Tutorial: Sigma notation, Riemann sums, definite integrals, popcubes, and all
Math 1310: CSM
SOLUTIONS IN RED
that
Part I: sigma notation and popcubes
Consider the sums
Sn =
n
X
i and
i=1
Tn =
n
X
i3 .
i=1
1. (a) Find S1 , S2 , S3 , S4 , and S5 .
P1
P2
P3
S
=
i
=
1,
S
=
i
=
1
+
2
=
3,
S
=
1
2
3
i=1
i=1
i=1 i = 1 + 2 + 3 = 6, S4 =
P4
P5
i=1 i = 10, S5 =
i=1 i = 15.
P
P
3
3
3
(b) Find T1 ,P
T2 , T3 , T4 , and T5 . T1 = 1i=1 i3 P
= 13 = 1, T2 = 2i=1
Pi5 = 31 + 2 =
3
4
3
3
3
3
3
9, T3 = i=1 i = 1 + 2 + 3 = 36, T4 = i=1 i = 100, T5 = i=1 i = 225.
(c) Do you see a relationship between Sn and Tn ? If so, what is this relationship?
Express you answer as an equation involving Sn and Tn .
It appears as though Tn = (Sn )2 for any n; certainly, this is true for n = 1, 2, 3, 4, 5.
2. We are now going to try and model the relationship that you noticed above GEOMETRICALLY.
(a) Using your popcubes, construct a linear (that is, one-dimensional) model of Sn .
(Hint: each di↵erent i should correspond to a di↵erent color pop cube.)
(b) Now, building up from your linear model in the previous part of this problem,
construct an area (that is, a two-dimensional) model of (Sn )2 .
(c) Can you now take apart your area model for (Sn )2 , and reassemble it (without
adding or removing any popcubes) into a volume model for Tn ? (Answer: yes,
you can!) What formula are you illustrating by doing so? (Hint: see above.)
Once you have done this reconstruction, for the value of n you’ve been working
with so far, do it for the n that’s one larger (if you have enough popcubes), to
make sure you REALLY get the idea.
1
Tutorial: Sigma notation, Riemann sums, definite integrals, popcubes, and all
Math 1310: CSM
SOLUTIONS IN RED
that
Part II: Riemann sums and integrals
3. On the axes below is the graph of y = x, from x = 0 to x = b.
y
b
(4b/5,4b/5)
(3b/5,3b/5)
(2b/5,2b/5)
(b/5,b/5)
b
x
(a) Draw, directly on top of this graph, rectangles representing a left endpoint RieRb
mann sum approximation, with n = 5, to 0 x dx. See above.
Rb
(b) Approximate 0 x dx, using the left endpoint approximation (with n = 5) that
you represented graphically in the previous part of this problem. Express your
answer in terms of b.
b a
b
ib
We have x =
= and xi = a + i x = , so
n
5
5
Z b
n 1
X
b b b 2b b 3b b 4b b
x dx ⇡
f (xi ) x = 0 · + · +
· +
· +
·
5
5
5
5
5
5
5
5
5
0
i=0
✓
◆
b2
10b2
2b2
=
0+1+2+3+4 =
=
.
25
25
5
2
Tutorial: Sigma notation, Riemann sums, definite integrals, popcubes, and all
Math 1310: CSM
SOLUTIONS IN RED
that
(c) Compute
Rb
0
x dx explicitly (by antidi↵erentiating).
Z
b
0
x2
x dx =
2
b
=
0
b2
.
2
(d) Is your Riemann sum approximation smaller or larger than the actual value of the
integral? If you had done a right endpoint approximation instead of a left endpoint
approximation, would you have gotten a number smaller than, or larger than, the
actual value of the integral? Please explain. The approximation underestimates
the integral. This is because, as we can see from the picture above, the sum of
the areas of the red rectangles is smaller than the area under the curve.
(e) Now suppose that, in part (b) above, you had used a general value of n, instead
of the particular value n = 5. What would your left endpoint approximation
look like? Express your answer in terms of n: YOU MAY USE THE FACT,
DEMONSTRATED GEOMETRICALLY IN CLASS, THAT
n
X
i=
i=1
Z
0
b
x dx ⇡
=
n 1
X
f (xi ) x =
i=0
2
n 1
X
i=0
b (n 1)
.
2n
n(n + 1)
.
2
n 1
n 1
ib b X ib b
b2 X
b2 (n 1)n
f( ) =
· = 2
i= 2 ·
n n
n n
n i=0
n
2
i=0
(f) Using your answer from part (e) above, demonstrate that, in the situation at
hand, the Riemann sum approximations really DO converge to the integral as
n ! 1.
Z
0
b
x dx = lim
n!1
n 1
X
i=0
✓
b2 (n 1)
b2
f (xi ) x = lim
= lim
1
n!1
n!1 2
2n
3
1
n
◆
=
b2
.
2
Tutorial: Sigma notation, Riemann sums, definite integrals, popcubes, and all
Math 1310: CSM
SOLUTIONS IN RED
that
4. On the axes below is the graph of y = x3 , from x = 0 to x = b.
3
b
y
(5b/5, (5b/5)3)
(4b/5, (4b/5)3)
(3b/5, (3b/5)3)
(2b/5, (2b/5)3)
(b/5, (b/5)3)
b
x
(a) Draw, directly on top of this graph, rectangles representing a right endpoint RieRb
mann sum approximation, with n = 5, to 0 x3 dx. See above.
