Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 1
1. Evaluate the improper integralZ ∞4
1
(x − 2)(x − 3)dx .
I Using a partial fraction expansionR1
(x−2)(x−3)dx =
R h1
x−3− 1
x−2
idx = ln | x−3
x−2|+ C .
I ThereforeR∞
41
(x−2)(x−3)dx = limt→∞ ln | t−3
t−2| − ln | 1
2| =
lnh
limt→∞ | t−3t−2|i
+ ln 2 = ln(1) + ln 2 = 0 + ln 2.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 1
1. Evaluate the improper integralZ ∞4
1
(x − 2)(x − 3)dx .
I Using a partial fraction expansionR1
(x−2)(x−3)dx =
R h1
x−3− 1
x−2
idx = ln | x−3
x−2|+ C .
I ThereforeR∞
41
(x−2)(x−3)dx = limt→∞ ln | t−3
t−2| − ln | 1
2| =
lnh
limt→∞ | t−3t−2|i
+ ln 2 = ln(1) + ln 2 = 0 + ln 2.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 1
1. Evaluate the improper integralZ ∞4
1
(x − 2)(x − 3)dx .
I Using a partial fraction expansionR1
(x−2)(x−3)dx =
R h1
x−3− 1
x−2
idx = ln | x−3
x−2|+ C .
I ThereforeR∞
41
(x−2)(x−3)dx = limt→∞ ln | t−3
t−2| − ln | 1
2| =
lnh
limt→∞ | t−3t−2|i
+ ln 2 = ln(1) + ln 2 = 0 + ln 2.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 2
2.What can be said about the integrals
(i)
Z 1
0
ex
x2dx ;
(ii)
Z ∞1
cos2 x
x2dx?
I Integral (i) diverges by the Comparison Theorem since the integrand isgreater than 1
x2 .
I Integral (ii) converges by the Comparison Theorem since the integrand isless than 1
x2 .
I Therefore (i) diverges and (ii) converges.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 2
2.What can be said about the integrals
(i)
Z 1
0
ex
x2dx ;
(ii)
Z ∞1
cos2 x
x2dx?
I Integral (i) diverges by the Comparison Theorem since the integrand isgreater than 1
x2 .
I Integral (ii) converges by the Comparison Theorem since the integrand isless than 1
x2 .
I Therefore (i) diverges and (ii) converges.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 2
2.What can be said about the integrals
(i)
Z 1
0
ex
x2dx ;
(ii)
Z ∞1
cos2 x
x2dx?
I Integral (i) diverges by the Comparison Theorem since the integrand isgreater than 1
x2 .
I Integral (ii) converges by the Comparison Theorem since the integrand isless than 1
x2 .
I Therefore (i) diverges and (ii) converges.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 2
2.What can be said about the integrals
(i)
Z 1
0
ex
x2dx ;
(ii)
Z ∞1
cos2 x
x2dx?
I Integral (i) diverges by the Comparison Theorem since the integrand isgreater than 1
x2 .
I Integral (ii) converges by the Comparison Theorem since the integrand isless than 1
x2 .
I Therefore (i) diverges and (ii) converges.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 3
3. Which of the following is an expression for the arclength of the curvey = cos x between x = −π
2and x = π
2?
(a)
Z π2
−π2
p1 + sin2 x dx . (b)
Z π2
−π2
p1− cos2 x dx .
(c) 2
Z π2
0
p1 + 2 sin2 x dx . (d)
Z π2
−π2
p1− sin2 x dx .
(e) π2
2
I The arclength formula gives the answer asR b
a
p1 + (f ′(x))2dx .
I =R π
2−π
2
p1 + (−sinx)2dx =
R π2−π
2
p1 + sin2 x dx .
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 3
3. Which of the following is an expression for the arclength of the curvey = cos x between x = −π
2and x = π
2?
(a)
Z π2
−π2
p1 + sin2 x dx . (b)
Z π2
−π2
p1− cos2 x dx .
(c) 2
Z π2
0
p1 + 2 sin2 x dx . (d)
Z π2
−π2
p1− sin2 x dx .
(e) π2
2
I The arclength formula gives the answer asR b
a
p1 + (f ′(x))2dx .
I =R π
2−π
2
p1 + (−sinx)2dx =
R π2−π
2
p1 + sin2 x dx .
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 3
3. Which of the following is an expression for the arclength of the curvey = cos x between x = −π
2and x = π
2?
(a)
Z π2
−π2
p1 + sin2 x dx . (b)
Z π2
−π2
p1− cos2 x dx .
(c) 2
Z π2
0
p1 + 2 sin2 x dx . (d)
Z π2
−π2
p1− sin2 x dx .
(e) π2
2
I The arclength formula gives the answer asR b
a
p1 + (f ′(x))2dx .
I =R π
2−π
2
p1 + (−sinx)2dx =
R π2−π
2
p1 + sin2 x dx .
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 4
4. Use Euler’s method with step size 0.1 to estimate y(1.2) where y(x) is thesolution to the initial value problem
y ′ = xy + 1 y(1) = 0.
I x0 = 1, y0 = 0
I x1 = x0 + h = 1.1, y1 = y0 + h(x0y0 + 1) = 0 + (0.1)(1 · 0 + 1) = 0.1
I x2 = x1 + h = 1.2, y2 = y1 + h(x1y1 + 1) = 0.1 + (0.1)((1.1) · (0.1) + 1)
I = 0.1 + 0.1(0.11 + 1) = 0.1 + 0.1(1.11) = 0.1 + 0.111 = 0.211
I y(1.2) ≈ 0.211.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 4
4. Use Euler’s method with step size 0.1 to estimate y(1.2) where y(x) is thesolution to the initial value problem
y ′ = xy + 1 y(1) = 0.
I x0 = 1, y0 = 0
I x1 = x0 + h = 1.1, y1 = y0 + h(x0y0 + 1) = 0 + (0.1)(1 · 0 + 1) = 0.1
I x2 = x1 + h = 1.2, y2 = y1 + h(x1y1 + 1) = 0.1 + (0.1)((1.1) · (0.1) + 1)
I = 0.1 + 0.1(0.11 + 1) = 0.1 + 0.1(1.11) = 0.1 + 0.111 = 0.211
I y(1.2) ≈ 0.211.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 4
4. Use Euler’s method with step size 0.1 to estimate y(1.2) where y(x) is thesolution to the initial value problem
y ′ = xy + 1 y(1) = 0.
