MA 104 Graded Homework 2 Solutions
Dr. L. Zhao, Sections B7, C7, D7
Issued: Monday 23 February 2004
Due: Beginning of class, Monday 8 March 2004
Z
0
(2 +
(1) (5 Points) Evaluate the integral by interpreting it in terms of areas:
−3
p
Consider the graph of y = f (x) = 2 + 9 − x2 . We have
p
y = 2 + 9 − x2 =⇒ (y − 2)2 = 9 − x2 =⇒ x2 + (y − 2)2 = 32 ,
p
9 − x2 )dx.
which is the equation of a circle of radius 3 centered at (0, 2). The integral can be interpreted
as the area of a quarter of the said circle and the area of the rectangle whose vertices are
(0, 0), (0, 2), (−3, 2) and (−3, 0). Therefore, we have
Z 0
p
1
9
(2 + 9 − x2 )dx = × π × 32 + 2 × 3 = π + 6.
4
4
−3
A plot of the figure and the numerical computations can be done on Mathematica with the
following commands.
Plot[2 + Sqrt[9 - x^2], {x, -3, 0}]
Integrate[2 + Sqrt[9 - x^2], {x, -3, 0}]
Z 4 √x
e
√ dx.
(2) (5 Points) Evaluate the following definite integral
x
0
This problem is done with Mathematica using the following command.
Integrate[Exp[Sqrt[x]]/Sqrt[x], {x, 0, 4}]
√
dx
dx
Manually, we set u = x and get du = √ or 2du = √ . We have
x
2 x
Z x=4
Z 4 √x
x=4
√ 4
e
√ dx =
= 2e x = 2(e2 − 1).
2eu du = 2eu x
x=0
0
x=0
0
(3) The following table gives the values of electric intensity I(t) at a point on a circuit at time
t seconds.
t(seconds)
0
0.5 1.0 1.5 2.0 2.5 3.0
I(t) (Amperes) 19.3 19.0 18.3 16.5 12.3 3.3 2.6
(a) (5 Points) Estimate the amount of electric charge passed through the point from t = 0
to t = 3 with the right endpoints as the sample points.
The total amount of electric charging passing through the point is
0.5 × 19.0 + 0.5 × 18.3 + 0.5 × 16.5 + 0.5 × 12.3 + 0.5 × 3.3 + 0.5 × 2.6 = 36 columb.
(b) (3 Points) Suppose it is known that I(t) is a decreasing function, is our estimate in
part (a) an over-estimate or under-estimate? Why?
If I(t) is a decreasing function, then for each of the subintervals from 0 to 3, I(t) at the
right-hand end point is less than all values of I(t) in the sub-interval. Hence, the approximating rectangles lie below the graph of I(t) and that is to say the approximation
we have in part (a) is an under-estimate.
1
2
MA 104 HOMEWORK 2 SOLUTIONS
(c) (2 Points) Had we used the left end point as the sample points and supposed that I(t)
is a decreasing function, would our estimate be an over-estimate?
Had we use the left-hand end points as the sample points, we would have gotten an
over-estimate via similar reasoning as in part (b).
Z 2p
4 − x2 dx.
(4) (10 Points) Use the mid-point and trapezoidal rules with n = 8 to approximate
−2
2 − (−2)
1
The interval of interest is [−2, 2]. We have ∆x =
= . The end points of the
8
2
i
subintervals are xi = −2 + i∆x = −2 + . The mid-point of the i-th subinterval is
2
1
1
i
i−1
2i − 1
x̄i = (xi + xi−1 ) =
.
−2 +
+ −2 +
= −2 +
2
2
2
2
4
Hence, for the mid-point rule
s
Z 2p
8
8
X
X
1
2i − 1 2
4 − x2 dx ≈
∆xf (x̄i ) =
= 6.36786.
4 − −2 +
2
4
−2
i=1
i=1
This can be done on Mathematica with the following command.
Sum[1/2*Sqrt[4 - (-2 + (2i - 1)/4)^2], {i, 1, 8}] // N
For the trapezoidal rule, we have
Z 2p
∆x
4 − x2 dx ≈
[f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (x6 ) + 2f (x7 ) + f (x8 )]
2
−2


s
2
7
p
X
i
1 p

4 − (−2)2 + 4 − 22 +
2 × 4 − −2 +
=
4
2
i=1
= 5.99142.
The above can be done on Mathematica with the following command.
1/4*(Sqrt[4-(-2)^2]+Sqrt[4-2^2]+Sum[2*Sqrt[4 - (-2 + i/2)^2], {i, 1, 7}]) // N
For the cognoscenti, this integral can be viewed as the area of a semi-circle centered at the
origin with radius 2. Hence the exact value of this integral is 2π. The value of this integral
and that of the integral in Problem (1) can be computed algebraically without interpreting
them as areas via a somewhat involved substitution.
(5) (5 Points) A uniform cable hanging over the edge of a tall building is 40 meters long and
has mass 60 kilograms. How much work is require to pull the cable completely to the top?