Rb
(b) Approximate 0 x3 dx, using the right endpoint approximation (with n = 5) that
you represented graphically in the previous part of this problem. Express your
answer in terms of b.
b a
b
ib
We have x =
= and xi = a + i x = , so
n
5
5
Z b
n
X
x3 dx ⇡
f (xi ) x
0
i=1
✓ ◆3
✓ ◆3
✓ ◆3
✓ ◆3
✓ ◆3
b
b
2b
b
3b
b
4b
b
5b
b
=
· +
· +
· +
· +
·
5
5
5
5
5
5
5
5
5
5
✓
◆
4
4
4
b
225b
9b
=
13 + 23 + 33 + 43 + 53 =
=
.
625
625
25
4
Tutorial: Sigma notation, Riemann sums, definite integrals, popcubes, and all
Math 1310: CSM
SOLUTIONS IN RED
that
Rb
(c) Compute 0 x3 dx explicitly (by antidi↵erentiating). What is the relationship
Rb
between this integral and the integral 0 x dx? (Write down a simple formula
relating these two integrals.)
Z
0
b
x4
x dx =
4
b
3
=
0
b4
.
4
Rb
(d) Is your Riemann sum approximation to 0 x3 dx smaller or larger than the actual
value of the integral? If you had done a left endpoint approximation instead of a
right endpoint approximation, would you have gotten a number smaller than, or
larger than, the actual value of the integral? Please explain.
The approximation overestimates the integral. This is because, as we can see
from the picture above, the sum of the areas of the red rectangles is larger than
the area under the curve.
(e) Now suppose that, in part (b) above, you had used a general value of n, instead
of the particular value n = 5. What would your right endpoint approximation
look like? Express your answer in terms of n: YOU MAY USE THE RESULTS
OF PROBLEM 1(c) ABOVE.
Z
0
b
n ✓ ◆3
ib b X ib
b
x dx ⇡
f (xi ) x =
f( ) =
·
n n
n
n
i=1
i=1
i=1
✓
◆2
n 1
b4 X 3
b4
n(n + 1)
= 4
i = 4·
n i=0
n
2
n
X
3
=
n
X
b4 (n + 1)2
.
4n2
(f) Using your answer from part (e) above, demonstrate that, in the situation at
hand, the Riemann sum approximations really DO converge to the integral as
n ! 1.
Z
0
b
x dx = lim
n!1
n
X
i=1
✓
◆2
b4 (n + 1)2
b4
1
b4
f (xi ) x = lim
= lim
1+
= .
n!1
n!1 4
4n2
n
4
5
Tutorial: Sigma notation, Riemann sums, definite integrals, popcubes, and all
Math 1310: CSM
SOLUTIONS IN RED
that
Part III: Density and infinitesimal arguments
5. We are now going to compute the total mass of a square sheet of side length b, and
density
⇢(x, y) = xy,
3:15:11 PM
in two di↵erent ways: first, by imagining the sheet is made up of infinitely many,
infinitesimally thin, horizontal rods of length b and width dy (see Figure 1, below),
and second, by imagining that the sheet is made up of infinitely many, infinitesimally
thin, L-shaped regions of length s and width ds (see Figure 2, below).
4:24:42 PM
Nov 7, 2010
Nov 7, 2010
b
s
b
Figure 1.
b
Figure 2.
b2
(a) Show that the bar in Figure 1 has mass equal to
y dy. Hint: integrate with
2
respect to x.
The total mass of this bar is
Z b
Z b
Z b
b
x2
b2
⇢(x, y) dx dy =
xy dx dy = y dy
x dx = y dy
= y dy.
2 0
2
0
0
0
b4
. Hint: integrate your
4
answer from part (a) with respect to y. The total mass of this sheet is
(b) Show that the square sheet in Figure 1 has mass equal to
Z
0
b
b2
b2
y dy =
2
2
6
Z
0
b
b2 y 2
y dy =
2 2
b
=
0
b4
.
4
Tutorial: Sigma notation, Riemann sums, definite integrals, popcubes, and all
Math 1310: CSM
SOLUTIONS IN RED
that
(c) Show that the shaded L-shaped region in Figure 2 has mass equal to s3 ds. Hint:
to get the mass of the horizontal portion of this region, integrate with respect to
x from 0 to s. Then, by symmetry, the mass of the entire region is just twice that
of its horizontal portion.
Much as in part (a) above, the total mass of the horizontal portion of this corner
is
Z s
Z s
Z s
s
x2
s3
⇢(x, s) dx ds =
xs dx ds = s ds
x dx = s ds
= ds.
2 0
2
0
0
0
Doubling this, we find that the entire mass of the L-shaped region is s3 ds.
b4
(d) Show that the square sheet in Figure 2 has mass equal to . Hint: integrate your
4
answer from part (c) with respect to s.
The total mass of this sheet is
Z b
b4
s3 ds = .
4
0
6. What does problem 5 above have to do with the rest of this worksheet?
We computed the mass of the sheet by thinking ofRit in two di↵erent ways. The first
b
way amounted, in essence, to squaring the integral 0 x dx. The second way amounted
Rb
to evaluating the integral 0 x3 dx. Both ways gave the same answer, this reconfirming
that
✓Z b
◆2 Z b
x dx =
x3 dx.
0
0
7