I x0 = 1, y0 = 0
I x1 = x0 + h = 1.1, y1 = y0 + h(x0y0 + 1) = 0 + (0.1)(1 · 0 + 1) = 0.1
I x2 = x1 + h = 1.2, y2 = y1 + h(x1y1 + 1) = 0.1 + (0.1)((1.1) · (0.1) + 1)
I = 0.1 + 0.1(0.11 + 1) = 0.1 + 0.1(1.11) = 0.1 + 0.111 = 0.211
I y(1.2) ≈ 0.211.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 4
4. Use Euler’s method with step size 0.1 to estimate y(1.2) where y(x) is thesolution to the initial value problem
y ′ = xy + 1 y(1) = 0.
I x0 = 1, y0 = 0
I x1 = x0 + h = 1.1, y1 = y0 + h(x0y0 + 1) = 0 + (0.1)(1 · 0 + 1) = 0.1
I x2 = x1 + h = 1.2, y2 = y1 + h(x1y1 + 1) = 0.1 + (0.1)((1.1) · (0.1) + 1)
I = 0.1 + 0.1(0.11 + 1) = 0.1 + 0.1(1.11) = 0.1 + 0.111 = 0.211
I y(1.2) ≈ 0.211.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 4
4. Use Euler’s method with step size 0.1 to estimate y(1.2) where y(x) is thesolution to the initial value problem
y ′ = xy + 1 y(1) = 0.
I x0 = 1, y0 = 0
I x1 = x0 + h = 1.1, y1 = y0 + h(x0y0 + 1) = 0 + (0.1)(1 · 0 + 1) = 0.1
I x2 = x1 + h = 1.2, y2 = y1 + h(x1y1 + 1) = 0.1 + (0.1)((1.1) · (0.1) + 1)
I = 0.1 + 0.1(0.11 + 1) = 0.1 + 0.1(1.11) = 0.1 + 0.111 = 0.211
I y(1.2) ≈ 0.211.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 4
4. Use Euler’s method with step size 0.1 to estimate y(1.2) where y(x) is thesolution to the initial value problem
y ′ = xy + 1 y(1) = 0.
I x0 = 1, y0 = 0
I x1 = x0 + h = 1.1, y1 = y0 + h(x0y0 + 1) = 0 + (0.1)(1 · 0 + 1) = 0.1
I x2 = x1 + h = 1.2, y2 = y1 + h(x1y1 + 1) = 0.1 + (0.1)((1.1) · (0.1) + 1)
I = 0.1 + 0.1(0.11 + 1) = 0.1 + 0.1(1.11) = 0.1 + 0.111 = 0.211
I y(1.2) ≈ 0.211.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 5
5. Which of the following gives the direction field for the differential equation
y ′ = y 2 − x2
Note the letter corresponding to each graph is at the lower left of the graph.
I We draw the direction vectors at a few points around the origin, say(0, 0), (0, 1), (1, 0), (1, 1), (−1, 0), (0,−1), (−1,−1) to get a feel for whatthe vector fiels might look like. (we will do this on paper)
I We note and sketch a few general patterns, if x = ±y , then y 2 − x2 = 0and the direction vector is horizontal.
I If |x | > |y | the slope is negative and if |y | > |x | the slope is positive.
I These and other observations allow us to eliminate some possibili-ties and form a picture to see that the direction field is the one shown below.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 5
5. Which of the following gives the direction field for the differential equation
y ′ = y 2 − x2
Note the letter corresponding to each graph is at the lower left of the graph.
I We draw the direction vectors at a few points around the origin, say(0, 0), (0, 1), (1, 0), (1, 1), (−1, 0), (0,−1), (−1,−1) to get a feel for whatthe vector fiels might look like. (we will do this on paper)
I We note and sketch a few general patterns, if x = ±y , then y 2 − x2 = 0and the direction vector is horizontal.
I If |x | > |y | the slope is negative and if |y | > |x | the slope is positive.
I These and other observations allow us to eliminate some possibili-ties and form a picture to see that the direction field is the one shown below.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 5
5. Which of the following gives the direction field for the differential equation
y ′ = y 2 − x2
Note the letter corresponding to each graph is at the lower left of the graph.
I We draw the direction vectors at a few points around the origin, say(0, 0), (0, 1), (1, 0), (1, 1), (−1, 0), (0,−1), (−1,−1) to get a feel for whatthe vector fiels might look like. (we will do this on paper)
I We note and sketch a few general patterns, if x = ±y , then y 2 − x2 = 0and the direction vector is horizontal.
I If |x | > |y | the slope is negative and if |y | > |x | the slope is positive.
I These and other observations allow us to eliminate some possibili-ties and form a picture to see that the direction field is the one shown below.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 5
5. Which of the following gives the direction field for the differential equation
y ′ = y 2 − x2
Note the letter corresponding to each graph is at the lower left of the graph.
I We draw the direction vectors at a few points around the origin, say(0, 0), (0, 1), (1, 0), (1, 1), (−1, 0), (0,−1), (−1,−1) to get a feel for whatthe vector fiels might look like. (we will do this on paper)
I We note and sketch a few general patterns, if x = ±y , then y 2 − x2 = 0and the direction vector is horizontal.
I If |x | > |y | the slope is negative and if |y | > |x | the slope is positive.
I These and other observations allow us to eliminate some possibili-ties and form a picture to see that the direction field is the one shown below.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 5
5. Which of the following gives the direction field for the differential equation
y ′ = y 2 − x2
Note the letter corresponding to each graph is at the lower left of the graph.
I We draw the direction vectors at a few points around the origin, say(0, 0), (0, 1), (1, 0), (1, 1), (−1, 0), (0,−1), (−1,−1) to get a feel for whatthe vector fiels might look like. (we will do this on paper)
I We note and sketch a few general patterns, if x = ±y , then y 2 − x2 = 0and the direction vector is horizontal.
