Practice problems
Sequences
Exercise 1. You make an initial deposit of $100 into your bank account that earns an annual percentage rate of 8%, compounded annually. Find the future value of your account after 5 years.
Solution. After 5 years the future value is
F V(5) = 100·(1.08)5.
Exercise 2. A deposit of $100 is made each month into a bank account that pays an annual interest rate of 11%, compounded monthly. Find the future value of this account at the end of 10 years.
Solution. Since the interest is compounded monthly, at the end of the first month the future value is
F(1) = 100
1 + 0.11 12
.
The deposit is made each month, so at the end of the second month the future value is F(2) = 100
1 + 0.11 12
+ 100
1 + 0.11 12
2
.
Therefore, at the end of 10 years, that is at the end of 120 months, the future value is F(120) = 100
1 + 0.11 12
+ 100
1 + 0.11 12
2
+. . .+ 100
1 + 0.11 12
120
,
106
that is
Exercise 3. If the interest rate is 7% compounded annually, after how many years will the deposit be doubled?
Solution. After n years the future value of the deposit xis F V(n) =x·(1.07)n which is the double of the deposit if
F V(n) = 2x .
So, it takes 11 years to more than double the deposit.
Exercise 4. If the annual percentage rate is 9% compounded monthly, what is the effective interest?
Solution. When the interest is compounded monthly, the future value of depositxafter 1 year is The effective interest r is compounded annually, so
F V(1) =x· 1 + r
100
.
Therefore,
The effective interest is 9.38%.
Exercise 5. Find the following limits.
(a) lim
(a) Calculating the first several members of the sequence, we have
−2,0,−2,0,−2,0,−2,0, . . . This sequence has no limit, it is divergent.
(b) Like we calculated the function limit at infinity, basically we just write 0 instead ofc/np, for any positive p and non-zero constant c.
nlim→∞
(d) After factoring out the term with the highest exponent and simplifying, we have
nlim→∞
(f) The geometric sequence qn converges to 0 if and only if |q|<1. Therefore, factoring out the biggest term, we have
n→∞lim
Exercise 6. Write out the first four terms of each series to show how the series starts. Then find the sum of the series.
(a)
This is a geometric series ∞
With the substitution m =n−1, we have n = m+ 1 and using the fact
This series diverges, since the ratio 7 4 >1. both converge, we have
∞
Exercise 7. Calculate the sums of the following series.
(a)
(b) However, the series converges, the series
∞
diverges, because −8
5 ≤ −1. Therefore, the series
Exercise 8. Give the following repeating decimals as a fraction of two integers.
(a) 0.727272. . .= 0.˙7˙2 (b) 2.494494494. . .= 2.˙49˙4
Exercise 9. The government passes a $20 billion tax cut. If an average person spends 90% of any extra income earned and saves the other 10% how much spending would be created by this tax cut?
Solution. The 90% of the $20 billion, that is 20·0.9 = 18 billion dollars will be spent in the economy. This will then become new income for someone else, creating additional spending of 18·0.9 billion, and so on. Hence, the total amount of spending is
T S = 20·0.9 + 20·0.92+ 20·0.93+. . .=
Thus a $20 billion tax cut will create additional spending of $180 billion.
Exercise 10. A person must take 100 mg per day of a certain drug, but each day the body eliminates 20% of the amount of the drug present. Estimate the long-term level of the drug present in the patient.
Solution. On each subsequent day, the drug level is 80% of the level on the previous day plus the 100 mg dose of the actual day, that is
LtL= 100 + 100·0.8 + 100·0.82+ 100·0.83 +. . .
= ∞ n=0
100·0.8n= 100· 1
1−0.8 = 500.
The long-term level of the drug is 500 mg.
Supplementary exercises
Sequences
Exercise 1. We make an initial deposit of $100 into your bank account that earns an annual percentage rate of 9%, compounded annually. Find the future value of your account after 7 years.
Solution.
F V(7) = 100·(1.09)7
Exercise 2. A deposit of $200 is made each month into a bank account that pays an annual interest rate of 6%, compounded monthly. Find the future value of this account at the end of 12 years.
Exercise 3. If the interest rate is 5%, compounded annually, after how many years will the deposit be tripled?
Solution. n = ln 3 ln 1.05
Exercise 4. If the annual percentage rate is 8% compounded monthly, what is the effective interest?
Exercise 5. Calculate the limit of the following sequences.
(a) 3n−2n2
(h) n+ (−1)n
Exercise 6. Calculate the sums of the following series.
(a)
Exercise 7. Give the following repeating decimals as a fraction of two integers.
(a) 0.1212121212. . . (b) 2.125125125. . .
(c) 4.1111111111. . . (d) −3.5858585858. . .
(e) −2.817817817. . . (f) 1.0505050505. . . Solution.
(a) 12 99 (b) 2123
999
(c) 37 9 (d) −355
99
(e) −2815 999 (f) 104
99
Exercise 8. The government passes a $10 billion tax cut. If the average person spends 80%
of any extra income earned and saves the other 20% how much spending would be created by this tax cut?
