Math 152.02 Calculus with Analytic Geometry II · 2011-01-24 · Math 152.02 Lecture 7 - 1/19 Basic integration rules Di erential equations Motion More basic integrals You also know

Math 152.02

Lecture 7 - 1/19

Basic integrationrules

Differentialequations

Motion

Math 152.02Calculus with Analytic Geometry II

January 24, 2011

Math 152.02

Lecture 7 - 1/19



Motion

1 Lecture 7 - 1/19Basic integration rulesDifferential equationsMotion

Math 152.02

Lecture 7 - 1/19



Motion

Basic integration rules

Each rule for differentiation gives us a rule for integration

Fromc d

dx F (x) = ddx

(cF (x)

)we get

Theorem 52 (Constant rule for integration)

∫cf (x) dx = c

∫f (x) dx

Math 152.02

Lecture 7 - 1/19



Motion

Proof of Theorem 52.

Suppose ddx F (x) = f (x).

We have the derivative rule

c ddx F (x) = d

dx

(cF (x)

)Reinterpreting this rule as an antiderivative gives∫

c ddx F (x) dx = cF (x) + C .

Thus we may conclude∫cf (x) dx =

∫c d

dx F (x) dx

= cF (x) + C

= c(F (x) + C2)

= c

∫f (x) dx .

Math 152.02

Lecture 7 - 1/19



Motion From ddx F (x) + d

dx G (x) = ddx

(F (x) + G (x)

)we get

Theorem 53 (Sum rule for integration)

∫f (x) + g(x) dx =

∫f (x) dx +

∫g(x) dx

Math 152.02

Lecture 7 - 1/19



Motion

Proof of Theorem 53.

Suppose ddx F (x) = f (x) and d

dx G (x) = g(x).

We have the derivative rule

ddx F (x) + d

dx G (x) = ddx

(F (x) + G (x)

)Reinterpreting this rule as an antiderivative gives∫

ddx F (x) + d

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

Thus we may conclude∫f (x) + g(x) dx =

∫d

dx F (x) + ddx G (x) dx

= F (x) + G (x) + C

=

∫f (x) dx +

∫g(x) dx

Note: We drop constants when we have integrals on both sides of anequation.

Math 152.02

Lecture 7 - 1/19



Motion

Basic Integrals

Each basic derivative gives us a basic integral

Table 1: Basic integrals to memorize

differentiation integrationrule rule

ddx x r+1 = (r + 1)x r

∫x r dx = 1

r+1 x r+1 + C if r 6= −1

ddx ln |x | = 1

x

∫1x dx = ln |x |+ C

ddx cos x = − sin x

∫sin x dx = − cos x + C

ddx sin x = cos x

∫cos x dx = sin x + C

ddx ex = ex

∫ex dx = ex + C

ddx arctan x = 1

1+x2

∫1

1+x2 dx = arctan x + C

ddx arcsin x = 1√

1−x2

∫1√

1−x2dx = arcsin x + C

Math 152.02

Lecture 7 - 1/19



Motion

More basic integrals

You also know a few more derivative rules

Table 2: More basic integrals to memorize

differentiation integrationrule rule

ddx tan x = sec2 x

∫sec2 x dx = tan x + C

ddx cot x = − csc2 x

∫csc2 x dx = − cot x + C

ddx sec x = sec x tan x

∫sec x tan x dx = sec x + C

ddx csc x = − csc x cot x

∫csc x cot x dx = − csc x + C

ddx ax = (ln a)ax

∫ax dx = ax

ln a + C

Math 152.02

Lecture 7 - 1/19



Motion

Techniques of integration

Advanced derivative rules give us techniques of integration

differentiation technique ofrule integration

chain rule u-substitution (§7.1)

product rule integration by parts (§7.2)

We will return to these integration techniques later.