I If |x | > |y | the slope is negative and if |y | > |x | the slope is positive.
I These and other observations allow us to eliminate some possibili-ties and form a picture to see that the direction field is the one shown below.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 6
6. Find the solution of the differential equation
dy
dx=
p1− y 2
1 + x2
with initial condition y(0) = 0.
I Separating variables here givesR
dy√1−y2
=R
dx1+x2 .
I Integrating both sides gives arcsin y = arctan x + C .
I substituting y(0) = 0 we find C = 0.
I Therefore y = sin(arctan x) = x√1+x2
(from triangle).
x1 + x2
1
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 6
6. Find the solution of the differential equation
dy
dx=
p1− y 2
1 + x2
with initial condition y(0) = 0.
I Separating variables here givesR
dy√1−y2
=R
dx1+x2 .
I Integrating both sides gives arcsin y = arctan x + C .
I substituting y(0) = 0 we find C = 0.
I Therefore y = sin(arctan x) = x√1+x2
(from triangle).
x1 + x2
1
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 6
6. Find the solution of the differential equation
dy
dx=
p1− y 2
1 + x2
with initial condition y(0) = 0.
I Separating variables here givesR
dy√1−y2
=R
dx1+x2 .
I Integrating both sides gives arcsin y = arctan x + C .
I substituting y(0) = 0 we find C = 0.
I Therefore y = sin(arctan x) = x√1+x2
(from triangle).
x1 + x2
1
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 6
6. Find the solution of the differential equation
dy
dx=
p1− y 2
1 + x2
with initial condition y(0) = 0.
I Separating variables here givesR
dy√1−y2
=R
dx1+x2 .
I Integrating both sides gives arcsin y = arctan x + C .
I substituting y(0) = 0 we find C = 0.
I Therefore y = sin(arctan x) = x√1+x2
(from triangle).
x1 + x2
1
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 6
6. Find the solution of the differential equation
dy
dx=
p1− y 2
1 + x2
with initial condition y(0) = 0.
I Separating variables here givesR
dy√1−y2
=R
dx1+x2 .
I Integrating both sides gives arcsin y = arctan x + C .
I substituting y(0) = 0 we find C = 0.
I Therefore y = sin(arctan x) = x√1+x2
(from triangle).
x1 + x2
1
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 7
7. Find a general solution, valid for −π2
< x < π2
, of the differential equation
dy
dx− (tan x)y = 1.
I The integrating factor here is I = eR− tan xdx = e ln(cos x) = cos x
I Multiplying the equation by the integrating factor, we get(cos x) dy
dx− (cos x)(tan x)y = cos x or
d(y cos x)
dx= cos x .
I so a general solution is given by y cos x =R
cos xdx = sin x + C
I Therefore y = sin x+Ccos x
.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 7
7. Find a general solution, valid for −π2
< x < π2
, of the differential equation
dy
dx− (tan x)y = 1.
I The integrating factor here is I = eR− tan xdx = e ln(cos x) = cos x
I Multiplying the equation by the integrating factor, we get(cos x) dy
dx− (cos x)(tan x)y = cos x or
d(y cos x)
dx= cos x .
I so a general solution is given by y cos x =R
cos xdx = sin x + C
I Therefore y = sin x+Ccos x
.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 7
7. Find a general solution, valid for −π2
< x < π2
, of the differential equation
dy
dx− (tan x)y = 1.
I The integrating factor here is I = eR− tan xdx = e ln(cos x) = cos x
I Multiplying the equation by the integrating factor, we get(cos x) dy
dx− (cos x)(tan x)y = cos x or
d(y cos x)
dx= cos x .
I so a general solution is given by y cos x =R
cos xdx = sin x + C
I Therefore y = sin x+Ccos x
.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 7
7. Find a general solution, valid for −π2
< x < π2
, of the differential equation
dy
dx− (tan x)y = 1.
I The integrating factor here is I = eR− tan xdx = e ln(cos x) = cos x
I Multiplying the equation by the integrating factor, we get(cos x) dy
dx− (cos x)(tan x)y = cos x or
d(y cos x)
dx= cos x .
I so a general solution is given by y cos x =R
cos xdx = sin x + C
I Therefore y = sin x+Ccos x
.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 7
7. Find a general solution, valid for −π2
< x < π2
, of the differential equation
dy
dx− (tan x)y = 1.
I The integrating factor here is I = eR− tan xdx = e ln(cos x) = cos x
I Multiplying the equation by the integrating factor, we get(cos x) dy
dx− (cos x)(tan x)y = cos x or
d(y cos x)
dx= cos x .
I so a general solution is given by y cos x =R
cos xdx = sin x + C
I Therefore y = sin x+Ccos x
.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 8
8. Find the limit: limn→∞
(−1)nn2
n2 + n + 1
I limn→∞n2
n2+n+1= limn→∞
11+1/n+1/n2 = 1 6= 0
I Therefore
limn→∞
(−1)nn2
n2 + n + 1does not exist.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 8
8. Find the limit: limn→∞
(−1)nn2
n2 + n + 1
I limn→∞n2
n2+n+1= limn→∞
11+1/n+1/n2 = 1 6= 0
I Therefore
limn→∞
(−1)nn2
n2 + n + 1does not exist.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 8
8. Find the limit: limn→∞
(−1)nn2
n2 + n + 1
I limn→∞n2
n2+n+1= limn→∞
11+1/n+1/n2 = 1 6= 0
I Therefore
limn→∞
(−1)nn2
n2 + n + 1does not exist.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 9
9. Calculate
limn→∞
(ln n)2
n.
I Use L’Hospital’s rule (twice):
I limx→∞(ln x)2
x= limx→∞
((ln x)2)′
x′
I = limn→∞2 1
xln x
1
I = limx→∞(2 ln x)′
x′
I = limx→∞2x1
= 0.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 9
9. Calculate
limn→∞
(ln n)2
n.
I Use L’Hospital’s rule (twice):
I limx→∞(ln x)2
x= limx→∞
((ln x)2)′
x′
I = limn→∞2 1
xln x
1
I = limx→∞(2 ln x)′
x′
I = limx→∞2x1
= 0.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 9
9. Calculate
limn→∞
(ln n)2
n.
