Math 132
Fall 2006 Exam I
N
∑ f( ξ ) ∆ x for a function f on an interval [ a, b ] is said to be a lower
j
j=1
Riemann sum if, for each j,
th
subinterval is chosen so that f( ξ ) is minimized. Calculate the
the point ξ in the j
j
j
2
lower Riemann sum for f( x ) = x ,
 3 5
[ a, b ] =  − ,  , and N = 4. (Use a partition of [a,b] into equal length subintervals.)
 2 2
1. A Riemann sum
a) 7/42
f) 3
b) 2
g) 13/4
c) 9/4
h) 7/2
d) 5/2
i) 15/4
e) 11/4
j) 4
Solution: (e)
> f := x -> x^2;
f := x → x2
> plot(f(x),x=-3/2..5/2);
> a := -3/2: b := 5/2:
> Delta := (b-a)/N;
N := 4:
∆ := 1
> with(plottools): with(plots):
Warning, the names arrow and changecoords have been redefined
> r1 := rectangle([-3/2,f(-1/2)], [-1/2,0], color=COLOR(RGB,
0.94,0.94,0.65)):
r2 := rectangle([1/2,f(1/2)], [3/2,0], color=COLOR(RGB,
0.94,0.94,0.65)):
r3 := rectangle([3/2,f(3/2)], [5/2,0], color=COLOR(RGB,
0.94,0.94,0.65)): fnPlot := plot(f(x),x=-3/2..5/2,
thickness=2,color=MAROON):
display(r1,r2,r3,fnPlot);
> (f(-1/2)+f(0)+f(1/2)+f(3/2))*Delta;
11
4
π
3
⌠
2
2. Calculate 
 sec( θ ) dθ.
⌡
0
a) 1
f)
3 3
2
b) 2
c)
3
g) 2 2
h) 2 3
3 2
2
d) 2
e)
i) 3 2
j) 3 3
Solution: ( c )
> int(sec(theta)^2,theta = 0 .. Pi/3);
3
> J := Int(sec(theta)^2,theta);
antiderivative := value(J);
definiteIntegral :=
subs(theta=Pi/3, antiderivative) - subs(theta=0,
antiderivative);
⌠
J := sec( θ )2 dθ
⌡
antiderivative :=
definiteIntegral :=
sin( θ )
cos( θ )
π
sin 
3
π
cos 
3
−
sin( 0 )
cos( 0 )
> simplify( definiteIntegral );
3
3. An antiderivative of f is the function x →
3 x2 + 4 x + 2
x2 + x + 1
is b?
a) 1
f) 7/4
b) 4/3
g) 7/3
c) 5/4
h) 9/4
d) 5/3
i) 8/3
e) 3/2
j) 5/2
Solution: (a)
> restart;
> F := x -> (3*x^2+4*x+2)/(x^2+x+1);
b
. If
⌠
 f( x ) dx = 1, then what
⌡
0
F := x →
3 x2 + 4 x + 2
x2 + x + 1
> eqn := F(b) - F(0) = 1;
eqn :=
3 b2 + 4 b + 2
b2 + b + 1
−2=1
> solve(eqn, b);
1
4. Calculate
3
⌠
2



 d 7x +9

 dx .


 dx  2

 x +3 
⌡
0
a) 1
b) 2
c) 3
d) 4
e) 5
f) 6
g) 7
h) 8
i) 9
j) 10
Solution: ( c )
> F := x -> (7*x^2+9)/(x^2+3); #This is an antiderivative of the
integrand
F := x →
7 x2 + 9
x2 + 3
> F(3) - F(0);
3
4
4
⌠
⌠
5. Suppose that  f( x ) dx = 8 and  ( 2 + f( x ) ) dx = 22.
⌡
⌡
1
−1
a) 1
f) 6
b) 2
g) 7
c) 3
h) 8
d) 4
i) 9
e) 5
j) 10
1
⌠
What is  f( x ) dx?
⌡
−1
Solution: ( d )
> restart;
> eqn := int(2+f(x),x = -1 .. 4) = 22;
4
⌠
eqn :=  2 + f( x ) dx = 22
⌡
-1
> eqn2 := 10 + int(f(x),x = -1 .. 4) = 22;
4
⌠
eqn2 := 10 +  f( x ) dx = 22
⌡
-1
> eqn3 := 10 + int(f(x),x = -1 .. 1) + int(f(x),x = 1 .. 4) = 22;
1
4
⌠
⌠
eqn3 := 10 +  f( x ) dx +  f( x ) dx = 22
⌡
⌡
-1
1
> eqn3 := 10 + int(f(x),x = -1 .. 1) + 8 = 22;
1
⌠
eqn3 := 18 +  f( x ) dx = 22
⌡
-1
> solve( eqn3, int(f(x),x = -1 .. 1));
4
2
6. Suppose that f( x ) = x + 1 . The Mean Value Theorem for Integrals asserts that there is a
point c in
the interval [1,7] such that f( c ) = f
where f
is the average value of f( x ) for x in the
ave
ave
interval [1,7].
What is c?
a)
5
b)
6
c) 3
d)
10
e) 2 3
f)
14
g)
15
h) 4
i)
19
j)
Solution: ( i )
21
> f := x -> x^2+1;
f := x → x2 + 1
> m := int(f(x),x=1..7)/(7-1);
m := 20
> solve(f(c) = m, c);
19 , − 19
x
⌠
2

