Math 132
Summer 2015 Exam 1
Formulas
ln( 1 ) = 0, ln( e ) = 1,
ln( x y ) = ln( x ) + ln( y ) , ln( xp ) = p ln( x )
3
2
π 1
π
π
π
π
π
, sin  = cos  =
,
sin  = cos  = , sin  = cos  =
2
2
3 2
3
6
4
4
6
π
π
3π
π
sin( 0 ) = sin( π ) = cos  = 0 , sin  = cos( 0 ) = tan  = 1, sin
 = cos( π ) = −1
2
4
 2 
2
⌠
1
⌠
⌠
⌠
 dx = ln( x ) + C,
ln( x ) dx = x ln( x ) − x + C,
 u dv = u v −  v du

⌡
⌡
⌡
x
⌡
⌠
⌠

 1
x
x
1
1


  + C,
arctan
dx = arcsin  + C,
x
=
d


2
2

a
a
a
a + x
 a2 − x2
⌡
⌡
⌠

x
1
1

  + C
arcsec
d
=
x

2
2
a
a
 x x −a
⌡
⌠
⌠
2
2
sec( x ) dx = tan( x ) + C,
csc( x ) dx = −cot( x ) + C
⌡
⌡
⌠
⌠
csc( x ) cot( x ) dx = −csc( x ) + C
sec( x ) tan( x ) dx = sec( x ) + C,
⌡
⌡
⌠
⌠
tan( x ) dx = ln( sec( x ) ) + C,
sec( x ) dx = ln( sec( x ) + tan( x ) ) + C
⌡
⌡
1. Let P be the point with Cartesian coordinates ( −2 3 , 6 ). Find
polar coordinates of P with polar angle θ satisfying −π < θ < 0 .
Solution
> P := [-2*sqrt(3),6];
P := [ −2
3, 6]
First the radius:
> r := sqrt(P[1]^2 + P[2]^2);
r := 4 3
Next the angle
> theta := arctan(P[2]/P[1]);
θ := −
π
3
Observe that this angle satisfies the requirements that −π < θ < 0 .
However, it is a 4th-quadrant angle and P is in the second quadrant, which is opposite
to the 4th quadrant. So we must use a negative value of r. Thus,
> r := -r;
r := −4
3
Let us verify our assignment of polar coordinates for P:
> x = r*cos(theta), y = r*sin(theta);
x = −2
3, y = 6
Success.
π
π
and θ = .
8
4
π
Sketch the polar curve r = cos( 4 θ ) for 0 < θ < .
4
2. Draw the lines with polar equations θ =
Solution
> Line1 := plot(tan(Pi/8)*x, x = -0.75 .. 1, color = GREY,
thickness = 2, scaling = constrained, view =
[-0.75..1,-0.75..0.75]):
Line2 := plot(tan(Pi/4)*x, x = -0.75 .. 1, color = GREY,
thickness = 2, scaling = constrained, view =
[-0.75..1,-0.75..0.75]):
roseArc := plots[polarplot](cos(4*theta), theta = 0 .. Pi/4,
color = BLUE, thickness = 3, scaling = constrained,
tickmarks=[3,3], view = [-0.75..1,-0.75..0.75]):
plots[display](Line1, Line2, roseArc);
−2
⌠
 1
3. Calculate 
dx

 x
⌡
−10
Solution
( −2 )
⌠
1


dx = ln( |-2| ) - ln ( |-10| ) = ln(2) - ln(10) = ln( 2/10) = ln( 1/3) =

x

⌡
−10
ln( 5
( −1 )
) = −ln( 5 )
Remark: Observe that −ln( 5 ) < 0 . Because the integrand 1/x is negative on the
entire interval [-10 , -2], and because we are integrating from left to right, we expect a negative
value
for the integral.
2
⌠
3
2