I Use L’Hospital’s rule (twice):
I limx→∞(ln x)2
x= limx→∞
((ln x)2)′
x′
I = limn→∞2 1
xln x
1
I = limx→∞(2 ln x)′
x′
I = limx→∞2x1
= 0.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 9
9. Calculate
limn→∞
(ln n)2
n.
I Use L’Hospital’s rule (twice):
I limx→∞(ln x)2
x= limx→∞
((ln x)2)′
x′
I = limn→∞2 1
xln x
1
I = limx→∞(2 ln x)′
x′
I = limx→∞2x1
= 0.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 9
9. Calculate
limn→∞
(ln n)2
n.
I Use L’Hospital’s rule (twice):
I limx→∞(ln x)2
x= limx→∞
((ln x)2)′
x′
I = limn→∞2 1
xln x
1
I = limx→∞(2 ln x)′
x′
I = limx→∞2x1
= 0.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 9
9. Calculate
limn→∞
(ln n)2
n.
I Use L’Hospital’s rule (twice):
I limx→∞(ln x)2
x= limx→∞
((ln x)2)′
x′
I = limn→∞2 1
xln x
1
I = limx→∞(2 ln x)′
x′
I = limx→∞2x1
= 0.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 10
10. Find∞Xn=1
22n
3 · 5n−1.
I This is a geometric series of forP∞
n=1 arn−1 = a + ar + . . . .
IP∞
n=122n
3·5n−1 = 43
+ 1615
+ . . .
I a = 43
and r = 1615
‹43
= 45.
I ThereforeP∞
n=122n
3·5n−1 = a1−r
= 4/31−(4/5)
= 4/31/5
= 203.
I AlternativelyP∞
n=122n
3·5n−1 =P∞
n=14n
3·5n−1 =P∞
n=14·4n−1
3·5n−1 =43
P∞n=1
`45
´n−1= 4
3
“1
1− 45
”= 20
3.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 10
10. Find∞Xn=1
22n
3 · 5n−1.
I This is a geometric series of forP∞
n=1 arn−1 = a + ar + . . . .
IP∞
n=122n
3·5n−1 = 43
+ 1615
+ . . .
I a = 43
and r = 1615
‹43
= 45.
I ThereforeP∞
n=122n
3·5n−1 = a1−r
= 4/31−(4/5)
= 4/31/5
= 203.
I AlternativelyP∞
n=122n
3·5n−1 =P∞
n=14n
3·5n−1 =P∞
n=14·4n−1
3·5n−1 =43
P∞n=1
`45
´n−1= 4
3
“1
1− 45
”= 20
3.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 10
10. Find∞Xn=1
22n
3 · 5n−1.
I This is a geometric series of forP∞
n=1 arn−1 = a + ar + . . . .
IP∞
n=122n
3·5n−1 = 43
+ 1615
+ . . .
I a = 43
and r = 1615
‹43
= 45.
I ThereforeP∞
n=122n
3·5n−1 = a1−r
= 4/31−(4/5)
= 4/31/5
= 203.
I AlternativelyP∞
n=122n
3·5n−1 =P∞
n=14n
3·5n−1 =P∞
n=14·4n−1
3·5n−1 =43
P∞n=1
`45
´n−1= 4
3
“1
1− 45
”= 20
3.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 10
10. Find∞Xn=1
22n
3 · 5n−1.
I This is a geometric series of forP∞
n=1 arn−1 = a + ar + . . . .
IP∞
n=122n
3·5n−1 = 43
+ 1615
+ . . .
I a = 43
and r = 1615
‹43
= 45.
I ThereforeP∞
n=122n
3·5n−1 = a1−r
= 4/31−(4/5)
= 4/31/5
= 203.
I AlternativelyP∞
n=122n
3·5n−1 =P∞
n=14n
3·5n−1 =P∞
n=14·4n−1
3·5n−1 =43
P∞n=1
`45
´n−1= 4
3
“1
1− 45
”= 20
3.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 10
10. Find∞Xn=1
22n
3 · 5n−1.
I This is a geometric series of forP∞
n=1 arn−1 = a + ar + . . . .
IP∞
n=122n
3·5n−1 = 43
+ 1615
+ . . .
I a = 43
and r = 1615
‹43
= 45.
I ThereforeP∞
n=122n
3·5n−1 = a1−r
= 4/31−(4/5)
= 4/31/5
= 203.
I AlternativelyP∞
n=122n
3·5n−1 =P∞
n=14n
3·5n−1 =P∞
n=14·4n−1
3·5n−1 =43
P∞n=1
`45
´n−1= 4
3
“1
1− 45
”= 20
3.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 10
10. Find∞Xn=1
22n
3 · 5n−1.
I This is a geometric series of forP∞
n=1 arn−1 = a + ar + . . . .
IP∞
n=122n
3·5n−1 = 43
+ 1615
+ . . .
I a = 43
and r = 1615
‹43
= 45.
I ThereforeP∞
n=122n
3·5n−1 = a1−r
= 4/31−(4/5)
= 4/31/5
= 203.
I AlternativelyP∞
n=122n
3·5n−1 =P∞
n=14n
3·5n−1 =P∞
n=14·4n−1
3·5n−1 =43
P∞n=1
`45
´n−1= 4
3
“1
1− 45
”= 20
3.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 11
11. Calculate the arc length of the curve if y = x2
4− ln(
√x), where 2 ≤ x ≤ 4.
I Recall
L =
Z b
a
p1 + (y ′)2dx .
I Note
y ′ =x
2− 1√
x· 1
2√
x=
1
2
“x − 1
x
”.
I Thus
1 + (y ′)2 = 1 +1
4
“x − 1
x
”2
= 1 +1
4
“x2 − 2x
1
x+
1
x2
”= 1 +
1
4
“x2 − 2 +
1
x2
”= 1 +
1
4x2 − 1
2+
1
4x2=
1
4x2 +
1
2+
1
4x2=
1
4
“x2 + 2x
1
x+
1
x2
”=
1
4
“x +
1
x
”2
.
