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
Basic integrationrules
Differentialequations
Motion
1 Lecture 7 - 1/19Basic integration rulesDifferential equationsMotion
Math 152.02
Lecture 7 - 1/19
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
Motion
Problem 58
Find an antiderivative G (x) of g(x) = sin x + 7 satisfyingG (π) = −20.
Solution to Problem 58
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
Basic integrationrules
Differentialequations
Motion
Problem 59
The average value of h(x) = x3 − 3x2 on [−a, a] is −8 solve for a.
Solution to Problem 59
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
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
Motion
Problem 61
Show that y = e7x is a solution to the differential equation
y ′ = 7y
Solution to Problem 61
If y = e7x theny ′ = 7e7x
and7y = 7e7x
soy ′ = 7e7x = 7y
Math 152.02
Lecture 7 - 1/19
Basic integrationrules
Differentialequations
Motion
Problem 62
Show that y = x3 + 3x − 2 is a solution to the differential equation
6xy − 6y ′ = x3y ′′ − 12x − 18
Solution to Problem 62
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
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
Motion
Problem 64
Find the general solution to the differential equation
y ′ = csc2 t − 20 · 6t + 12
Solution to Problem 64
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
Basic integrationrules
Differentialequations
Motion
Problem 65
Find the solution to the differential equation
y ′′ = −8
satisfying the initial conditions
y(0) = 200, y ′(0) = 10
Solution to Problem 65
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
Basic integrationrules
Differentialequations
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
Basic integrationrules
Differentialequations
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.
Solution to Problem 68
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
Basic integrationrules
Differentialequations
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?
Solution to Problem 68
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