 2x −x −3
4. Calculate 
dx

2

x

⌡
1
Solution
> J := Int( (2*x^3-x^2-3)/(x^2), x = 1 .. 2);
2
⌠
 2 x3 − x2 − 3
J := 
dx

x2

⌡
1
> J := Int( expand(integrand(J)), x = 1 .. 2);
2
⌠

3
J := 
2 x − 1 − 2 dx


x
⌡
1
> int( integrand(J), x);
x2 − x +
3
x
> F := x -> x^2-x+3/x;
F := x → x2 − x +
> F(2) - F(1);
1
2
Verification using Maple's builtin integrator:
> value(J);
1
2
4
⌠
2
5. Calculate 
 21 x ( x − 1 ) dx
⌡
1
3
x
Solution
> J1 := Int(21*sqrt(x)*(x-1)^2,x = 1 .. 4);
#
# J1 is the inert (unevaluated) form of the given integral
# In Maple, "Int" prevents the evaluation of the integral,
# whereas "int" calls for an evaluation, if possible
4
⌠
J1 :=  21 x ( x − 1 )2 dx
⌡
1
> eqn1 := J1 = map(expand,J1);
#
# "expand" does what it says it does to the integrand of J1
# First it expand the square (x-3)^2, then it multiplies
each
# term of the exapnded square by sqrt(x)
4
4
⌠
(5 / 2)
(3 / 2)
⌠
eqn1 :=  21 x ( x − 1 )2 dx = 
− 42 x
+ 21 x dx
 21 x
⌡
⌡
1
1
> J1 = value(rhs(eqn1));
#
# "value" forces the evaluation of an inert integral
4
1696
⌠
 21 x ( x − 1 )2 dx =
⌡
5
1
x
2
⌠
 π 


 cot


3
t

6. Let F( x ) = 
dt . Calculate F' (2), the derivative of F( x ) at x = 2.


3

17 + t


⌡
1
Solution
> F := x -> Int(cot(Pi/(3*t))^2/sqrt(17+t^3),t = 1 .. x);
x
2
⌠
 cot π 

3t

F := x → 
dt

3
 17 + t

⌡
1
> D(F)(x);
#
# This is calculated using the Fundamental Theorem of
Calculus - without any integration
2
 π 

cot
3x
17 + x3
> derivative := D(F)(2);
Answer = simplify( derivative );
derivative :=
3 25
25
Answer =
⌠
7. Let F( x ) = 

⌡
x
3
5
2.5
3
17 − t dt . Calculate F' (2), the derivative of F( x ) at x = 2.
Solution
> F := x -> Int(sqrt(17-t^3), t = x .. 2.5);
⌠
F := x → 

⌡
2.5
17 − t3 dt
x
> F := x -> -Int(sqrt(17-t^3), t = 2.5 .. x);
#
# Before applying the Fundamental Theorem of Calculus we
reverse the direction of integration
# so that x is the upper limit of integration. (Maple would
have done this without our intervention.)
⌠
F := x → −

⌡
x
2.5
> D(F)(x);
17 − t3 dt
#
# This is calculated using the Fundamental Theorem of
Calculus - without any integration
− 17 − x3
> derivative := D(F)(2);
Answer = simplify( derivative );
derivative := − 9
Answer = -3
ln( x )
⌠

8. Let F( x ) = 


⌡
π
 π 
 dt . Calculate F' (e), the derivative of F( x ) at x = e.
sec
3t
Solution
> F := (x) -> Int(sqrt(sec(Pi/3/t)),t = sqrt(Pi) .. ln(x));
⌠

F := x → 


⌡
ln( x )
 π 
 dt
sec
3t
π
> D(F)(x);
#
# Note the factor 1/x, or D(ln)(x). It arises from the Chain
Rule
1 π 

sec
 3 ln( x ) 
x
> D(F)(exp(1));
2
e
3
x
⌠
 112 + t
9. Let F( x ) = 
dt . Calculate F' (2), the derivative of F( x ) at x = 2.

4+t

⌡
2x
Solution
> F := (x)-> Int( (112+t)/(4+t) , t = 2*x .. x^3);
x
3
⌠
 112 + t
F := x → 
dt

 4+t
⌡
2x
> D(F)(2);
91
To get this answer, Maple has done something like the following:
> F := (x)-> Int( (112+t)/(4+t) , t = 0 .. x^3) - Int(
(112+t)/(4+t) , t = 0 .. 2*x);
x
3
2x
⌠
⌠
 112 + t
 112 + t

F := x → 
d
−
dt
t


 4+t
 4+t
⌡
⌡
0
0
> derivative := D(F)(x);
derivative :=
3 x2 ( 112 + x3 )
4+x
3
> subs(x=2, derivative);
91
1
1
⌠
 

