Math 132
Fall 2006 Exam I
1 . A Riemann sum
for a function
Riemann sum if, for each
, the point
in the
minimized. Calculate the lower Riemann sum for
(Use a partition of [a,b] into equal length subintervals.)
a) 7/42
b) 2
c) 9/4
f) 3
g) 13/4
h) 7/2
Solution : (e)
> f := x -> x^2;
> plot(f(x),x=-3/2..5/2);
d) 5/2
i) 15/4
on an interval
e) 11/4
j) 4
is said to be a lower
subinterval is chosen so that
,
, and
is
.
> a := -3/2: b := 5/2: N := 4:
> Delta := (b-a)/N;
> 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;
2. Calculate
a) 1
f)
.
b)
c)
g)
d) 2
h)
e)
i)
j)
Solution : ( c )
> int(sec(theta)^2,theta = 0 .. Pi/3);
> J := Int(sec(theta)^2,theta);
antiderivative := value(J);
definiteIntegral :=
subs(theta=Pi/3, antiderivative) - subs(theta=0, antiderivative);
> simplify( definiteIntegral );
3. An antiderivative of
is the function
. If
, then
what is
?
a) 1
b) 4/3
c) 5/4
d) 5/3
e) 3/2
f) 7/4
g) 7/3
h) 9/4
i) 8/3
j) 5/2
Solution : (a)
> restart;
> F := x -> (3*x^2+4*x+2)/(x^2+x+1);
> eqn := F(b) - F(0) = 1;
> solve(eqn, b);
4. Calculate
.
a) 1
b) 2
c) 3
f) 6
g) 7
h) 8
d) 4
i) 9
e) 5
j) 10
Solution : ( c )
> F := x -> (7*x^2+9)/(x^2+3); #This is an antiderivative of the integrand
> F(3) - F(0);
5. Suppose that
and
a) 1
b) 2
c) 3
f) 6
g) 7
h) 8
d) 4
i) 9
.
e)
5
j) 10
Solution : ( d )
> restart;
> eqn := int(2+f(x),x = -1 .. 4) = 22;
> eqn2 := 10 + int(f(x),x = -1 .. 4) = 22;
> eqn3 := 10 + int(f(x),x = -1 .. 1) + int(f(x),x = 1 .. 4) = 22;
> eqn3 := 10 + int(f(x),x = -1 .. 1) + 8 = 22;
What is
?
> solve( eqn3, int(f(x),x = -1 .. 1));
6. Suppose that
. The Mean Value Theorem for Integrals asserts that there is a point
in the interval [1,7] such that
b)
f)
g)
is the average value of
?
the interval [1,7]. What is
a)
where
c) 3
h) 4
d)
e)
i)
j)
Solution : ( i )
> f := x -> x^2+1;
> m := int(f(x),x=1..7)/(7-1);
> solve(f(c) = m, c);
7. Calculate
at
a) 0
b) 1
c) 2
d) 3
e) 4
f) 5
g) 6
h) 7
i) 8
j) 9
Solution : ( e )
> subs(t=1,(7*t^2+9)/(t^2+3));
.
for
in
8. Suppose that
.
What is
? (The derivative of F(x) at
).
a) 1
b)
f)
c)
g)
d) 2
h)
e)
i)
j)
Solution : ( g )
> F := (x) -> int(sqrt(1+t^2),t = 0 .. tan(x));
> D(F)(Pi/4);
9 . Suppose that
. What is D(F)(1)? (The derivative of F(x) at x = 1).
a) 3
b) 4
c) 5
d) 6
e) 7
f) 8
g) 9
h) 10
i) 11
j) 12
Solution : ( j )
> F := (x) -> int(sqrt(8+t^2),t = x .. x^5);
> D(F)(1);
>
10. Calculate
.
a) 3
b) 4
c) 5
d) 6
e) 7
f) 8
g) 9
h) 10
i) 11
j) 12
Solution : ( c )
> int(4*x^2*(x^3+1)^3,x = 0 .. 1);
> 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
> value( % );
11. Calculate
.
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
Solution : ( f )
> int(x*sqrt(x-1),x = 1 .. 2);
> J := student[changevar](u=(x-1), Int(x*sqrt(x-1),x = 1 .. 2), u);
# This is the required change of variable
> J1 := Int(expand(student[integrand](J)), u = 0 .. 1);
> value(J1);
12. Calculate
a)
.
b)
c)
d)
e)
f)
g)
h)
i)
Solution : ( b )
> int(exp(x)/(1+exp(x)),x = 0 .. 1);
> J := student[changevar](u=1+exp(x), Int(exp(x)/(1+exp(x)),x = 0 .. 1), u);
# This is the required change of variable
> value(J);
13. Calculate the area between
a) 8/3
b) 3
f) 21/4
g) 14/3
and
c) 10/3
d) 7/2
h) 9/2
i) 5
e)
.
4
j) 16/3
Solution : ( h )
> solve( x^2-1 = x+1, x);
> plot([x+1,x^2-1], x = -1 .. 2, thickness=[2,2], color = [NAVY,PLUM]);
j)
> int((x+1)-(x^2-1), x = -1 .. 2);
14. Tle Lorenz function
of a certain country has the following values:
.
Using trapezoids and all the given data, obtain an estimate for the area under
a) 2910
b) 2920
f) 2960
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;
15. By applying Simpson's Rule with four subintervals, what approximation of the area under the graph of
and over the x-axis is obtained?
a) 301/6
b) 307/6
f) 252/5
g) 152/3
c) 158/3
d) 209/4
h) 201/4
e) 107/2
i) 103/2
j) 256/5
Solution : ( g )
> f := x -> 16 - x^4;
> (f(-2)+4*f(-1) + 2*f(0) + 4*f(1) + f(2))/3;
16. If
is the unique solution of the initial value problem
then what is
?
a)
b)
c)
f)
g)
h)
Solution : ( h )
> eqn := int(1/y,y) = int(2*x , x) + C;
> eqn2 := y = solve(eqn,y);
> eqn3 := subs({y= exp(1), x = 0}, eqn2);
d)
i)
e)
j)
,
> eqn4 := C = solve(eqn3, C);
> eqn5 := subs(eqn4, eqn2);
> eqn6 := simplify( subs(x=2,eqn5) );
17. The height
where
of water in a leaking tank is given by the differential equation
is the (constant) radius of the hole through which the water leaks. If the initial height (i.e., at
) of the water was 144 and if the tank becomes empty at
a) 0
f)
b) 1
d)
c)
g)
h)
Solution : ( i )
> eqn := dsolve( diff(y(t),t) = -r^2*sqrt(y(t)) , y(t) );
> eqn1 := subs( {t=0,y(t)=144}, eqn);
> eqn2 := _C1 = solve(eqn1, _C1);
> eqn3 := subs(eqn2, eqn);
, then what is the value of
e)
i)
j)
?
> eqn4 := y(t) = solve(eqn3, y(t));
> subs({y(t) = 0, t = 8}, eqn4);
> solve(%,r);
18. Calculate
.
a)
b)
c)
f)
g)
h)
Solution : ( b )
> int(1/x,x = -3 .. -1);
19. Calculate
.
d)
e)
i)
j)
a)
f)
b)
c)
g)
h)
d)
e)
i)
j)
Solution : ( a )
> int(tan(x),x = 0 .. Pi/4);
20. Calculate
.
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
Solution : ( d )
> int(sec(x),x = 0 .. Pi/3);