Math 16B - Short Calculus
Final Review Solutions
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52·2
.05
1. 10, 000 1 +
52
2. t =
1
ln(2)
.08
3. r =
1
ln(1.5)
3
4. y = 80e 10,000 ln( 2 )·150000 ≈ .002g
1
13
50
5. y =
· e 18 ln( 10 )·100 ≈ 208
13
1
ln(
)
e 9 10
1
6.
1
1 + e−2
2 − e−2
2
x −6x
x2 −6x 1
7. (a) sec e
ln(x) · e
(2x − 6) ln(x) + e
x
1
(b) − sin(log6 (x8 − 4x7 + 3x − π)) ·
· (8x7 − 28x6 + 3)
ln(6) · (x8 − 4x7 + 3x − π)
1
1
6
2 ln(x)
6
2
5
ln(x + 3x ) + ln(x) 6
· (6x − 6x)
(c) (x + 3x )
·
x
x + 3x2
2
(d) 2x
3
−πx
x2 −6x
· ln(2) · (3x2 − π) + 4x−2/3
8. Compute the following integrals:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
−1
1
(x + 1)−2 + (x + 1)−3 + C
2
3
− log7 (x)
1 1
−
+C
x
ln(7) x
1
x
sin4 ( ) + C
2
2
1 3 1 2
x − x − 2x − 2 ln |x − 2| + C
3
2
−x cos(x) + 2 sin(x) + C
−1
cos(x2 ) + C
2
3
1 3 x3
x e − ex + C
3
3 1
3 1
ln(x) +
−
+C
2x+3 2x−3
(x2 + x)ex − (2x + 1)ex + 2ex + C
or
1
2
1
− (1 − x)4 − (1 − x)5 + (1 − x)6 + C
4
5
6
ln | sec(x) + tan(x)| + C
√
√
√
−2 x cos( x) + 2 sin( x) + C
1
x2 ex − xex + ex + C
(m)
(n)
(o)
(p)
(q)
(r)
−1
1
1
ln |x| + ln |x + 1| + ln |x − 2| + C
2
3
6
2
x ln(x + x ) − 2x + ln |x + 1| + C
1 3 4x
3
6
6 4x
x e − x2 e4x + xe4x −
e +C
4
16
64
256
−1
1
1
ln |x − 1| − ln |x + 2| + ln |x − 2| + C
3
6
2
−1
3
cos (x) + C
3
−e1/x + C
(s) −(x − 1)2 e−x − 2(x − 1)e−x − 2e−x + C
34
1
1
+C
−
(t) 6 ln |x + 2| + 29
x+2
2 (x + 2)2
or
− x2 e−x − e−x + C
9. Determine the particular solution that satisfies the given conditions.
4 5 4 3
1
x − x +x+
5
3
15
2 3x
(b) f (x) = e + 1
3
(a) f (x) =
10. Compute the following definite integrals:
p
p
(a) ln(16) − ln(2)
(b) ∞ - Diverge
√
−1
(c) √ − e1/ 3 + 1 + e
3
2
(d) 6 +
3
10
(e)
3
4
4
(f) − (−3)3/4 + (12)3/4
3
3
(g) 1 − ln(3)
(h) 0
1
(i)
24
(j) 3
1
1
(k) ln(5) + ln(9)
2
4
11.
128
15
12. π − 1
13.
14.
7
6
• 1
•
15.
π
2
•
•
•
•
3
4
1
4
1
4
2
16.
17. k = 1
18. b =
1
3
• Show f is a pdf
19.
• 10
• V (x) = 100, σ = 10
• m = −10 ln(1/2)
• Show f is a pdf
20.
•
1
2
• V (x) =
• m=
21.
4
16
22.
10
52
3
10
− 14 , σ =
q
1
20
1
2
3