3

10. Calculate  28 ( 7 x + 1 )
dx.


⌡
0
Solution
−
2 ( 112 + 2 x )
4+2x
> J := Int(28*(7*x+1)^(1/3), x = 0 .. 1);
1
⌠
(1 / 3)
J := 
dx
 28 ( 7 x + 1 )
⌡
0
> F := Int(integrand(J), x);
⌠
(1 / 3)
F := 
dx
28 ( 7 x + 1 )
⌡
> G := changevar(u = 7*x+1, F, u);
⌠ (1 / 3)
G := 
du
4 u
⌡
> G := value(G);
G := 3 u
(4 / 3)
> F := subs(u = 7*x+1, G);
(4 / 3)
F := 3 ( 7 x + 1 )
> answer := subs(x = 1, F) - subs(x = 0, F);
answer := 24 8
(1 / 3)
−3
> simplify( answer );
45
4
e
⌠

1
11. Calculate 
dx.

 x ln( x )
⌡
e
Solution
> J1 := Int(1/x/sqrt(ln(x)), x = exp(1) .. exp(4));
#
# Gives the name J1 to the inert integral of the problem
# In Maple, "Int" tells Maple to set up an integral that is
not
# to be evaluated immediately. For the eventual evaluation,
# the command "value" will be used
e
4
⌠

1
J1 := 
dx

 x ln( x )
⌡
e
> J2 := changevar(u = ln(x), J1, u);
4
⌠
 1
J2 := 
du

 u
⌡
1
> J1 = value(J2);
#
# Forces the evaluation of the inert integral
e
4
⌠

1

dx = 2

 x ln( x )
⌡
e
1
⌠ 2
12. Calculate 
 x 1 − x dx.
⌡
0
Solution
> J1 := Int(x^2*sqrt(1-x),x = 0 .. 1);
1
⌠
J1 :=  x2
⌡
1 − x dx
0
> J2 := changevar(u = 1-x, J1, u);
#
# This line says, Make the substitution u = x+1 in J1 and set
J2 to be the resulting integral wrt u
# Note that the new limits of integration for u are
calculated
1
⌠
J2 :=  ( 1 − u )2 u du
⌡
0
> J3 := map(expand, J2);
#
# This expands everything in J2: First the square is
expanded, then each term is multiplied by sqrt(u)
⌠
J3 := 

⌡
1
u −2u
(3 / 2)
+u
(5 / 2)
du
0
The next line evaluates this integral:
> J1 = value(J3);
1
⌠ 2
x
⌡
1 − x dx =
0
16
105
13. Calculate the area of the region that is bounded above by y = 3 x − 2 and below by
2
y=x .
Solution
> f := x -> 3*x-2;
g := x -> x^2;
f := x → 3 x − 2
g := x → x2
> solve(f(x) = g(x) ,x);
#
# There are exactly two points of intersection
1, 2
> f(1.5), g(1.5);
#
# These evaluations at a point between x = 1 and x = 2 show
us that g(x) < f(x) for 1 < x < 2
2.5, 2.25
> plot([f(x),g(x)], x = 1..2, thickness=2, color = [NAVY,BLUE],
tickmarks = [3,3]);
#
# A graph also shows that g(x) < f(x) for 1 < x < 2
> Area := Int( f(x) - g(x), x = 1 .. 2);
2
⌠
Area :=  3 x − 2 − x2 dx
⌡
1
> Area = value(Area);
2
1
⌠
 3 x − 2 − x2 dx =
⌡
6
1
14. Functions f(x) and g(x) satisfy g( x ) ≤ f( x ) for 1 ≤ x ≤
5
and f( x ) ≤ g( x ) for
3
5
≤ x ≤ 2. In the table below, the number in an f(x) or g(x) cell is the value of
3
f or g at the value of x in the x cell above it. For example, f(7/6) = 3.1 and
g(11/6) = 2.6.
x || 1 | 7/6 | 8/6 | 9/6 | 10/6 | 11/6 | 2
_________________________________________
f(x) || 2.8 | 3.1 | 3.2 | 3.0 | 2.5 | 2.3 | 2.0
_________________________________________
g(x) || 1.7 | 1.8 | 2.0 | 2.2 | 2.5 | 2.6 | 2.8
Use a Riemann sum with midpoints to estimate the area between
the graphs of y = f( x ) and y = g( x ) for 1 ≤ x ≤ 2.
Solution
> Delta := ((2-1)/6)*2;
∆ :=
1
3
> ( (3.1-1.8) + (3.0-2.2) + (2.6-2.3) )*Delta;
0.8000000000
15. Let R be the region that lies below y = ln( x ), above
the x-axis, and to the left of x = e. Express as an integral the
volume of the solid that results when R is rotated about
the x-axis.
Solution
Method of Disks (Method of Discs in British Commonwealth countries) (Not
too bad a choice)
> Volume_by_disks := Pi*Int(ln(x)^2, x = 1 .. exp(1));
e
⌠
Volume_by_disks := π  ln( x )2 dx
⌡
1
> value(Volume_by_disks);
π (e − 2)
Note: the integral arising from the Method of Disks requires two integration by parts.
For the first, u = ln( x )2, and, for the second, u = ln( x ).
Method of Shells (An excellent choice)
> Volume_by_shells := 2*Pi*Int(y*(exp(1) - exp(y)), y = 0 ..
1);
1
⌠
Volume_by_shells := 2 π  y ( e − ey ) dy
⌡
0
> value(Volume_by_shells);

