LECTURE 26: REVIEW FOR MIDTERM 2
MINGFENG ZHAO
March 13, 2015
Example 1. Evaluate lim
n
X
6(k − 1)2
r
1+2
(k − 1)3
.
n3
n3
1
First, you should split off a ∆x = in the sum, then
n
r
r
n
n
X
1 X 6(k − 1)2
6(k − 1)2
(k − 1)3
(k − 1)3
=
.
1+2
1+2
3
3
2
n
n
n
n
n3
n→∞
k=1
k=1
k=1
k−1
(left Riemann sum) or nk (right Riemann sum), so we get
n
r
2 s
2
n
n
k−1
1 X 6(k − 1)2
1X
(k − 1)3
k−1
6
·
=
.
1
+
2
1
+
2
·
n
n3
n3
n
n
n
Then write down the resulting sum as a function of
k=1
k=1
k−1
k
Now substitute ∆x by dx, and substitute
or
by x, we get our integrand f (x). Then when n → ∞, the sum
n
n
Z 1
will converge to the definite integral
f (x) dx. In our case, we have
0
lim
n→∞
n
X
6(k − 1)2
k=1
n3
r
1+2
(k − 1)3
=
n3
Z
1
6x2
p
1 + 2x3 dx.
0
Z
sin3 (x) dx.
Z
Z
Z
Split off sin(x) first, then
sin3 (x) dx =
sin2 (x) · sin(x) dx = − sin2 (x) d cos(x). Notice that the power
Example 2. Evaluate
of sin(x) is even, then one can write the even power of sin(x) to a polynomial of cos2 (x). In our case, we have
Z
Z
− sin2 (x) d cos(x) = − [1 − cos2 (x)] d cos(x), then use the substitution u = cos(x).
We are luck that it works when we try splitting off sin(x). What if it could not work when trying splitting off sin(x),
Z
Z
sin3 (x)
sin3 (x)
sin3 (x)
let try to split off cos(x), that is,
· cos(x) dx =
d sin(x). From
, we see that the power of
cos(x)
cos(x)
cos(x)
cos(x) is odd, which can not be a polynomial of sin2 (x). So this does not work.
Z
Example 3. Evaluate
sec3 (x) dx.
1
2
MINGFENG ZHAO
Z
Let I =
sec3 (x) dx, then
Z
I
=
sec(x) · sec2 (x) dx
Z
sec(x) d tan(x) Since d tan(x) = sec2 (x) dx
Z
= sec(x) tan(x) − tan(x) d sec(x) Use the integration by parts
Z
= sec(x) tan(x) − tan2 (x) sec(x) dx Since d sec(x) = sec(x) tan(x) dx
Z
= sec(x) tan(x) − [sec2 (x) − 1] sec(x) dx Since tan2 (x) = 1 = sec2 (x)
Z
Z
3
= sec(x) tan(x) − sec (x) dx + sec(x) dx
=
=
sec(x) tan(x) − I + ln | sec(x) + tan(x)| + C.
So we get
2I = sec(x) tan(x) + ln | sec(x) + tan(x)| + C.
Hence
Z
I=
sec3 (x) dx =
sec(x) tan(x) ln | sec(x) + tan(x)|
+
+ C.
2
2
x4
dx.
(x − 1)2 (x − 2)
Since the numerator has higher degree than the denominator, then we should use the long division first so that we will
Z
Example 4. Evaluate
get a polynomial plus some rational function. After the long division, we should do the partial fraction decomposition
for the rational part which will be:
A
B
C
+
+
.
2
x − 1 (x − 1)
x−2
Z 1
1
dx converge or diverge?
Example 5. Does the integral
x
Z 1 −1
1
1
You should realize that
dx is an improper integral, since −1 < 0 < 1 and = ∞. Then
x
0
−1
Z 1
Z a
Z 1
1
1
1
dx = lim
dx + lim
dx.
a→0
b→0
x
x
x
−1
b
−1
Z a
1
Look at
dx, then
−1 x
Z a
1
a
dx = ln |x||−1 = ln |a| − ln |1| = ln |a|.
x
−1
LECTURE 26: REVIEW FOR MIDTERM 2
Z
a
So when a → 0, we have ln |a| → −∞, that is, lim
a→0
−1
1
dx = −∞, which implies that
x
3
Z
1
−1
1
dx diverges.
x
t
.
y
In this course, we only can solve the separable equations of the form y 0 = f (x) · g(y). So our first step is to write the
Example 6. Solve e−t y 0 =
equation to a separable equation. In our case, we should move e−t to the left hand side, we get
y0 =
tet
1
= tet · .
y
y
Then solve this separable equation.
Z
1
Example 7. solve f (x) = 2 +
f (t) dt.
x
For an integral equation, one should take the derivative on the both side to get a separable differential equation. In
our case, we get f 0 (x) = −f (x). Let y = f (x), then y 0 = −y.
Then solve this separable differential equation with the initial value which can be seen that from the point such that
the lower limit is the same as the upper limit. In our case, we take x = 1, then y(1) = f (1) = 2.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca