Math 132
Fall 2005 Exam I
1 . A Riemann sum
for a function
Riemann sum if, for each
, the point
on an interval
in the
maximized. Calculate the upper Riemann sum for
. (Use a partition of [a,b] into equal length subintervals.)
a) 14
f) 24
b) 16
c) 18
g) 26
h) 28
d) 20
e) 22
i) 30
Solution : ( i )
The nodes (points of the uniform partition) are
> a := -1; b := 3; N := 4; Delta := (b-a)/N;
> for j from 0 to N do
x[j] := a + j*Delta;
od;
j) 32
is said to be a n upper
subinterval is chosen so that
,
is
, and
> f := x -> x^3-3*x+2;
> Diff('f(x)',x)=factor(D(f)(x));
This calculation tells us that
increases for
decreases from -1 to 1 (
= ( - )( + ) so
> 1 . Therefore:
> xi[1] := x[0]; xi[2] := x[1]; xi[3] := x[3]; xi[4] := x[4];
> fnGraph := plot(f(x),x=-1..3, color=PLUM, thickness=2):
> nodes := plot( [ seq([xi[j],f(xi[j])], j = 1..N)], style = POINT, symbol = CIRCLE, color = NAVY):
> plots[display](fnGraph, nodes);
) and
> rect[1] := plottools[rectangle]([x[0],f(x[0])], [x[1],0], color = wheat):
rect[2] := plottools[rectangle]([x[1],f(x[1])], [x[2],0], color = wheat):
rect[3] := plottools[rectangle]([x[2],f(x[3])], [x[3],0], color = wheat):
rect[4] := plottools[rectangle]([x[3],f(x[4])], [x[4],0], color = wheat):
> plots[display](fnGraph, nodes,seq(rect[j],j=1..N));
> ANSWER := (f(xi[1])+f(xi[2])+f(xi[3])+f(xi[4]))*Delta;
2. Calculate
a) 1
.
b)
f)
c)
g)
d) 2
h)
e)
i)
j)
Solution : ( a )
> F := unapply(int(sec(theta)*tan(theta), theta), theta);
> F(Pi/3) - F(0);
3. An antiderivative of
. If
is the function
?
a)
f)
b)
g) 1
c)
h)
Solution : ( i )
> restart;
> F := x -> (x+exp(x))/(1+exp(x));
d)
i) 2
e)
j) 2
, what is
> eqn1 := Int(f(x)+c,x = 0 .. 1) = F(1) - F(0) + c;
> eqn2 := 5/2 = rhs(eqn1);
> solve(eqn2,c);
4. Calculate
a) 1
f)
.
b) 2
g)
c)
d)
h)
e)
i)
Solution : ( c )
> F := x -> x*ln(x); #This is an antiderivative of the integrand
> F(exp(1)) - F(1);
j)
5. Suppose that
and
a) 1
b) 2
c) 3
f) 6
g) 7
h) 8
d) 4
.
e)
i) 9
What is
?
5
j) 10
Solution : ( d )
> restart;
> eqn1 := int(x^2+f(x),x = 0..3) = 17;
> eqn2 := int(x^2,x = 0..3) + int(f(x),x = 0..3)= 17;
> eqn3 := 9 + int(f(x),x = 0 .. 2) + int(f(x),x = 2 .. 3) = 17;
> eqn4 := 9 + 4 + int(f(x),x = 2 .. 3) = 17;
> solve( eqn4, int(f(x),x = 2 .. 3));
6. Suppose that
. The Mean Value Theorem for Integrals asserts that there is a point
in the interval [-1,2] such that
where
in the interval [1,7]. What is
a)
b)
f)
g)
is the average value of
?
c)
d)
h)
e)
i)
j)
Solution : ( f )
> f := x -> x^2+6*x; interval := -1 .. 2;
> Ave := int(f(x), x=interval)/3;
> solve(f(x) = Ave, x);
7. Calculate
at
a) 0
b) 1
c) 2
d) 3
e) 4
f) 5
g) 6
h) 7
i) 8
j) 9
Solution : ( c )
> J := Int((9*t+tan(Pi*t/4))/(t^2+4),t = -1 .. x);
.
for
> Integrand := student[integrand](J);
> simplify(subs(t=1, Integrand));
8. Suppose that
.
What is
d) 2
e)
? (The derivative of F(x) at
).
a) 1
f)
b)
c)
g)
h)
Solution : ( b )
> restart: with(student):
> F := (x) -> Int(sqrt(7/2+t^2),t = 0..sin(x));
i)
j)
> D(F)(x);
> simplify(D(F)(Pi/4));
Where answer comes from:
> derivative := subs(t=sin(x), integrand(F(x)))*D(sin)(x);
> simplify(subs(x=Pi/4, derivative));
9 . Suppose that
. What is D(F)(2)? (The derivative of F(x) at x =
5).
a) 13
b) 20
c) 27
d) 33
e) 40
f) 47
g) 53
h) 60
i) 67
j) 73
Solution : ( f )
> F := (x) -> Int(sqrt(144+t^2),t = x .. (3*x+1));
> simplify(D(F)(5));
Where answer comes from:
> h := x -> 3*x+1; g := x -> x;
> derivative := subs(t=h(x), integrand(F(x)))*D(h)(x) - subs(t=g(x), integrand(F(x)))*D(g)(x);
> simplify(subs(x=5, derivative));
10. Calculate
.