1
2 π  e − 1 

2
Note: only one integration by parts is needed in this calculation.
16. Let R be the region that lies below y = 5 x − 6 and above
2
y = x for x between 2 and 3. Express as an integral the
volume of the solid that results when R is rotated about
the y-axis.
Solution
Method of Washers (a fine choice)
> Volume_by_washers := Pi*Int( (sqrt(y))^2 - ((y+6)/5)^2, y = 4
.. 9);
9
⌠
2

 y 6

Volume_by_washers := π  y −  +  dy
 5 5

⌡
4
> value(Volume_by_washers);
5π
6
Method of Shells (another fine choice)
> Volume_by_shells := 2*Pi*Int(x*( (5*x-6) - x^2), x = 2 .. 3);
3
⌠
Volume_by_shells := 2 π  x ( 5 x − 6 − x2 ) dx
⌡
2
> value(Volume_by_shells);
5π
6
17. Let R be the region that lies to the right of the
2
arc of the parabola y = x , 1 ≤ x ≤ 2 , to the left
of the line segment y = x − 2, 3 ≤ x ≤ 4, above y = 1,
and below y = 2. Express as an integral the volume of the
solid that results when R is rotated about the axis y = 5.
Solution
The figure below shows the region R and the axis of rotation.
As a visual aid, the region is divided into three parts by means of
two vertical line segments: one at x =
2 and one at x = 3.
Method of Washers (not enthusiastically recommended here)
The volume can be calculated by the Method of Washers, but that is not an efficient choice.
The integration must be divided into three parts: one with [ 1,
2 ] as the interval of integration,
one with [ 2 , 3 ] as the interval of integration, and one with [ 3, 4 ] as the interval of
integration.
> Volume_by_washers := Pi*Int( (5-1)^2 - (5-x^2)^2, x = 1 ..
sqrt(2))
+ Pi*Int( (5-1)^2 - (5-2)^2, x =
sqrt(2) .. 3)
+ Pi*Int( (5-(x-2))^2 (5-2)^2, x = 3 .. 4)
;
2
4
3
⌠
2
⌠
⌠
2
Volume_by_washers := π 
7 dx + π  ( 7 − x )2 − 9 dx
 16 − ( 5 − x ) dx + π 
⌡
⌡
⌡
3
2
1
> value(Volume_by_washers);
evalf(value(Volume_by_washers));
 47 2 88 
π  −
+  + π ( 21 − 7

15
15 
2)+
10 π
3
49.85488439
> Volume_by_shells := 2*Pi*Int((5-y)*((y+2) - sqrt(y)), y = 1
.. 2);
2
⌠
Volume_by_shells := 2 π  ( 5 − y ) ( y + 2 −
⌡
y ) dy
1
> value(Volume_by_shells);
evalf(value(Volume_by_shells));
 76 2 151 

+
2 π  −

15
10 
49.85488438
Method of Shells (the recommended method)
The Method of Shells provides a more efficient calculation:
> Volume_by_shells := 2*Pi*Int((5-y)*((y+2) - sqrt(y)), y = 1
.. 2);
2
⌠
Volume_by_shells := 2 π  ( 5 − y ) ( y + 2 −
⌡
y ) dy
1
> value(Volume_by_shells);
evalf(value(Volume_by_shells));
 76 2 151 

+
2 π  −
15
10 

49.85488438
1 8. Let R be the "triangular" region that is bounded above
4
for 1 ≤ x ≤ 2, below by y = 3, and on the left by x = 1.
by y = x +
2
x
Express as an integral the volume of the solid that results when R
is rotated about the axis x = −2.
Solution
> ;
x3 − 8
x3
> ;
>
Code