Math 237. Calculus II
Solutions to the Homework for Section 5.1
Assigned: 11, 13, 14, 17, 19, 20, 23, 25, 26, 28, 31, 32, 33, 34
Selected for Grading: 13, 23, 28, 34
Solutions.
11. The region R is bounded by
and y = 0, between x = 0 and x = 3. Here is a sketch.
4
2
Approximating:
ΔAi = (height)(base) = (yi)Δxi = (3 – (xi2)/3)Δxi
So
and so
Integrating:
13. For this one R is the region bounded by y= (x – 4)(x+2) and y = 0, between x = 0 and x = 3. Here is a
sketch.
y


x







Approximating:
ΔAi = [0 – (x – 4)(x + 2)] Δx = – (x2 – 2x – 8)Δx = (8 + 2x – x2) Δx
So
Integrating:
14. Here is a sketch of the region R bounded by y = x2 – 4x – 5 and y = 0, from x = –1 to x = 4.
y










Approximating:
ΔAi = – (xi2 – 4xi – 5)Δxi
Integrating:
and the x-axis from x = –2 to x = 2.
17. Here, to get started, is a sketch of the region R bounded by

y

x






I'll have to slice this up twice: once for the interval [–2, 0], and once for the interval [0, 2].
On [–2, 0]:
On [0, 2]:
Integrating:
19. Here is a sketch of the given curves, y = (x – 3)(x – 1) and y = x.

y

x
a

b

I need to find the x-values a and b.
(x – 3)(x – 1) = x
x2 – 4x + 3 = x
x2 – 5x + 3 = 0
Approximating:
ΔAi = [xi – (xi – 3)(xi – 1)] Δxi = [xi – (xi2 – 4xi + 3)] Δxi = [–xi2 + 5xi – 3] Δxi
So
.
Integrating:
And you can factor out
from all three terms:
20. The region bounded by
, y = x – 4, and x = 0 (the y-axis) looks like this:
y


x


 b



The two curves intersect at . . .
x – 4 = x1/2
x2 – 8x + 16 = x
x2 – 9x + 16 = 0
Approximating:
23. Here is a sketch of the region bounded by x = 8y – y2 and x = 0.
y




x

ΔA = (8y – y2)Δy




25. Here are two sketches of the region bounded by x = –6y2 + 4y and x + 3y – 2 = 0, one a "blow-up" of the
other.
(0, 2/3)










(0.5, 0.5)

ΔA = [(–6y2 + 4y) – (2 – 3y)] Δy
26. The region R bounded by the curves x = y2 – 2y and x – y – 4 = 0 looks like the following.

y




x









I need to know the limits of integration:
y2 – 2y = y + 4
y2 – 3y – 4 = 0
(y – 4)(y + 1) = 0
y = –1, 4.
Approximating:
ΔA = (xright – xleft)Δy = [(y + 4) – (y2 – 2y)] Δy = [–y2 +3y + 4] Δy
Integrating:

28. The region bounded by x = 4y4 and x = 8 – 4y4 looks like . . .
y

x









First, I need to verify that the limits of integration are ±1.
4y4 = 8 – 4y4
8y4 = 8
y4 = 1
y = ±1.
Approximating:
ΔA = (xright – xleft)Δy = [(8 – 4y4) – 4y4] Δy = 8(1 – y4)Δy
Integrating:
31. To find displacement we merely integrate the velocity, but to find the total distance traveled we integrate the
absolute value of the velocity. What I'll do first is to find the places where the velocity changes sign.
v(t) = 3t 2 – 24t + 36 = 3(t 2 – 8t + 12) = 3(t – 2)(t – 6).
For –1 ≤ t < 2, the velocity is positive,
for 2 < t < 6, the velocity is negative, and
for 6 < t ≤ 9, the velocity is positive.
So to find the total distance, I'll evaluate the following three integrals.
To find the displacement we integrate only once.
32. This time we are given a different function for velocity:
and the interval is [0, 3π/2].
The steps are the same. I'll begin by determining where the velocity changes sign.
so
But only the first two of these are in the given interval.
v(t) is positive for 0 ≤ t < 7π/12, negative for 7π/12 < t < 11π/12, and positive for 11π/12 < t ≤ 3π/2.
To find the displacement, we integrate only once.
But to find total distance, we integrate three times.
33. To determine how long it takes for the object to reach s = 12, integrate the velocity from zero to an unknown
upper limit of integration, set that integral equal to 12, and solve for that upper limit.
b 2 – 4b – 12 = 0
(b – 6)(b + 2) = 0
b = –2, 6 (but the negative value couldn't be an answer to "how long")
b = 6/
It takes six seconds for the object to be displaced by 12 cm.
To determine how long it takes for the object to travel a total distance of 12 cm, integrate the absolute value
of the velocity until some unknown time b, set that equal to 12, and solve for b.
b 2 – 4b + 8 = 12
b 2 – 4b – 4 = 0
but subtracting the square root would give a negative time which makes no
sense in this context.
It takes
seconds for the object to move a total of 12 cm.
34. For this exercise we are dealing with the function f (x) = 1/x2 = x –2 on the interval [1, 6].
(a) The area under this curve is given by the single integral
.
(b) For the line x = c to bisect that area we'd need
(c) This one is a little bit more tricky. See the figure.
y

x









We'd need for the integral of
from x = 1 to x = (the x-coördinate where y = d intersects y = 1/x2) to
be equal to 5/12. That point of intersection is . . .
d = 1/x2
x2 = 1/d
So we'd need to get 5/12 when we integrate
Setting this equal to 5/12 and solving for d:
(This is "quadratic in
", so I can use the quadratic formula.)
Adding the square root would produce a value for d that is above 1/x2, which would not work for us. So
our answer is