Name:
Q UIZ 2
Please write your solutions in complete sentences, as simply as you can, and justify each step.
1. Find
d
dx
2. Find
0
2sin t dt.
x
1
dx.
1 x2
3. It looks as if we can integrate the function 2 sin x cos x in two different ways:
(a) Let u = s
H OMEWORK 8
6.3.22. We can approximate this by
1 + (e0 e1 )2 +
1 + (e1 e0 )2 +
1 + (e2 e1 )2 +
1 + (e3 e2 )2 .
6.3.32. Since f (x) = 1, the arclength is given by
1 + (1)2 dx = 3 2.
5
2
6.3.38. Since f (x) = (1/2)(1 x2 )1/2 (2x), the arclength is given by
H OMEWORK 6
5.7.10. In this situation, we can rewrite the integral as
b
c
b
f (x) dx.
f (x) dx +
f (x) dx =
c
a
a
Each of the integrands on the right side is monotonic, so we can use LEF T (n) or RIGHT (n)
sums to evaluate each integral. On the rst integr
H OMEWORK 10
7.2.34. We have
k2 2
2k
2
= lim
= lim = 1
2 + 2k + 2
k k
k 2k + 2
k 2
by LHopitals rule. Since the sequence converges, it is bounded above and below.
To see this, let = 1. Then the limit statement tells us that there exists an N > 0 such that
H OMEWORK 7
1
b
6.2.12. Fitting 0 2xx2 dx to the general formula a (radius)(height) dx, we nd that the
radius of the shell is x and the height is x2 . Hence, the region is bounded by y = x2 and the
x-axis, from x = 0 to x = 1, revolved around the y -axis.
Name:
Q UIZ 5
Please write your solutions in complete sentences, as simply as you can, and justify each step.
1. Let h and r be positive constants. Consider the region R bounded by y = h x, y = h,
r
and the y -axis. Use the shell method to nd the volume o
H OMEWORK 9
6.5.26. We have
1 dy
= sin x
y dx
so integrating both sides with respect to x, we obtain
1 dy
dx = sin x dx
y dx
ln |y | = cos x + C
|y | = e cos x eC
y = eC e cos x .
If we let A represent the constant, we obtain y (x) = Ae cos x .
6.5.34. Se
H OMEWORK 4
5.3.24. The integrand is a proper rational function so we can apply partial fractions. The
rational function has the form
25x2
B
C
A
+
+
=
.
2
(x + 3)(x 2)
x + 3 x 2 (x 2)2
Clearing denominators, we obtain
25x2 = A(x 2)2 + B (x + 3)(x 2) + C (
H OMEWORK 1
4.3.26. The integrand is the upper half circle of radius r centered at 0. Here, r is a constant
and x is the variable of integration since it appears in the differential dx. Hence,
r
r2 x2 dx
r
is
1
2
of the area of a circle of radius r. This
H OMEWORK 2
4.7.2 (c) We cannot anti-differentiate this function directly. However, we can use the second
Fundamental Theorem of Calculus to write
x
d
2
2
esin t dt = esin x ,
dx 0
x
d
2
2
esin t dt = esin x ,
dx 1
x
d
2
2
esin t dt = esin x ,
dx 2
x sin2
H OMEWORK 3
1
1
5.2.24. If we let u = ln x and dv = x dx then we have du = x dx and v = ln x. Integration
by parts gives
ln x
ln x
dx = (ln x)2
dx.
x
x
If we rearrange by combining the integrals, then we get
ln x
dx = (ln x)2 + C
x
2
so
ln x
(ln x)2
dx =
Name:
Q UIZ 1
Please write your solutions in complete sentences, as simply as you can, and justify each step.
1. Evaluate
2
4 |2x| dx
2
using geometry (not the Fundamental Theorem of Calculus).
2. Find
4x3
dx.
x4 5
3. Use the Fundamental Theorem of Calcul
Name:
D IAGNOSTIC Q UIZ
1. Factor x4 16 as much as possible.
2. Prove that x = 1 is the only real solution of x3 1 = 0.
3. Simplify
1
2x3 x2
+
4 (3x1/2 )3
x
x7
p/q
to show that f (x) is a function of the form Ax .
f (x) =
4. List all values of x in [0, ]
H OMEWORK 5
5.5.44. To solve
becomes
(x2 + 4)3/2 dx we let x = 2 tan so dx = 2 sec2 d. The integral
1
1
(sec )1 d =
4
4
(2 sec )3 (2 sec2 d) =
=
5.5.50. To solve
becomes
cos d =
1
sin + C
4
1
1
x
sin(tan1 (x/2) + C =
+ C.
4
4 x2 + 4
x2 4
2x
dx we let x =