Calculus 1 Assignment 7 Solutions
Alex Cowan
[email protected]
1. Partition the interval [0, π] into n pieces of equal length. The endpoints of this interval are given by k nπ
for k going from 0 to n. For example, if you split [0, π] into 5 equal parts, then the endpoints of the pieces
are 0, π5 , 2 π5 , 3 π5 , 4 π5 , π. Above each of these mini-intervals, draw a rectangle with height equal to the height
of the curve above the right endpoint (other choices of how to draw the rectangle will work too). The sum
of the areas of these rectangles will be a reasonable estimate of the area under the curve. The sum of these
areas is length of base of the first rectangle·height of the first rectangle+length of base of the second rectangle·
height of the second rectangle + · · · , which here is
π π
π
π
π
π
· sin
+ · sin 2
+ · · · + · sin n
n
n
n
n
n
n
We can write this more compactly using summation notation:
Area ≈
n
X
π
k=1
π
sin k
n
n
To get the area exactly, we need to take the limit as n goes to infinity. This corresponds to making each individual
rectangle smaller, and filling the area under the curve with more of them, to get rid of excess bits. The area is
exaclty
n
π
X
π
lim
sin k
n→∞
n
n
k=1
It was given that this limit is equal to
π π
cot
n→∞ n
2n
π
As n goes to infinity, 2n
gets very small, so the cos(π/2n) in the numerator goes to 1. The denominator sin(π/2n)
can be approximated as π/2n, since sin x ≈ x for small x. Hence the limit is
lim
π · 1 · lim
n→∞
1
π =2
n 2n
The area under the curve is exactly 2.
2. In a short time ∆t near a time t, the ball will travel a distance approximately v(t)∆t. For example, if
at t = 1s the ball is travelling at a speed 1 m
s , then between t = 0.99 s and t = 1 s, the ball will travel roughtly
0.01 s · 1 m
=
0.01
m.
s
For the first n1 of a second, the ball is travelling
at roughly v n1 s . In this time it travels v n1 s n1 s. In the
second n1 s, the ball travels roughly v n2 s n1 s. Overall the ball is going to travel a distance
n
X
1
1
v j s
s
n
n
j=1
Substituting v(t) = 1 sm3 t2 gives
2
n
n
X
m
1
1
1 X 2
1 3 j s
s = 1m 3
j
s
n
n
n j=1
j=1
1
Using the given sum, this is
n(n + 1)(2n + 1)
m
6n3
This is an approximation, but if we let n go to infinity, then we will be exacty right. The limit of this quantity
as n goes to infinity is 13 m.
p
3. The distance between two points (x0 , y0 ) and (x1 , y1 ) is (x1 − x0 )2 + (y1 − y0 )2 (using pythagoras). Cut
the interval [0, 1] into n equal pieces. The x-coordinates of the ends of the pieces are j/n
r for j going from 0 to n,
2
2 2
and the corresponding y coordinates are (j/n)2 . Thus the length of each short line is n12 + nj − j−1
n
To estimate the total length of the curve, add these up, and to get the length exactly, take the limit as n goes
to infinity:
v
u
2 2 !2
n
Xu 1
j−1
j
t
+
−
lim
n→∞
n2
n
n
j=1
4.
a) C
b) x + C
c) 21 x2√+ C
d) −6 x
a
e) n+1
xn+1 + C if n 6= −1 and log |x| + C if it is.
f) log |x| + C
g) − cos(x) + C
h) 12 (sin x)2 + C
i) 1c ecx + C
x2
j) e√
+C
k) x2 + 1 + C
l) xf (x) + C
m) x log x − x + C
2