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Math 106: Review for Exam I
- Find the following. [Substitution tip: usually let u = a function that’s “inside” another function,
especially if du (possibly off by a multiplying constant) is also present in the integrand.]
(a) Let u =
x, so du =
dx
x
and 2 du =
dx
x
4
1
e
√
x
x
dx
4
1
e
√
x
x
dx
4
1
e
√
x
x
dx =
x=
x=
e
u
· 2 du If you prefer to switch the limits, use u = 1 to u = 2.
= 2e
u
x=
x=
= 2e
√
x
4
1
= 2e
2
− 2 e (≈ 9 .342)
(b) Let u = cos(5x), so du = −5 sin(5x) and −du/5 = sin(5x).
cos
7
(5x) sin(5x) dx
cos
7
(5x) sin(5x) dx
cos
7
(5x) sin(5x) dx =
u
7
−du
u
7
du
u
8
+ C
cos
8
(5x) + C
(c) Use u = x
4
, so du = 4x
3
dx and 3 du = 12x
3
dx.
3
2
12 x
3
1 + x
8
dx
3
2
12 x
3
1 + x
8
dx
3
2
12 x
3
1 + x
8
dx =
81
16
3 du
1 + u
2
This time we have switched the limits.
= 3 arctan u
81
16
= 3 arctan(81) − 3 arctan(16)(≈ 0 .150)
(d) Use #42 from the table of integrals with n = 3 and a = 5.
cos
3
(5x) dx
cos
3
(5x) dx
cos
3
(5x) dx =
cos
2
(5x) sin(5x)
cos(5x) dx
cos
2
(5x) sin(5x)
cos(u)
du
Subsitute u = 5x, so du = 5 dx.
cos
2
(5x) sin(5x)
sin(u) + C
cos
2
(5x) sin(5x)
sin(5x) + C
- If f(x)
f(x) f(x) is decreasing and concave up, put the following quantities in ascending order.
L
100
, R
100
, T
100
, M
100
b
a
f(x) dx
L
100
, R
100
, T
100
, M
100
b
a
f(x) dx L 100
, R
100
, T
100
, M
100
b
a
f(x) dx R 100
< M
100
b
a
f(x) dx < T 100
< L
100
What can you say with certainty about where S 200
S
200
S
200
would fit into your list above?
It would be somewhere between M 100
and T 100
but we don’t know how it compares to
b
a
f(x) dx.
- Suppose f(t)
f(t) f(t) is the rate of change (in animals per month) of a population P (t)
P (t) P (t).
(a) What does
12
4
f(t) dt
12
4
f(t) dt
12
4
f(t) dt represent in this problem?
It represents the total (or net) change in the number of animals during the time period [4, 12].
(b) Find the best possible left, right, midpoint, trapezoidal, and Simpson’s approxima-
tions to
12
4
f(t) dt
12
4
f(t) dt
12
4
f(t) dt given the data in the table below.
ttt 4 6 8 10 12
fff(((ttt))) 15 11 8 4 3
L
4
= (15 + 11 + 8 + 4)(2) = 76 R
4
= (11 + 8 + 4 + 3)(2) = 52 T
4
L
4
+ R
4
We cannot compute M
4
because it requires the values of f at x = 5, 7, 9, and 11. Instead, we do
M
2
M
2
= (11 + 4)(4) = 60 Now, to find S
4
, we need T
2
L
2
+ R
2
S
4
2 M
2
+ T
2
- Find bounds for each of the following errors if I =
7
2
I = ln x dx
7
2
I = ln x dx
7
2
ln x dx.
(a) |I − L 100
K
1
(b − a)
2
2 n
1
2
2
K
1
= max of |f
′
(x)| on [2, 7] = max of
x
on [2, 7] =
(occurs at x = 2)
(b) |I − T 100
K
2
(b − a)
3
12 n
2
1
4
3
2
K
2
= max of |f
′′
(x)| on [2, 7] = max of
x
2
on [2, 7] =
(occurs at x = 2)
(c) |I − M 100
K
2
(b − a)
3
24 n
2
1
4
3
2
K
2
= same as in previous part
- Use Euler’s method with three steps on the differential equation
dy
dt
= y − t
dy
dt
= y − t
dy
dt
= y − t to estimate
y(2.5)
y(2.5) y(2.5) if y(1) = 0
y(1) = 0 y(1) = 0.
t y
dy
dt
· ∆t = ∆y
So, y(2.5) ≈ − 3 .25 (or -13/4).
- A pyramid has a square base 30 feet to a side and a height of 10 feet. Write integrals
equal to
(a) the volume of the pyramid
We slice horizontally, so each slice is a “box”
with a square top and bottom and a height
(thickness) of ∆h, as shown to the right.
s
s
∆h
The picture shown below is a vertical cross-section through the center of the pyramid.
A
A
A
A
A
A
10 − h
s
∆h
h
Similar triangles:
10 − h
s
⇒ s = 3(10−h).
volume of slice ≈ s
2
∆h ≈ [3(10 − h)]
2
∆h
total volume =
10
0
[3(10 − h)]
2
dh
(b) the work done in pumping all the fluid to a point 5 feet above the pyramid if the
pyramid is filled to a height of 8 feet with water (62.4 pounds per cubic foot)
We use the same sketch as in the previous part.
volume of slice ≈ s
2
∆h ≈ [3(10 − h)]
2
∆h From above.
weight of slice ≈ 62 .4[3(10 − h)]
2
∆h Weight=(density)(volume).
work to lift slice ≈ 62 .4[3(10 − h)]
2
∆h(15 − h) Work=(force)(distance); here, force=weight.
total volume = 62. 4
8
0
[3(10 − h)]
2
(15 − h) dh
