Review Problem - Math 222

 Z 3
e
x2 ln(x)dx 




 e2







Z

 1


1
2n3 + 4n




Let v = 
,w = 
. Find v · w.

dx
lim
2
3


 −1 x + 6x + 8 
n→∞ (3n + 6)









 Z 1


(n − 2)! + n + (−1)n


1
lim
√
dx
n→∞
n!
x2 + 1
0

5n + 2
lim n
n→∞ 3 − 3(5n )

5n + 2
n→∞ 3n − 3(5n )
1. lim
Solution: We factor out a 5n from the numerator and denominator.
5n 1 + 52n
5n + 2
=
lim
n
n→∞ 5n 3n − 3
n→∞ 3n − 3(5n )
5
lim
1 + 52n
n→∞ ( 3 )n − 3
5
1
=
−3
= lim
This last step follows since limn→∞
2
5n
= 0, limn→∞ ( 53 )n = 0 as | 35 | < 1.
2n3 + 4n
n→∞ (3n + 6)3
2. lim
Solution: Expanding the denominator and using the function rule, we have:
2n3 + 4n
2n3 + 4n
=
lim
n→∞ (3n + 6)3
n→∞ 27n3 + . . . + 63
2x3 + 4x
= lim
x→∞ 27x3 + . . . + 63
lim
Here, . . . means terms I was too lazy to write out. This is a ratio of polynomials of degree 3, so
2
the limit is the ratio of the leading coefficients, which is 27
.
(n − 2)! + n + (−1)n
n→∞
n!
Solution: We look at the three limits being summed here.
3. lim
1
2
(n − 2)!
(n − 2)!
= lim
n→∞
n→∞ (n − 2)!(n − 1)(n)
n!
1
= lim
n→∞ (n − 1)n
=0
n
1
lim
= lim
n→∞ (n − 1)!
n→∞ n!
=0
n
(−1)
1
0 ≤ lim
≤ lim
=0
n→∞
n→∞
n!
n!
lim
The last limit used the sandwich theorem. Since all three of these is 0, the limit of their sum is
0.
Z e3
4.
x2 ln(x)dx
e2
Solution: We use integration by parts. Let F = ln(x), G0 = x2 . So F 0 = x1 , G =
Z
e3
e2
Z
1
5.
−1
x3
3 .
We get:
3 Z e3
x3 e
1 x3
x ln(x)dx = ln(x) −
dx
3 e2
e2 x 3
3 Z e3 2
x
x3 e
dx
= ln(x) −
3 e2
3
2
e
3
3
x3 e
x3 e
= ln(x) − 3 e2
9 e2
9
6
9
e
e6
2 e
3 e
−
−
= ln(e ) − ln(e )
3
3
9
9
6
9
6
2e
e
e
= e9 −
−
+
3
9
6
2
1
dx
x2 + 6x + 8
Solution: Note that the denominator factors as (x + 2)(x + 4). Therefore, this function has
A
B
partial fraction expansion x+2
+ x+4
. Finding A, B we have:
1
A
B
=
+
(x + 2)(x + 4)
x+2 x+3
=⇒ 1 = A(x + 4) + B(x + 2)
=⇒ 1 = x(A + B) + 4A + 2B
=⇒ A + B = 0, 4A + 2B = 1
−1
1
=⇒ A = , B =
2
2
1
1
1
=⇒
=
−
(x + 2)(x + 4)
2(x + 2) 2(x + 4)
3
Integrating, we get:
Z
1
−1
1
dx =
2
x + 6x + 8
Z
1
1
1
−
dx
2(x + 4)
−1 2(x + 2)
1
1
1
= ln |x + 2| − ln |x + 4|
2
2
−1
1
1
1
1
= ln |3| − ln |5| − ln |1| + ln |3|
2
2
2
2
Z
6.
0
1
√
1
x2
+1
dx
Solution: We use a trigonometric substitution. Because it is x2 +1, we let x = tan(θ), 0 ≤ θ < π2 .
Then dx = sec2 (θ)dθ. Also note that if x = tan(θ) = 0, then θ = 0. If x = tan(θ) = 1, then
θ = π4 , so we get:
Z
0
1
√
1
x2 + 1
Z
π/4
dx =
0
Z
sec2 (θ)dθ
p
tan2 (θ) + 1
π/4
sec(θ)dθ
=
0
π/4
= ln | sec(θ) + tan(θ)|
0
= ln | sec(π/4) + tan(π/4)| − ln | sec(0) + tan(0)|
2
= ln | √ + 1|
2