Math 152.02

Lecture 7 - 1/19



Motion

Problem 54

Find a formula for∫

10ex + 7 sin x dx

Solution to Problem 54∫10ex + 7 sin x dx =

∫10ex dx +

∫7 sin x dx (Sum rule)

= 10

∫ex dx + 7

∫sin x dx (Constant rule)

= 10ex − 7 cos x + C (Table 1)

Check your answer!

ddx (10ex − 7 cos x) = 10ex + 7 sin x X

Math 152.02

Lecture 7 - 1/19



Motion

Problem 55

Find a formula for∫

18√1−t2− 8t22 dt

Solution to Problem 55∫18√

1− t2− 8t22 dt =

∫18 · 1√

1− t2dt −

∫8 · t22 dt (Sum rule)

= 18

∫1√

1− t2dt − 8

∫t22 dt (Const. rule)

= 18 arcsin t − 8 · t23

23 + C (Table 1)

Check your answer!

ddt

(18 arcsin t − 8

23 · t23)

=18√

1− t2− 8

23 ·23t22 =18√

1− t2−8t22 X

Math 152.02

Lecture 7 - 1/19



Motion

Problem 56

Find a formula for∫ arcsin(3−π2)√

2+ sec2 u du

Solution to Problem 56

∫arcsin(3− π2)√

2+ cos u du

=

∫arcsin(3− π2)√

2du +

∫sec2 u du (Sum rule)

= arcsin(3−π2)√2

· u + tan u + C (Tables 1 and 2)

Check your answer!

ddt

(arcsin(3− π2)√

2· u + tan u

)=

arcsin(3− π2)√2

+ sec2 u X

Math 152.02

Lecture 7 - 1/19



Motion

Problem 57

Compute∫ 2

1

3√y(1−6 6

√y)2

3√y dy

Solution to Problem 57∫ 2

1

3√

y(1− 6 6√

y)2

3√

ydy =

∫ 2

1

3y12 (1− 12y

16 + 36y

26 )

y13

dy

=

∫ 2

1

3y12 − 36y

46 + 108y

56 )

y13

dy

=

∫ 2

1

3y16 − 36y

26 + 108y

36 dy

=

∫ 2

1

3y16 − 36y

13 + 108y

12 dy

= 3 · 67 y

76 − 36 · 3

4 y43 + 108 · 2

3 y32

∣∣∣21

= 187 y

76 − 27y

43 + 72y

32

∣∣∣21

=(

187 · 2

76 − 27 · 2 4

3 + 72 · 2 32

)−(

187 + 45

)

Math 152.02

Lecture 7 - 1/19



Motion

Problem 58

Find an antiderivative G (x) of g(x) = sin x + 7 satisfyingG (π) = −20.


G (x) =

∫sin x + 7 dx

= − cos x + 7x + C

Use fact that G (π) = −20 to solve for C .

−20 = G (π) = − cosπ + 7π + C

So

C = −20 + cosπ − 7π = −20 + (−1)− 7π = −21− 7π

G (x) = − cos x + 7x − 21− 7π

Math 152.02

Lecture 7 - 1/19



Motion

Problem 59

The average value of h(x) = x3 − 3x2 on [−a, a] is −8 solve for a.


have =1

a− (−a)

∫ a

−ax3 − 3x2 dx

=1

2a· ( 1

4 x4 − x3)∣∣∣a−a

=1

2a·(

[ 14 a4 − a3]− ( 1

4 (−a)4 − (−a)3))

=1

2a·(

14 a4 − a3 − 1

4 a4 + a3)

=1

2a·(− 2a3

)= −a2

Use fact that have = −8 to solve for a.

−8 = have = −a2 so a = ±√

8

Math 152.02

Lecture 7 - 1/19



Motion

Differential equations

A standard equation is an equation satisfied by a number

Example 60

x = 2 is solution to the equation

x3 − 4x2 + 7x + 1 = 7

since23 − 4 · 22 + 7 · 2 + 1 = 8− 16 + 14 + 1 = 7

A differential equation is an equation satisfied by a function.

Solving differential equation may be difficult but checking thesolution is easy.