a) 3
b) 4
c) 6
d) 8
e) 9
f) 8
g) 12
h) 15
i) 16
j) 20
Solution : ( h )
> J1 := Int(4*cos(x)*(sin(x)+1)^3,x = 0 .. Pi/2);
> J2 := changevar(u = sin(x) + 1, J1, u);
> value(J2);
11. Calculate
.
a)
b)
c)
d)
f)
g)
h)
i)
Solution : ( i )
> J1 := Int(x*sqrt(x-3),x = 3..4);
> J2 := changevar(u = x-3, J1, u);
> J3 := Int(expand(integrand(J2)), u = 0 .. 1);
> value(J3);
e)
j)
12. Calculate
.
a)
b)
c)
f)
g)
h)
d)
e)
i)
j)
Solution : ( e )
> J1 := Int(1/(x+x*ln(x)),x = 1 .. exp(1));
> J2 := Int(factor(integrand(J1)), x = 1 .. exp(1));
> J3 := changevar(u=1+ln(x), J2, u);
> value(J3);
13. Find the solutions x = a
and x = b of the equation sin(x) = cos(x) in the first and third quadrants
respectively. Calculate the area between
and
for
in [a,b].
a) 1
f)
b)
c) 2
g)
d)
h)
e)
i)
j) 4
Solution : ( e )
> solve( sin(x) = cos(x) );
> plot([sin(x) , cos(x)], x = Pi/4 .. 5*Pi/4, color = [NAVY,MAROON]);
> int(sin(x) - cos(x), x = Pi/4 .. 5*Pi/4);
>
14. At irregular intervals during the first 10 seconds of a race, a radar gun recorded the following speeds
(in m/s) of a runner:
.
Given that distance
at time
is expressed by the formula
, estimate the
distance (in m) the runner has covered during those 10 seconds. Use trapezoids and all the given data.
a) 99.4
b) 99.5
f) 99.9
g) 100.0
c) 99.6
d) 99.7
h) 100.1
i) 100.2
e) 99.8
j) 100.3
Solution : ( d )
> (0+6.8)/2*1 + (6.8+10.2)/2*2 + (10.2+11.0)/2*2 + (11.0+11.6)/2*1 + (11.6+11.8)/2*4;
15. A swimming pool has the shape of a rectanle with width 12 feet and length 20 feet. Measured at 5 foot
intervals along its length, starting at the shallow end and ending at the deepest end, the depths in feet are 1, 3, 7,
9, 10. By applying Simpson's Rule with four subintervals, what approximation to the volume of the pool (in
cubic feet) is obtained?
a) 1410
b) 1420
f) 1460
g) 1470
c) 1430
d) 1440
h) 1480
e) 1 450
i) 1490
j) 1500
Solution : ( f )
> 12*5/3*(1+4*3+2*7+4*9+1*10);
16. If
is the unique solution of the initial value problem
then what for what positive value
a)
b)
f)
g)
is
c)
h)
,
?
d)
i)
e)
j)
Solution : ( i )
> eqn1 := int(y,y) = int(2*x/(1+x^2) , x) + C;
> eqn2 := C = simplify(solve(subs({y=2,x=0}, eqn1),C));
> eqn3 := subs(eqn2, eqn1);
> eqn4 := x = solve(subs(y=4,eqn3), x);
17. The differential equation for the current
in a certain circuit is
the resistance (measured in ohms). If
and
a) 1
b) 2
c) 3
d) 4
e) 5
f) 6
g) 10
h) 12
i) 15
j) 20
Solution : ( h )
> eqn := dsolve( {diff(y(t),t) = 60-r*y(t), y(0)=0} , y(t) );
> eqn1 := 5*(1-1/exp(12)) = simplify(rhs(subs(t=1, eqn)));
where
, then what is
is
?
> testeq( subs(r=12,eqn1) );
18. Find the solutions a and b,
area under
with a < b < 0 , of the equation
and over
for
. Calculate the
.
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
Solution : ( e )
> solve(1/x = -(4*x+5), x );
> int( 1/x+(4*x+5), x=-1..-1/4);
19. Calculate
.
a)
f)
b)
g)
c)
h)
d)
i)
e)
j)
Solution : ( b )
> Integrand := int(cot(x),x);
> answer := subs(x=Pi/2, Integrand) - subs(x=Pi/6, Integrand);
> simplify(answer);
20. Calculate
.
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
Solution : ( j )
> J := Int((sin(x)+1)/cos(x),x = 0 .. Pi/3);
> eqn := J = Int(expand((sin(x)+1)/cos(x)),x = 0 .. Pi/3);
> K := rhs(eqn);
> L := Int(op(1, student[integrand](K)), x = 0 .. Pi/3) + Int(op(2, student[integrand](K)), x = 0 .. Pi/3);
> antiderivative := int( op(1,student[integrand](L)) , x)+int( op(2,student[integrand](L)) , x);
> answer := subs(x=Pi/3, antiderivative) - subs(x=0, antiderivative);
> simplify(answer);
> testeq(simplify(answer) = ln(4 + 2*sqrt(3)));