d  7t +9

dt
dx 
 t2 + 3
⌡
−1
7. Calculate
a) 0
f) 5
b) 1
g) 6
c) 2
h) 7
d) 3
i) 8
at x = 1.
e) 4
j) 9
Solution: ( e )
> subs(t=1,(7*t^2+9)/(t^2+3));
4
⌠
8. Suppose that F( x ) = 

⌡
0
π
at x = ).
4
a) 1
f)
3 3
2
tan( x )
2
1 + t dt.
b) 2
c)
3
g) 2 2
h) 2 3
π
What is D( F ) ? (The derivative of F(x)
4
3 2
2
d) 2
e)
i) 3 2
j) 3 3
Solution: ( g )
> F := (x) -> int(sqrt(1+t^2),t = 0 .. tan(x));
⌠
F := x → 

⌡
tan( x )
1 + t 2 dt
0
> D(F)(Pi/4);
2 2
⌠
9. Suppose that F( x ) = 

⌡
x
5
2
8 + t dt. What is D(F)(1)? (The derivative of F(x) at x =
x
1).
a) 3
f) 8
b) 4
g) 9
c) 5
h) 10
d) 6
i) 11
e) 7
j) 12
Solution: ( j )
> F := (x) -> int(sqrt(8+t^2),t = x .. x^5);
⌠
F := x → 

⌡
x
5
x
> D(F)(1);
12
>
8 + t 2 dt
1
3
⌠

2 3
10. Calculate  4 x ( x + 1 ) dx .

⌡
0
a) 3
f) 8
b) 4
g) 9
c) 5
h) 10
d) 6
i) 11
e) 7
j) 12
Solution: ( c )
> int(4*x^2*(x^3+1)^3,x = 0 .. 1);
5
> student[changevar](u=(x^3+1), Int(4*x^2*(x^3+1)^3,x = 0 .. 1),
u);
# This is the required change of variable
2
⌠
 4 u3

du

 3
⌡
1
> value( % );
5
2
⌠
11. Calculate  x x − 1 dx .
⌡
1
8
15
16
f)
15
a)
2
3
6
g)
5
b)
4
5
5
h)
3
c)
d) 1
i)
7
5
4
3
8
j)
5
e)
Solution: ( f )
> int(x*sqrt(x-1),x = 1 .. 2);
16
15
> J := student[changevar](u=(x-1), Int(x*sqrt(x-1),x = 1 .. 2),
u);
# This is the required change of variable
1
⌠
J :=  ( 1 + u ) u du
⌡
0
> J1 := Int(expand(student[integrand](J)), u = 0 .. 1);
⌠
J1 := 

⌡
1
u +u
(3 / 2)
du
0
> value(J1);
16
15
1
⌠
x

 e
12. Calculate 
dx .

 1 + ex
⌡
0
e
a) ln 
2
e
f)
( 1 + e )2
1+e

b) ln
 2 
e
g)
1+e

e
c) ln 1 + 
2

1
h)
1+e
1

d) ln + e 
2

e) ln( 2 )
i) 1 − ln( 2 )
j) e − 1
Solution: ( b )
> int(exp(x)/(1+exp(x)),x = 0 .. 1);
ln( 1 + e ) − ln( 2 )
> J := student[changevar](u=1+exp(x), Int(exp(x)/(1+exp(x)),x = 0
.. 1), u);
# This is the required change of variable
⌠