I Therefore
L =
Z 4
2
p1/4(x + 1/x)2dx =
Z 4
2
1
2
“x+
1
x
”dx =
1
2
»x2
2+ ln x
–4
2
= 3+1
2ln 2.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 11
11. Calculate the arc length of the curve if y = x2
4− ln(
√x), where 2 ≤ x ≤ 4.
I Recall
L =
Z b
a
p1 + (y ′)2dx .
I Note
y ′ =x
2− 1√
x· 1
2√
x=
1
2
“x − 1
x
”.
I Thus
1 + (y ′)2 = 1 +1
4
“x − 1
x
”2
= 1 +1
4
“x2 − 2x
1
x+
1
x2
”= 1 +
1
4
“x2 − 2 +
1
x2
”= 1 +
1
4x2 − 1
2+
1
4x2=
1
4x2 +
1
2+
1
4x2=
1
4
“x2 + 2x
1
x+
1
x2
”=
1
4
“x +
1
x
”2
.
I Therefore
L =
Z 4
2
p1/4(x + 1/x)2dx =
Z 4
2
1
2
“x+
1
x
”dx =
1
2
»x2
2+ ln x
–4
2
= 3+1
2ln 2.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 11
11. Calculate the arc length of the curve if y = x2
4− ln(
√x), where 2 ≤ x ≤ 4.
I Recall
L =
Z b
a
p1 + (y ′)2dx .
I Note
y ′ =x
2− 1√
x· 1
2√
x=
1
2
“x − 1
x
”.
I Thus
1 + (y ′)2 = 1 +1
4
“x − 1
x
”2
= 1 +1
4
“x2 − 2x
1
x+
1
x2
”= 1 +
1
4
“x2 − 2 +
1
x2
”= 1 +
1
4x2 − 1
2+
1
4x2=
1
4x2 +
1
2+
1
4x2=
1
4
“x2 + 2x
1
x+
1
x2
”=
1
4
“x +
1
x
”2
.
I Therefore
L =
Z 4
2
p1/4(x + 1/x)2dx =
Z 4
2
1
2
“x+
1
x
”dx =
1
2
»x2
2+ ln x
–4
2
= 3+1
2ln 2.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 11
11. Calculate the arc length of the curve if y = x2
4− ln(
√x), where 2 ≤ x ≤ 4.
I Recall
L =
Z b
a
p1 + (y ′)2dx .
I Note
y ′ =x
2− 1√
x· 1
2√
x=
1
2
“x − 1
x
”.
I Thus
1 + (y ′)2 = 1 +1
4
“x − 1
x
”2
= 1 +1
4
“x2 − 2x
1
x+
1
x2
”= 1 +
1
4
“x2 − 2 +
1
x2
”= 1 +
1
4x2 − 1
2+
1
4x2=
1
4x2 +
1
2+
1
4x2=
1
4
“x2 + 2x
1
x+
1
x2
”=
1
4
“x +
1
x
”2
.
I Therefore
L =
Z 4
2
p1/4(x + 1/x)2dx =
Z 4
2
1
2
“x+
1
x
”dx =
1
2
»x2
2+ ln x
–4
2
= 3+1
2ln 2.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 11
11. Calculate the arc length of the curve if y = x2
4− ln(
√x), where 2 ≤ x ≤ 4.
I Recall
L =
Z b
a
p1 + (y ′)2dx .
I Note
y ′ =x
2− 1√
x· 1
2√
x=
1
2
“x − 1
x
”.
I Thus
1 + (y ′)2 = 1 +1
4
“x − 1
x
”2
= 1 +1
4
“x2 − 2x
1
x+
1
x2
”= 1 +
1
4
“x2 − 2 +
1
x2
”= 1 +
1
4x2 − 1
2+
1
4x2=
1
4x2 +
1
2+
1
4x2=
1
4
“x2 + 2x
1
x+
1
x2
”=
1
4
“x +
1
x
”2
.
I Therefore
L =
Z 4
2
p1/4(x + 1/x)2dx =
Z 4
2
1
2
“x+
1
x
”dx =
1
2
»x2
2+ ln x
–4
2
= 3+1
2ln 2.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 12
12. (a) Circle the letter below alongside the trapezoidal approximation to
ln 3 =
Z 3
1
1
xdx using n = 8
A
Z 3
1
1
xdx ≈ 1
8
h1+2
“4
5
”+2“2
3
”+2“4
7
”+2“1
2
”+2“4
9
”+2“2
5
”+2“ 4
11
”+“1
3
”i
B
Z 3
1
1
xdx ≈ 1
12
h1+4
“4
5
”+2“2
3
”+4“4
7
”+2“1
2
”+4“4
9
”+2“2
5
”+4“ 4
11
”+“1
3
”i
C
Z 3
1
1
xdx ≈ 1
8
h1+“4
5
”+“2
3
”+“4
7
”+“1
2
”+“4
9
”+“2
5
”+“ 4
11
”+“1
3
”i
I ∆x = b−a8
= 28
= 14.
I T8 = ∆x2
[f (1) + 2f ( 54) + 2f ( 3
2) + 2f ( 7
4) + 2f (2) + 2f ( 9
4) + 2f ( 5
2) +
2f ( 114
) + f (3)] = A (f (x) = 1x
).
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 12
12. (a) Circle the letter below alongside the trapezoidal approximation to
ln 3 =
Z 3
1
1
xdx using n = 8
A
Z 3
1
1
xdx ≈ 1
8
h1+2
“4
5
”+2“2
3
”+2“4
7
”+2“1
2
”+2“4
9
”+2“2
5
”+2“ 4
11
”+“1
3
”i
B
Z 3
1
1
xdx ≈ 1
12
h1+4
“4
5
”+2“2
3
”+4“4
7
”+2“1
2
”+4“4
9
”+2“2
5
”+4“ 4
11
”+“1
3
”i
C
Z 3
1
1
xdx ≈ 1
8
h1+“4
5
”+“2
3
”+“4
7
”+“1
2
”+“4
9
”+“2
5
”+“ 4
11
”+“1
3
”i
I ∆x = b−a8
= 28
= 14.