Solution. 40 billion
Exercise 9. A person must take 50 mg per day of a certain drug, but each day the body eliminates 40% of the amount of the drug present. Estimate the long-term level of the drug present in the patient.
Solution. 125 mg
Exercise 10. A person requires a supplemental level of 50 mg of a certain drug. If the body eliminates 10% of the amount of the drug present on each day, find the daily dosage that would achieve the 50 mg supplemental level.
Solution. 5 mg
Quizzes
Version A
Exercise 1. Calculate the limit of the following sequences (a) 3n−2
1 +√
n3 (b) 4n−1−n3
5n+1−3n2 Exercise 2. Give the repeating decimals
1.˙72˙7 = 1.727727727. . . as a fraction of two integers.
Exercise 3. A person requires a supplemental level of 70 mg of a certain drug. If, each day, the body eliminates 20% of the amount of the drug present, find the daily dosage that would achieve the 70 mg supplemental level.
Version B
Exercise 1. Calculate the limit of the sequence 22n−1−n7 4n+1+ 1 Exercise 2. Calculate the sums of the series
∞ n=1
2·4n−1−6n+1 32n−1
Exercise 3. If the interest rate is 8% compounded annually, after how many years will the deposit be tripled?
Solution of Quizzes
Version A
Solution of Exercise 1.
(a) Solution of Exercise 2.
1.˙72˙7 = 1 + 727 Solution of Exercise 3.
LtL= 70 =d+d·0.8 +d·0.82+d·0.83+. . .
Version B
Solution of Exercise 1.
nlim→∞ Solution of Exercise 2. Since
∞ both converge, hence
∞ Solution of Exercise 3.
x·(1.08)n = 3x
Integral
Practice problems
Indefinite integrals
Exercise 1. Find the indefinite integral of the functions (a) f(x) = x4
(b) f(y) = 3y−2
(c) h(t) = 2t2−t+ 1 (d) C(q) = 1
q −3eq
(e) f(x) = √3
x+ 5
√x
Solution.
(a) The Power Rule
xα dx= xα+1 α+ 1 +C
for α= 4 gives
x4 dx= x5 5 +C.
(b) By the Coefficient Rule
cf(x)dx=c
f(x)dx
we have
3y−2 dy= 3·
y−2 dy = 3· y−1
−1 +C =−3 y +C.
(c) The Sum and Difference Rule
(f(x)±g(x))dx=
f(x)dx±
g(x)dx gives
(2t2−t+ 1) dt= 2
t2 dt−
t dt+
1dt
= 2· t3 3 − t2
2 +t+C.
119
(d) Since
x−1 dx = 1
x dx= ln|x|+C and
ex dx=ex+C we get
1 q −3eq
dq = q−1−3eq
dq= ln|q| −3eq+C.
(e) Rewriting the radicals to fractional exponents, we have
√3
x+ 5
√x
dx= x1/3+ 5x−1/2
dx = 3
4x4/3+ 10x1/2.
Exercise 2. Give the indefinite integrals of the following functions using the Linearity Rule.
(a) (4x−1)8 (b) √3
2x+ 1 (c) e3−x (d) 5
3x−1 Solution. The Linearity Rule says, if
f(x)dx=F(x) +C,
then
f(ax+b)dx= F(ax+b)
a +C.
(a) In this case the outer function is f(x) = x8, therefore
x8 dx = x9 9 +C
and
(4x−1)8 dx= (4x−1)9/9
4 +C = (4x−1)9 36 +C.
(b) By √3
2x+ 1 = (2x+ 1)1/3 we have that f(x) =x1/3, thus
x1/3 dx= x4/3
4/3 +C= 3
4x4/3+C.
Therefore,
(2x+ 1)1/3 dx=
3
4(2x+ 1)4/3
2 +C= 3
8(2x+ 1)4/3+C.
(c) Now f(x) = ex and
ex dx=ex+C
e3−x dx= e3−x
−1 +C =−e3−x+C.
(d) Byf(x) = 1
Exercise 3. Find the indefinite integral of the following product functions.
(a) (x3−1)8·3x2 (b) 10x·√3
5x2+ 1 (c) x·e3−x2 (d) 3x2 −2 x3−2x Solution. By the Chain Rule, we have
f(g(x))·g(x)dx=
f(u)du=F(u) +C=F(g(x)) +C.
(a) In this case the composite function is (x3−1)8, therefore, its inner function isx3−1 =u, the derivative of uis exactly the second factor 3x2. Applying the Chain Rule, we get
(b) First, let us rewrite the integral as manipulations, we can write
hence, by the Chain Rule, we have
−1
Exercise 4. Find the antiderivative of the function f(x) = 1
x+ 1, such thatF(0) = 2.
Solution. The indefinite integral of f is 1
x+ 1 dx= ln|x+ 1|+C =F(x),
this gives all the antiderivatives of f. The condition F(0) = 2 gives a unique value for C.
Substituting x= 0 intoF(x) = ln|x+ 1|+C gives
F(0) = ln 1 +C =C.
Since F(0) = 2, solving this equation forC gives C = 2. So F(x) = ln|x+ 1|+ 2 is the antiderivative satisfyingF(0) = 2.