J := 


⌡
1+e
1
u
du
2
> value(J);
ln( 1 + e ) − ln( 2 )
2
13. Calculate the area between y = x − 1
a) 8/3
f) 21/4
b) 3
g) 14/3
c) 10/3
h) 9/2
and
d) 7/2
i) 5
y = x + 1.
e) 4
j) 16/3
Solution: ( h )
> solve( x^2-1 = x+1, x);
2, -1
> plot([x+1,x^2-1], x = -1 .. 2, thickness=[2,2], color =
[NAVY,PLUM]);
> int((x+1)-(x^2-1), x = -1 .. 2);
9
2
14. Tle Lorenz function L of a certain country has the following values:
L ( 0 ) = 0,
L( 20 ) = 5,
L( 40 ) = 15,
L( 60 ) = 30,
L( 80 ) = 50,
L( 90 ) = 70,
L( 100 ) = 100.
Using trapezoids and all the given data, obtain an estimate for the area under y = L( x ).
a) 2910
f) 2960
b) 2920
g) 2970
c) 2930
h) 2980
d) 2940
i) 2990
e) 2950
j) 3000
Solution: ( e )
> (0+5)/2*20 + (5+15)/2*20 + (15+30)/2*20 + (30+50)/2*20 +
(50+70)/2*10+(70+100)/2*10;
2950
15. By applying Simpson's Rule with four subintervals, what approximation of the area under
the graph of y = 16 − x4
and over the x-axis is obtained?
a) 301/6
f) 252/5
b) 307/6
g) 152/3
c) 158/3
h) 201/4
d) 209/4
i) 103/2
Solution: ( g )
> f := x -> 16 - x^4;
f := x → 16 − x4
e) 107/2
j) 256/5
> (f(-2)+4*f(-1) + 2*f(0) + 4*f(1) + f(2))/3;
152
3
16. If
y( x ) is the unique solution of the initial value problem
d
y( x ) = 2 x y( x ) ,
dx
y( 0 ) = e
then what is
a) 2 e
4
f) e
b) e
y( 2 ) ?
2
g) 5 e
c) 3 e
5
h) e
d) e
3
i) 6 e
e) 4 e
6
j) e
Solution: ( h )
> eqn := int(1/y,y) = int(2*x , x) + C;
eqn := ln( y ) = x2 + C
> eqn2 := y = solve(eqn,y);
2
(x + C)
eqn2 := y = e
> eqn3 := subs({y= exp(1), x = 0}, eqn2);
eqn3 := e = eC
> eqn4 := C = solve(eqn3, C);
eqn4 := C = 1
> eqn5 := subs(eqn4, eqn2);
2
(x + 1)
eqn5 := y = e
> eqn6 := simplify( subs(x=2,eqn5) );
eqn6 := y = e5
17. The height y( t ) of water in a leaking tank is given by the differential equation
d
2
y( t ) = −r y( t ) where r
dt
is the (constant) radius of the hole through which the water leaks. If the initial height (i.e., at
t = 0) of the water
was 144 and if the tank becomes empty at t = 8, then what is the value of r ?
a) 0
f)
b) 1
2
g)
2 3
3
c)
2
2
d)
3
3
h)
2 2
i)
3
1
2
3
j)
2
e)
Solution: ( i )
> eqn := dsolve( diff(y(t),t) = -r^2*sqrt(y(t)) , y(t) );
eqn := y( t ) +
r2 t
− _C1 = 0
2
> eqn1 := subs( {t=0,y(t)=144}, eqn);
eqn1 := 144 − _C1 = 0
> eqn2 := _C1 = solve(eqn1, _C1);
eqn2 := _C1 = 12
> eqn3 := subs(eqn2, eqn);
eqn3 := y( t ) +
r2 t
2
− 12 = 0
> eqn4 := y(t) = solve(eqn3, y(t));
eqn4 := y( t ) =
r 4 t2
4
− 12 r2 t + 144
> subs({y(t) = 0, t = 8}, eqn4);
0 = 16 r4 − 96 r2 + 144
> solve(%,r);
3, − 3,
3, − 3
−1
⌠
 1
18. Calculate 
dx.

 x
⌡
−3
a) −ln( 2 )
b) −ln( 3 )
1
f) −ln 
3
1
g) −ln 
6
c) −ln( 6 )
d) −ln( 9 )
1
h) −ln 
9
i)
1
1
−
e
3
e
1
e) −ln 
2
1
1
j)
−
3 e
e
Solution: ( b )
> int(1/x,x = -3 .. -1);
−ln( 3 )
π
4
⌠
19. Calculate  tan( x ) dx.
⌡
0
1
ln( 2 )
2
f) 2 ln( 3 )
a)
b) ln( 2 )
c) 2 ln( 2 )
g) 3 ln( 3 )
h) 1
1
ln( 3 )
2
i) 2
d)
Solution: ( a )
> int(tan(x),x = 0 .. Pi/4);
1
2
ln( 2 )
e) ln( 3 )
j)
2
π
3
⌠
20. Calculate  sec( x ) dx.
⌡
0
a) ln( 2 )
b) ln( 3 )
f) ln( 2 + 2 ) g) ln( 3 + 2 )
c) ln( 1 + 3 )
h) ln( 2 2 )
Solution: ( d )
> int(sec(x),x = 0 .. Pi/3);
ln( 2 +
3)
d) ln( 2 + 3 )
e) ln( 1 + 2 )
i) ln( 2 3 )
j) ln( 2 + 3 )