I T8 = ∆x2
[f (1) + 2f ( 54) + 2f ( 3
2) + 2f ( 7
4) + 2f (2) + 2f ( 9
4) + 2f ( 5
2) +
2f ( 114
) + f (3)] = A (f (x) = 1x
).
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 12
12. (a) Circle the letter below alongside the trapezoidal approximation to
ln 3 =
Z 3
1
1
xdx using n = 8
A
Z 3
1
1
xdx ≈ 1
8
h1+2
“4
5
”+2“2
3
”+2“4
7
”+2“1
2
”+2“4
9
”+2“2
5
”+2“ 4
11
”+“1
3
”i
B
Z 3
1
1
xdx ≈ 1
12
h1+4
“4
5
”+2“2
3
”+4“4
7
”+2“1
2
”+4“4
9
”+2“2
5
”+4“ 4
11
”+“1
3
”i
C
Z 3
1
1
xdx ≈ 1
8
h1+“4
5
”+“2
3
”+“4
7
”+“1
2
”+“4
9
”+“2
5
”+“ 4
11
”+“1
3
”i
I ∆x = b−a8
= 28
= 14.
I T8 = ∆x2
[f (1) + 2f ( 54) + 2f ( 3
2) + 2f ( 7
4) + 2f (2) + 2f ( 9
4) + 2f ( 5
2) +
2f ( 114
) + f (3)] = A (f (x) = 1x
).
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 1212.(b) Recall that the error ET in the trapezoidal rule for approximatingR b
af (x)dx satisfies ˛̨̨ Z b
af (x)dx − Tn
˛̨̨= |ET | ≤
K(b − a)3
12n2
whenever |f ′′(x)| ≤ K for all a ≤ x ≤ b. Use the above error bound todetermine a value of n for which the trapezoidal approximation to
ln 3 =
Z 3
1
1
xdx has an error |ET | ≤ 1
310−4.
I f (x) = 1x, f ′(x) = −1
x2 , f ′′(x) = 2x3
I Since |f ′′(x)| = 2x3 is decreasing on the interval 1 ≤ x ≤ 2, we have
|f ′′(x)| ≤ f ′′(1) = 2 for 1 ≤ x ≤ 2. Hence, we can use K = 2 in the errorbound above.
I For the trapezoidal approximation Tn, we have
|ET | ≤K(b − a)3
12n2=
2(3− 1)3
12n2=
16
12n2=
4
3n2
I If we find a value of n for which 1310−4 ≥ 4
3n2 , then we will have
|ET | ≤ 1310−4.
I 1310−4 ≥ 4
3n2 → n2 ≥ 4 · 104 → n ≥ 2 · 102 = 200.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 1212.(b) Recall that the error ET in the trapezoidal rule for approximatingR b
af (x)dx satisfies ˛̨̨ Z b
af (x)dx − Tn
˛̨̨= |ET | ≤
K(b − a)3
12n2
whenever |f ′′(x)| ≤ K for all a ≤ x ≤ b. Use the above error bound todetermine a value of n for which the trapezoidal approximation to
ln 3 =
Z 3
1
1
xdx has an error |ET | ≤ 1
310−4.
I f (x) = 1x, f ′(x) = −1
x2 , f ′′(x) = 2x3
I Since |f ′′(x)| = 2x3 is decreasing on the interval 1 ≤ x ≤ 2, we have
|f ′′(x)| ≤ f ′′(1) = 2 for 1 ≤ x ≤ 2. Hence, we can use K = 2 in the errorbound above.
I For the trapezoidal approximation Tn, we have
|ET | ≤K(b − a)3
12n2=
2(3− 1)3
12n2=
16
12n2=
4
3n2
I If we find a value of n for which 1310−4 ≥ 4
3n2 , then we will have
|ET | ≤ 1310−4.
I 1310−4 ≥ 4
3n2 → n2 ≥ 4 · 104 → n ≥ 2 · 102 = 200.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 1212.(b) Recall that the error ET in the trapezoidal rule for approximatingR b
af (x)dx satisfies ˛̨̨ Z b
af (x)dx − Tn
˛̨̨= |ET | ≤
K(b − a)3
12n2
whenever |f ′′(x)| ≤ K for all a ≤ x ≤ b. Use the above error bound todetermine a value of n for which the trapezoidal approximation to
ln 3 =
Z 3
1
1
xdx has an error |ET | ≤ 1
310−4.
I f (x) = 1x, f ′(x) = −1
x2 , f ′′(x) = 2x3
I Since |f ′′(x)| = 2x3 is decreasing on the interval 1 ≤ x ≤ 2, we have
|f ′′(x)| ≤ f ′′(1) = 2 for 1 ≤ x ≤ 2. Hence, we can use K = 2 in the errorbound above.
I For the trapezoidal approximation Tn, we have
|ET | ≤K(b − a)3
12n2=
2(3− 1)3
12n2=
16
12n2=
4
3n2
I If we find a value of n for which 1310−4 ≥ 4
3n2 , then we will have
|ET | ≤ 1310−4.
I 1310−4 ≥ 4
3n2 → n2 ≥ 4 · 104 → n ≥ 2 · 102 = 200.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 1212.(b) Recall that the error ET in the trapezoidal rule for approximatingR b
af (x)dx satisfies ˛̨̨ Z b
af (x)dx − Tn
˛̨̨= |ET | ≤
K(b − a)3
12n2
whenever |f ′′(x)| ≤ K for all a ≤ x ≤ b. Use the above error bound todetermine a value of n for which the trapezoidal approximation to
ln 3 =
Z 3
1
1
xdx has an error |ET | ≤ 1
310−4.
I f (x) = 1x, f ′(x) = −1
x2 , f ′′(x) = 2x3
I Since |f ′′(x)| = 2x3 is decreasing on the interval 1 ≤ x ≤ 2, we have
|f ′′(x)| ≤ f ′′(1) = 2 for 1 ≤ x ≤ 2. Hence, we can use K = 2 in the errorbound above.
I For the trapezoidal approximation Tn, we have
|ET | ≤K(b − a)3
12n2=
2(3− 1)3
12n2=
16
12n2=
4
3n2
I If we find a value of n for which 1310−4 ≥ 4
3n2 , then we will have
|ET | ≤ 1310−4.
