MATH 101 V2A
January 30th – Practice problems
Hints and solutions
Practice Problems:
1. Explain why
1
,
(x − c1 )(x − c2 ) · · · (x − cn )
where c1 , c2 , . . . c2 are distinct constants, may always be written in the form
A2
An
A1
+
+ ··· +
,
x − c1
x − c2
x − cn
where A1 , A2 , . . . , An are constants. (Are the constants A1 , A2 , . . . , An uniquely determined?)
Solution: If x 6= ci for i = 1, . . . , n, then we can multiply both sides of the equation by (x−c1 ) · · · (x−cn )
to get,
1 = A1 (x − c2 ) · · · (x − cn ) + A2 (x − c1 )(x − c3 ) · · · (x − cn ) + . . . + An (x − c1 ) · · · (x − cn−1 ),
Qn
or, using the compact product notation j=1 aj = a1 a2 · · · an ,
1=
n
X
Ai
i=1
n
Y
(x − cj ).
(1)
j=1
j6=i
If we were to expand the righthand side of the equation, we’d get a polynomial of degree n − 1 of the
form
1 = b0 + b1 x + b2 x2 + . . . bn−1 xn−1 ,
where each bi is a linear equation of the constants A1 , A2 , . . . , An and c1 , c2 , . . . cn . However, such an
equation must be such that bi = 0 for i = 1, . . . , n and b0 = 1. This gives us n linear equations in n
unknowns, which has a solution. In fact, since the ci are distinct, there is a unique solution and we
can solve for the Ai explicitly. Note that, by continuity of polynomials, equation (1) must hold for all
x (including x = c1 , c2 , . . . cn ). Plugging x = ci into (1) gives us
1 = Ai
n
Y
(ci − cj ),
j=1
j6=i
so
Ai =
1
n
Y
.
(ci − cj )
j=1
j6=i
(Note that c1 , c2 , . . . cn being distinct constants means that ci 6= cj for i 6= j, so (ci − cj ) 6= 0.)
Z
2. Evaluate
x
sec3 (t)dt. Can you think of a second way to evaluate this integral?
0
Solution: There are at least a few ways to calculate this. Here’s one: let u = sec(t) and dv = sec2 (t)dt.
Then du = sec(t) tan(t)dt and v = tan(t)dt, so using integration by parts gives us that
Z x
x Z x
sec(t) tan2 (t)dt
sec3 (t)dt = sec(t) tan(t) −
0
0
0
Z x
= sec(x) tan(x) −
sec(t) tan2 (t)dt
0
Now, since sin2 (t)+cos2 (t) = 1, dividing both sides of this equation by cos(t) gives us that tan2 (t)+1 =
sec2 (t), or tan2 (t) = sec2 (t) − 1. Substituting this into the above, we get
Z x
Z x
3
sec (t)dt = sec(x) tan(x) −
sec(t)(sec2 (t) − 1)dt
0
0
Z x
Z x
3
sec (t)dt +
sec(t)dt.
= sec(x) tan(x) −
0
Therefore
Z
2
0
x
sec3 (t)dt = sec(x) tan(x) +
Z
0
Now
Z
x
x
Z
sec(t)dt =
0
0
x
sec(t)dt.
0
1
cos(t)
·
dt =
cos(t) cos(t)
x
Z
0
cos(t)
dt =
cos2 (t)
Z
x
=
0
cos(t)
dt.
1 − sin2 (t)
If we let u = sin(t), then du = cos(t) and the integral becomes
Z
x
Z
sec(t)dt =
0
Since
0
sin(x)
1
du.
1 − u2
1
1
1
+
,
=
1 − u2
2(1 − u) 2(1 + u)
we get that
Z
x
sin(x)
Z
sec(t)dt =
0
0
sin(x)
1
1
1
1
+
du = − log |1 − u| + log |1 + u|
2(1 − u) 2(1 + u)
2
2
0
1
1
= − log |1 − sin(x)| + log |1 + sin(x)|
2
2
1 + sin(x) 1
= log 2
1 − sin(x) (1 + sin(x))2 1
= log 2
1 − sin2 (x) (1 + sin(x)) = log cos(x) = log | sec(x) + tan(x)|
Therefore
Z
0
x
sec3 (t)dt =
1
(sec(x) tan(x) + log | sec(x) + tan(x)|) .
2
2
Z
3. Evaluate
e
t log(t2 )dt. Can you think of a second way to evaluate this integral?
1
Hint: There are a couple ways to do this. One way is to use integration by parts with u = log(t2 )
2
and dv = tdt. Another way is to notice that log(t
) = 2 log(t), and then use integration by parts with
u = 2 log(t) and dv = tdt. Answer: 21 e2 + 1 .
Z x 3
t
4. Suppose we want to find
dt. This can be done using the method of parts repeatedly. An alternate
t
e
0
way is to set
Z x n
t
In =
dt,
t
0 e
and then find a formula for In in terms of In−1 .
(a) Prove that, for n ≥ 1, In = nIn−1 −
xn
ex .
Solution: Let u = tn and dv = e−t dt. Then du = ntn=1 dt and v = −e−t , and using integration by
parts we get
Z x
x Z x
n −t
n −t In =
t e dt = −t e +
ntn−1 e−t dt = −xe−x + nIn−1 .
0
0
0
(b) Find I0 , I1 , I2 and I3 .
Hint: Answer:
I0 = 1 − e−x
x1
= 1 − e−x − xe−x
ex
x2
I2 = 2 · I1 − x = 2 − 2e−x − 2xe−x − x2 e−x
e
x3
I3 = 3 · I2 − x = 6 − 6e−x − 6xe−x − 3x2 e−x − x3 e−x .
e
I1 = 1 · I0 −
3