(You may assume gravitational acceleration is 9.8m/s2 = 9.8N/Kg.)
588
= 14.7
The rope weighs 60×9.8 = 588 Newtons. Hence, the weight density of the rope is
40
Newtons per meter. Divide the rope into n equal pieces, each of length ∆x. The weight of
each piece is 14.7∆x and the work needed to lift that piece to the top is 14.7∆xi , where xi
is the position of the section in the rope. The total work needed to lift the rope to the roof
is
Z 40
n
40
X
14.7xdx = 7.35x2 = 7.35 × 402 = 11760 J.
∆x14.7xi =
Total Work = lim
n→∞
i=1
0
0
SECTIONS B14, C14, D14
3
In Mathematica, this is done via the following command.
Integrate[14.7x, {x, 0, 40}]
(6) (5 Points) If 10 J of work is needed to stretch a spring from 10 cm to 11 cm and another
15 J to stretch it from 11 cm to 12 cm, what is the natural length of the spring?
Let L meters be the natural length of the spring and k be its spring constant. We have
Z 0.12−L
Z 0.11−L
kxdx = 15.
kxdx = 14 and
(i)
0.1−L
0.11−L
Note that 10 centimeters is 0.1 meters and the same goes for the other quantities.This
problem can now be solved using Mathematica via the following command.
Solve[{Integrate[k*x, {x, 0.1-L, 0.11-L}] == 14,
Integrate[k*x, {x, 0.11-L, 0.12-L}] == 15}, {L, k}]
The answer is that L = 0.035 meters and the natural length of the spring is 3.5 centimeters.
Manually, the equations in (i) can be re-written as
k
k
(0.11 − L)2 − (0.10 − L)2 = 14, and
(0.12 − L)2 − (0.11 − L)2 = 15.
2
2
Expanding the left-hand sides of both the above equations and simplify, we get
k
k
(0.0021 − 0.02L) = 14 and (0.0023 − 0.02L) = 15.
2
2
Divide the first equation by the second, we get
14
0.0021 − 0.02L
=
=⇒ 15(0.0021 − 0.02L) = 14(0.0023 − 0.02L)
0.0023 − 0.02L
15
=⇒ −2L = 14 × 0.23 − 15 × 0.21 =⇒ L = −0.035 meters.
It is of course absurd to have a negative length, but the my negligence has caused this
mistake to happen.
√
x
(7) (5 Points) A supply curve is given by p = 15 +
. Find the producer surplus when the
5
selling price is $10.
First, setting the price to $16, we have
√
√
√
x
x
=⇒ 1 =
=⇒ 5 = x =⇒ x = 25.
16 = 15 +
5
5
Hence, the producer surplus is
√ Z 25
x
2 3 25 25
= 8.33333.
dx = x − x 2
=
16 − 15 +
5
15
3
0
0
In Mathematica, this can be done via the following commands.
price[x_] := 15 + Sqrt[x]/5
Solve[price[x] == 16, x]
Integrate[Abs[16 - price[x]], {x, 0, 25}] // N
4
MA 104 HOMEWORK 2 SOLUTIONS
(8) (5 Points) For a constant c given, find all functions f (x) that satisfy the equation
Z x
(ii)
f (t)dt = [f (x)]2 + c.
0
(iii)
By the fundamental theorem of calculus, the left-hand side of (ii) can be re-written as
Z x
f (t)dt = F (x) − F (0) =⇒ F (x) − F (0) = [f (x)]2 + c,
0
(iv)
where F (x) is an anti-derivative of f (x). Differentiating with respect to x on both sides of
the last equality (iii), we have
o
d
df (x)
d n
[f (x)]2 + c =⇒ f (x) = 2f (x)
(F (x) − F (0)) =
.
dx
dx
dx
df (x)
1
1
Hence it must be that
= or f (x) = x + c1 , for some constant c1 . Inserting this
dx
2
2
back into (ii) in order to solve for c1 , we have
Z x
i2
hx
t
+ c1 dt =
+ c1 + c.
2
0 2
The left-hand side of this above is
x
2
x2
t
+ c1 t =
+ c1 x,
4
4
0
while expanding the right-hand side yields
(v)
x2
+ c1 x + c21 + c.
4
√
If the expressions in (iv) and (v) are to be the same, we must have c21 + c = 0 or c1 = ± −c.
Therefore, if c < 0, then the functions that satisfy (ii) are
√
1
f (x) = x ± −c.
2
Note that, there is no real-valued function satisfying (ii) if c > 0. Finally, for the special
case of c = 0, there are two functions that satisfy the equation in question, namely
1
f (x) = x, and f (x) = 0.
2
For the cognoscenti, the statement regarding c > 0 can be seen more readily by setting
x = 0 in (ii). In that case, the left-hand side of (ii) is zero, while and right-hand side of (ii)
is positive due to the fact that c > 0 and a square of a real number is always non-negative.
Moreover, I doubt one can incorporate the use of Mathematica into this problem.