I 1310−4 ≥ 4
3n2 → n2 ≥ 4 · 104 → n ≥ 2 · 102 = 200.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 1212.(b) Recall that the error ET in the trapezoidal rule for approximatingR b
af (x)dx satisfies ˛̨̨ Z b
af (x)dx − Tn
˛̨̨= |ET | ≤
K(b − a)3
12n2
whenever |f ′′(x)| ≤ K for all a ≤ x ≤ b. Use the above error bound todetermine a value of n for which the trapezoidal approximation to
ln 3 =
Z 3
1
1
xdx has an error |ET | ≤ 1
310−4.
I f (x) = 1x, f ′(x) = −1
x2 , f ′′(x) = 2x3
I Since |f ′′(x)| = 2x3 is decreasing on the interval 1 ≤ x ≤ 2, we have
|f ′′(x)| ≤ f ′′(1) = 2 for 1 ≤ x ≤ 2. Hence, we can use K = 2 in the errorbound above.
I For the trapezoidal approximation Tn, we have
|ET | ≤K(b − a)3
12n2=
2(3− 1)3
12n2=
16
12n2=
4
3n2
I If we find a value of n for which 1310−4 ≥ 4
3n2 , then we will have
|ET | ≤ 1310−4.
I 1310−4 ≥ 4
3n2 → n2 ≥ 4 · 104 → n ≥ 2 · 102 = 200.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 1212.(b) Recall that the error ET in the trapezoidal rule for approximatingR b
af (x)dx satisfies ˛̨̨ Z b
af (x)dx − Tn
˛̨̨= |ET | ≤
K(b − a)3
12n2
whenever |f ′′(x)| ≤ K for all a ≤ x ≤ b. Use the above error bound todetermine a value of n for which the trapezoidal approximation to
ln 3 =
Z 3
1
1
xdx has an error |ET | ≤ 1
310−4.
I f (x) = 1x, f ′(x) = −1
x2 , f ′′(x) = 2x3
I Since |f ′′(x)| = 2x3 is decreasing on the interval 1 ≤ x ≤ 2, we have
|f ′′(x)| ≤ f ′′(1) = 2 for 1 ≤ x ≤ 2. Hence, we can use K = 2 in the errorbound above.
I For the trapezoidal approximation Tn, we have
|ET | ≤K(b − a)3
12n2=
2(3− 1)3
12n2=
16
12n2=
4
3n2
I If we find a value of n for which 1310−4 ≥ 4
3n2 , then we will have
|ET | ≤ 1310−4.
I 1310−4 ≥ 4
3n2 → n2 ≥ 4 · 104 → n ≥ 2 · 102 = 200.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 13
13. Find the family of orthogonal trajectories to the family of curves given by
y = kx2.
I For the family of curves described above dydx
= 2kx .
I Also y = kx2 giving k = yx2 .
I Therefore the given family satisfy the differential equationdydx
= 2 yx2 x = 2 y
x.
I The family of orthogonal trajectories must satisfy the differential equationdydx
= −x2y
.
I Separating the variables, we get 2ydy = −xdx
I and
2
Zydy = −
Zxdx , or y 2 =
−x2
2+ C .
I Hence our family of orthogonal trajectories is a family of curves of theform
y 2 +x2
2= C ,
a family of ellipses.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 13
13. Find the family of orthogonal trajectories to the family of curves given by
y = kx2.
I For the family of curves described above dydx
= 2kx .
I Also y = kx2 giving k = yx2 .
I Therefore the given family satisfy the differential equationdydx
= 2 yx2 x = 2 y
x.
I The family of orthogonal trajectories must satisfy the differential equationdydx
= −x2y
.
I Separating the variables, we get 2ydy = −xdx
I and
2
Zydy = −
Zxdx , or y 2 =
−x2
2+ C .
I Hence our family of orthogonal trajectories is a family of curves of theform
y 2 +x2
2= C ,
a family of ellipses.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 13
13. Find the family of orthogonal trajectories to the family of curves given by
y = kx2.
I For the family of curves described above dydx
= 2kx .
I Also y = kx2 giving k = yx2 .
I Therefore the given family satisfy the differential equationdydx
= 2 yx2 x = 2 y
x.
I The family of orthogonal trajectories must satisfy the differential equationdydx
= −x2y
.
I Separating the variables, we get 2ydy = −xdx
I and
2
Zydy = −
Zxdx , or y 2 =
−x2
2+ C .
I Hence our family of orthogonal trajectories is a family of curves of theform
y 2 +x2
2= C ,
a family of ellipses.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 13
13. Find the family of orthogonal trajectories to the family of curves given by
y = kx2.
I For the family of curves described above dydx
= 2kx .
I Also y = kx2 giving k = yx2 .
I Therefore the given family satisfy the differential equationdydx
= 2 yx2 x = 2 y
x.
I The family of orthogonal trajectories must satisfy the differential equationdydx
= −x2y
.
I Separating the variables, we get 2ydy = −xdx
I and
2
Zydy = −
Zxdx , or y 2 =
−x2
2+ C .
I Hence our family of orthogonal trajectories is a family of curves of theform
y 2 +x2
2= C ,
a family of ellipses.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 13
13. Find the family of orthogonal trajectories to the family of curves given by
y = kx2.
I For the family of curves described above dydx
= 2kx .
I Also y = kx2 giving k = yx2 .
I Therefore the given family satisfy the differential equationdydx
= 2 yx2 x = 2 y
x.
I The family of orthogonal trajectories must satisfy the differential equationdydx
= −x2y
.
I Separating the variables, we get 2ydy = −xdx
I and
2
Zydy = −
Zxdx , or y 2 =
−x2
2+ C .
I Hence our family of orthogonal trajectories is a family of curves of theform
y 2 +x2
2= C ,
a family of ellipses.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 13
13. Find the family of orthogonal trajectories to the family of curves given by
y = kx2.
I For the family of curves described above dydx
= 2kx .
I Also y = kx2 giving k = yx2 .
I Therefore the given family satisfy the differential equationdydx
= 2 yx2 x = 2 y
x.
