values of f on

[a, b] so that

m ≤ f(x) ≤M, for a ≤ x ≤ b,

then

m(b− a) ≤∫ b

a

f(x)dx ≤M(b− a).


Theorem 31. The Mean Value theorem for Integrals

If the function f is continuous on the closed interval [a, b],

there exists a number c in [a, b] such that∫ b

a

f(x)dx = f(c)(b− a).

Definition 10. If the function f is integrable on the closed

interval [a, b], then the average value of f on [a, b] is∫ baf(x)dx

b− a


The Fundamental Theorems of Calculus

Theorem 32. The First Fundamental Theorem of the

Calculus

Let f be continuous on the closed interval [a, b] and let x

be any number in [a, b]. If F is the function defined by

F (x) =

∫ x

a

f(t)dt


then

F ′(x) = f(x) (2)

⇔ d

dx

∫ x

a

f(t)dt = f(x) (3)

(If x = a, the derivative in (2) may be a derivative from the

right, and if x = b, it may be a derivative from the left.)


Example 10. Compute the following derivatives:

1.d

dx

∫ x

1

1

t3 + 1dt;

2.d

dx

∫ x2

3

√cos tdt


Theorem 33. The Second Fundamental Theorem of

the Calculus

Let the function f be continuous on the closed interval

[a, b] and let F be a function such that

F ′(t) = f(x)

for all x in [a, b]. Then,∫ b

a

f(t)dt = F (b)− F (a).


Example 11. Evaluate the following definite integrals.

1.

∫ 3

−1

(3x2 + 5x− 1)dx 5.

∫ 5

4

x2√x− 4dx

2.

∫ 4

1

√x(2 + x)dx 6.

∫ 4

−4

|x− 2|dx

3.

∫ 3

−1

1

(y + 2)3dy 7.

∫ π/6

0

(sin 2x+ cos 3x)dx

4.

∫ 3

1

x

(3x2 − 1)3dx 8.

∫ 1/2

0

sec2 1

2πt tan

1

2πtdt


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