MATH 18.01, FALL 2014 – PROBLEM SET #7
1. We have
E[(x − x̄)2 ] =
Z
1
(x − x̄)2 f (x)dx =
Z
1
x2 f (x)dx + x̄2
Z
Z
1
f (x)dx − 2x̄
xf (x)dx
0
2
0
0
2
0
1
= E(x2 ) + x̄ E(1) − 2x̄E(x) = E(x2 ) + x̄2 − 2x̄ = E(x2 ) − x̄2 .
2. (a) Let R be the region of the unit square where xy ≤ c. The probability that xy ≤ c
is then equal to the area of R divided by the total area of the square, which is 1. The region
R is the part of the unit square which lies below the graph of y = c/x. If c ≤ 1, then the
whole square is below that graph, so R is the whole square and the probability is 1. If c < 1,
the graph intersects the top side of the square at x = c, so the area of the region R is
Z 1
c
c+
dx = c + c ln(1) − c ln(c) = c(1 − ln(c)).
c x
Z N
1 −at t=N
1
1
−at
(b) W = lim
e dt = lim
e = lim (e0 − e−aN ) = .
N →∞ 0
N →∞ −a
N
→∞
a
a
t=0
(c) By (b), we have
Z T
P ([0, T ]) = a
e−at dt = 1 − e−aT .
0
e−aT
If P ([0, T ]) = 1/2, then
= 1/2, so the half-life is T =
half-life of 1 hour, then a = ln(2), so
1
a
ln(2) hours. If a particule has a
2T − 1
.
2T
In particular, P ([0, 10]) = 1023/1024 = 0.9990234375 and
1, 267, 650, 600, 228, 229, 401, 496, 703, 205, 375
P ([0, 100]) =
1, 267, 650, 600, 228, 229, 401, 496, 703, 205, 376
≈ 0.999999999999999999999999999999.
Z N
1
(d) W = lim
dt = lim arctan(N ) = π/2.
N →∞ 0 1 + t2
N →∞
(e) By (d),
Z
2 T 1
2
P ([0, T ]) =
dt = arctan(T ).
2
π 0 1+t
π
Thus, P ([0, 10]) ≈ 0.9365 and P ([0, 100]) ≈ 0.9936.
3. Assume without loss of generality that x0 = −h, x1 = 0, and x2 = h. The number C is
the intersection of the parabola with the y-axis, so C = y1 . The area beneath the parabola
is
Z h
A 3 B 2
2A 3
h
2
(Ax + Bx + C)dx =
x + x + Cx =
h + 2hy1 .
3
2
3
−h
−h
P ([0, T ]) = 1 − 2−T =
We have y0 = Ah2 − Bh + C and y2 = Ah2 + Bh + C, whence
2Ah2 = y0 + y2 − 2y1 .
1
2
MATH 18.01, FALL 2014 – PROBLEM SET #7
Thus, the area is
h
h
(y0 + y2 − 2y1 ) + 2hy1 = (y0 + y2 + 4y1 ).
3
3
4.
x 0
1/8
1/4
3/8
1/2
5/8
3/4
7/8
1
sinc(x) 1 .997398 .989616 .976727 .958851 .936156 .908851 .877193 .841471
Recall that Simpson’s rule with 2n intervals of length h is
Z b
n
hX
f (x)dx ≈
(y2i−2 + 4y2i−1 + y2i ).
3
a
i=1
R1
Using 4 intervals of length 1/4, we get 0 sinc(x)dx ≈ .946087.
R1
Using 8 intervals of length 1/8, we get 0 sinc(x)dx ≈ .946083.
The latter matches the actual value.
5. (a) Let u = tan x, du = sec2 x dx. Then
Z
Z
n+2
tan
x dx = tann x(sec2 x − 1)dx
Z
Z
Z
1
n
n
n+1
= u du − tan x dx =
tan
x − tann x dx.
n+1
(b) Applying (a) for n = 2 and for n = 0, we get:
Z
Z
Z
1
1
3
2
3
4
tan x dx = tan x − tan x dx = tan x − tan x + tan0 x dx
3
3
1
= tan3 x − tan x + x.
3
6. It is possible to find the area of the segment by elementary means, by taking the area
of the whole sector and subtracting the area of the triangle. This gives:
p
area(segment) = area(sector) − area(triangle) = a2 arccos(b/a) − b a2 − b2 .
We can give a different proof using integration. If we place the x-axis on the chord with
the origin at the closest√point from the center of
the region
√ the circle, the segment becomes
√
below the graph of y = a2 − x2 − b between ± a2 − b2 . Let’s abbreviate a2 − b2 to c and
let
α = arccos(b/a) = arcsin(c/a)
be the half-angle of the sector. Using the substituion x = a sin θ, we have
p
a2 − x2 = a cos θ and dx = a cos θ dθ,
whence
Z α
p
2
2
area(segment) =
( a − x − b)dx =
(a2 cos2 θ − ba cos θ)dθ
−c
−α
α
2θ
2 2 sin θ cos θ
= a +a
− ba sin θ
= a2 α + bc − 2bc = a2 α − bc.
2
4
−α
Z
c
Thus we recover the above formula.