I The family of orthogonal trajectories must satisfy the differential equationdydx
= −x2y
.
I Separating the variables, we get 2ydy = −xdx
I and
2
Zydy = −
Zxdx , or y 2 =
−x2
2+ C .
I Hence our family of orthogonal trajectories is a family of curves of theform
y 2 +x2
2= C ,
a family of ellipses.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 13
13. Find the family of orthogonal trajectories to the family of curves given by
y = kx2.
I For the family of curves described above dydx
= 2kx .
I Also y = kx2 giving k = yx2 .
I Therefore the given family satisfy the differential equationdydx
= 2 yx2 x = 2 y
x.
I The family of orthogonal trajectories must satisfy the differential equationdydx
= −x2y
.
I Separating the variables, we get 2ydy = −xdx
I and
2
Zydy = −
Zxdx , or y 2 =
−x2
2+ C .
I Hence our family of orthogonal trajectories is a family of curves of theform
y 2 +x2
2= C ,
a family of ellipses.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 13
13. Find the family of orthogonal trajectories to the family of curves given by
y = kx2.
I For the family of curves described above dydx
= 2kx .
I Also y = kx2 giving k = yx2 .
I Therefore the given family satisfy the differential equationdydx
= 2 yx2 x = 2 y
x.
I The family of orthogonal trajectories must satisfy the differential equationdydx
= −x2y
.
I Separating the variables, we get 2ydy = −xdx
I and
2
Zydy = −
Zxdx , or y 2 =
−x2
2+ C .
I Hence our family of orthogonal trajectories is a family of curves of theform
y 2 +x2
2= C ,
a family of ellipses.
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 14
14.
Solve the initial value problem
8><>:x2y ′ + 2xy = 1,
y(1) = 2.
I We put the equation in standard form: dydx
+ 2xy = 1
x2 .
I The integrating factor is given by I (x) = eR
(2/x)dx = e2 ln |x| = x2.
I Multiplying by the integrating factor, we get x2 dydx
+ 2xy = 1.
I Therefore ddx
x2y = 1 and x2y =R
1dx = x + C
I Dividing both sides by x2, we get y = x+Cx2 .
I The initial condition gives y(1) = 2 or 1 + C = 2 and C = 1. Hencey = x+1
x2 .
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 14
14.
Solve the initial value problem
8><>:x2y ′ + 2xy = 1,
y(1) = 2.
I We put the equation in standard form: dydx
+ 2xy = 1
x2 .
I The integrating factor is given by I (x) = eR
(2/x)dx = e2 ln |x| = x2.
I Multiplying by the integrating factor, we get x2 dydx
+ 2xy = 1.
I Therefore ddx
x2y = 1 and x2y =R
1dx = x + C
I Dividing both sides by x2, we get y = x+Cx2 .
I The initial condition gives y(1) = 2 or 1 + C = 2 and C = 1. Hencey = x+1
x2 .
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 14
14.
Solve the initial value problem
8><>:x2y ′ + 2xy = 1,
y(1) = 2.
I We put the equation in standard form: dydx
+ 2xy = 1
x2 .
I The integrating factor is given by I (x) = eR
(2/x)dx = e2 ln |x| = x2.
I Multiplying by the integrating factor, we get x2 dydx
+ 2xy = 1.
I Therefore ddx
x2y = 1 and x2y =R
1dx = x + C
I Dividing both sides by x2, we get y = x+Cx2 .
I The initial condition gives y(1) = 2 or 1 + C = 2 and C = 1. Hencey = x+1
x2 .
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 14
14.
Solve the initial value problem
8><>:x2y ′ + 2xy = 1,
y(1) = 2.
I We put the equation in standard form: dydx
+ 2xy = 1
x2 .
I The integrating factor is given by I (x) = eR
(2/x)dx = e2 ln |x| = x2.
I Multiplying by the integrating factor, we get x2 dydx
+ 2xy = 1.
I Therefore ddx
x2y = 1 and x2y =R
1dx = x + C
I Dividing both sides by x2, we get y = x+Cx2 .
I The initial condition gives y(1) = 2 or 1 + C = 2 and C = 1. Hencey = x+1
x2 .
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 14
14.
Solve the initial value problem
8><>:x2y ′ + 2xy = 1,
y(1) = 2.
I We put the equation in standard form: dydx
+ 2xy = 1
x2 .
I The integrating factor is given by I (x) = eR
(2/x)dx = e2 ln |x| = x2.
I Multiplying by the integrating factor, we get x2 dydx
+ 2xy = 1.
I Therefore ddx
x2y = 1 and x2y =R
1dx = x + C
I Dividing both sides by x2, we get y = x+Cx2 .
I The initial condition gives y(1) = 2 or 1 + C = 2 and C = 1. Hencey = x+1
x2 .
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 14
14.
Solve the initial value problem
8><>:x2y ′ + 2xy = 1,
y(1) = 2.
I We put the equation in standard form: dydx
+ 2xy = 1
x2 .
I The integrating factor is given by I (x) = eR
(2/x)dx = e2 ln |x| = x2.
I Multiplying by the integrating factor, we get x2 dydx
+ 2xy = 1.
I Therefore ddx
x2y = 1 and x2y =R
1dx = x + C
I Dividing both sides by x2, we get y = x+Cx2 .
I The initial condition gives y(1) = 2 or 1 + C = 2 and C = 1. Hencey = x+1
x2 .
Annette Pilkington Improper Integrals
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14
Question 14
14.
Solve the initial value problem
8><>:x2y ′ + 2xy = 1,
y(1) = 2.
I We put the equation in standard form: dydx
+ 2xy = 1
x2 .
I The integrating factor is given by I (x) = eR
(2/x)dx = e2 ln |x| = x2.
I Multiplying by the integrating factor, we get x2 dydx
+ 2xy = 1.
I Therefore ddx
x2y = 1 and x2y =R
1dx = x + C
I Dividing both sides by x2, we get y = x+Cx2 .
I The initial condition gives y(1) = 2 or 1 + C = 2 and C = 1. Hencey = x+1
x2 .
Annette Pilkington Improper Integrals