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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
2
x
and 2 du =
dx √ x
1
e
√ x √ x
dx
1
e
√ x √ x
dx
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).
This time, we’ll change the limits:
x = π ⇒ u = cos(5 · π) = −1 and x = 2π ⇒ u = cos(5 · 2 π) = 1
∫ (^2) π
π
cos 7 (5x) sin(5x) dx
∫ (^2) π
π
cos 7 (5x) sin(5x) dx
∫ (^2) π
π
cos 7 (5x) sin(5x) dx =
− 1
u 7 ·
−du
5
− 1
u 7 du
u^8
8
1
− 1
[
8
8
]
(c) Use u = x 3 , so du = 3x 2 dx and
du
3
= x 2 dx.
7 x 2
1 + x^6
dx
7 x 2
1 + x^6
dx
7 x 2
1 + x^6
dx = 7
∫ (^) du 3 1 + u^2
arctan u + C
arctan(x 3 ) + C
(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)
3 · 5
cos(5x) dx
cos^2 (5x) sin(5x)
15
cos(u)
du
5
Subsitute u = 5x, so du = 5 dx.
cos 2 (5x) sin(5x)
15
sin(u) + C
cos^2 (5x) sin(5x)
sin(5x) + C
(e) Use u = 10 − x, so du = −dx and dx = −du.
We’ll change the limits:
x = 6 ⇒ u = 10 − x = 4 and x = 10 ⇒ u = 10 − 10 = 0
∫ (^10)
6
x
10 − x dx
6
x
10 − x dx
6
x
10 − x dx =
4
(10 − u)
u(−du) Since u = 10 − x, we know x = 10 − u.
4
(u − 10)
u du
4
(u 3 / 2 − 10 u 1 / 2 )du
[
u 5 / 2 −
u 3 / 2
] 0
4
5 / 2 −
3 / 2
- If fff(((xxx))) is decreasing and concave up, put the following quantities in ascending order.
L 100 , R 100 , T 100 , M 100 ,
∫ (^) b
a
L 100 , R 100 , T 100 , M 100 , f(x) dx
∫ (^) b
a
L 100 , R 100 , T 100 , M 100 ,^ f(x)^ dx
∫ (^) 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 SSS 200200200 would fit into your list above? [8: and 9:30 sections may omit this part.]
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 fff(((ttt))) is the rate of change (in animals per month) of a population PPP (((ttt))).
(a) What does
4
f(t) dt
4
f(t) dt
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
4
f(t) dt
4
f(t) dt
4
f(t) dt given the data in the table below.[8:00 and 9:30 sections may omit
Simpson’s approximation.]
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 =
I = ln x dx
I = ln x dx
ln x dx.
(b) the area bounded by y = x 2 y = x − 8 x + 24 2 y = x^2 −−^88 xx^ + 24+ 24 and yyy = 3= 3= 3xxx
0 2 4 6
y
30
25
20
15
10
x
5
0 8 10
First, find where the curves intersect.
x 2 − 8 x + 24 = 3 x
x 2 − 11 x + 24 = 0
(x − 3)(x − 8) = 0
⇒ x = 3, x = 8
Between x = 3 and x = 8, y = 3x is above y = x 2 − 8 x + 24. (Plug in x = 5 or graph to check.)
So, the area between them is ∫ (^8)
3
[3x − (x
2 − 8 x + 24)] dx.
[This equals 125/6.]
- Consider the region bounded by y =
y = x
y = x
x, yyy = 0= 0= 0, and xxx = 9= 9= 9. Write an integral equal to the volume generated if this region is rotated about
(a) the xxx-axis
x=
x
volume of slice ≈ πr 2 ∆x
= πy 2 ∆x
= π(
x) 2 ∆x
= πx∆x
total volume = π
0
x dx
(b) the line xxx === −−− 111
r = inner radius = 1+x
y
x=−1 R = outer radius = 10
volume of slice ≈ πR 2 ∆y − πr 2 ∆y
= π( 2 )∆y − π(1 + x) 2 ∆y
= π[100 − (1 + y 2 ) 2 ]∆y
total volume = π
[100 − (1 + y 2 ) 2 ] dy
- 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 [8:00 and 9:30 sections may omit this part, though much of it will be repeated in part(b).]
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 10
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 =
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
[3(10 − h)]
2 (15 − h) dh