Math 152.02

Lecture 7 - 1/19



Motion

Problem 61

Show that y = e7x is a solution to the differential equation

y ′ = 7y


If y = e7x theny ′ = 7e7x

and7y = 7e7x

soy ′ = 7e7x = 7y

Math 152.02

Lecture 7 - 1/19



Motion

Problem 62

Show that y = x3 + 3x − 2 is a solution to the differential equation

6xy − 6y ′ = x3y ′′ − 12x − 18


If y = x3 + 3x − 2 theny ′ = 3x2 + 3

andy ′′ = 6x

so 6xy − 6y ′ = 6x(x3 + 3x − 2)− 6(3x2 + 3)

= 6x4 + 18x2 − 12x − 18x2 − 18

= 6x4 − 12x − 18

andx3y ′′ − 12x − 18 = x3(6x)− 12x − 18

= 6x4 − 12x − 18

Math 152.02

Lecture 7 - 1/19



Motion

Terminology

Differential equations usually have multiple solutions. The set of allsolutions is the general solution of the differential equation.

Further constraints on solutions to different equations called initialconditions are sometimes imposed

Example 63

The general solution to y ′ = 5y is y = Ae5t (we will prove this usingsubstitution)

If we further impose the intial condition y(0) = 20 then the uniquesolution is y = 20e5t

Integration allows us to solve simple differential equations of the form

y ′ = f (x)

Math 152.02

Lecture 7 - 1/19



Motion

Problem 64

Find the general solution to the differential equation

y ′ = csc2 t − 20 · 6t + 12


y =

∫csc2 t − 20 · 6t + 12 dt

= − cot t − 20ln 6 · 6

t + 12t + C

Check your answer!

y ′ = ddt

(− cot t − 20

ln 6 · 6t + 12t + C

)= −(− csc2 t)− 20

ln 6 · ln 6 · 6t + 12

= csc2 t − 20 · 6t + 12 X

Math 152.02

Lecture 7 - 1/19



Motion

Problem 65

Find the solution to the differential equation

y ′′ = −8

satisfying the initial conditions

y(0) = 200, y ′(0) = 10


y ′ =

∫−8 dt

= −8t + C1

10 = y ′(0) = −8 · 0 + C1

so C1 = 10

y ′ = −8t + 10

y =

∫−8t + 10 dt

= −4t2 + 10t + C2

200 = −4 · 02 + 10 · 0 + C2

so C2 = 200

y = −4t2 + 10t + 200

Math 152.02

Lecture 7 - 1/19



Motion

Motion

Definition 66

If the position of an object at time t is given by the function

s(t)

then its velocity isv(t) = ds

dt

and its acceleration is

a(t) = dvdt = d2s

dt2

Math 152.02

Lecture 7 - 1/19



Motion

Problem 67

An object is dropped from rest at an initial height of 17m withconstant acceration of −9.81m/s. Give a differential equation andinitial conditions satisfied by the position function.


Constant acceleration of −9.81m/s2 gives us the differential equation

s ′′ = −9.81

Initial height of 17m gives the initial condition

s(0) = 17

The object is dropped from rest giving the initial condition

s ′(0) = 0

s ′′ = −9.81, s ′(0) = 0, s(0) = 17

Math 152.02

Lecture 7 - 1/19



Motion

Problem 68

An object is dropped from rest on the moon from a height of 2m andtakes 1.57s to hit the ground. What is the (constant) acceleration ofgravity at the moon’s surface?


We have the following diff. eq.and initial conditions

s ′′ = a, s ′(0) = 0,

s(0) = 2, s(1.57) = 0

s ′ =

∫a dt

= at + C1

0 = s ′(0) = a · 0 + C1

so C1 = 0

s =

∫at dt

= at2

2 + C2

2 = a·02

2 + C2

so C2 = 2

s(t) = at2

2 + 2

0 = s(1.57) = a·(1.57)2

2 + 2

a = 2·(−2)(1.57)2

a ≈ −1.62m/s2

Documents

Math 152.02 Calculus with Analytic Geometry II · 2011-01-24 · Math 152.02 Lecture 7 - 1/19 Basic integration rules Di erential equations Motion More basic integrals You also know