MATH 141 – Tutorial 1 (with solutions)
1. Antiderivatives
Recall from Calculus 1 that we are interested in finding the derivative of a given function, that
is a function that describes the evolution of the slope, where we have by definition for the
one-sided limit (using Leibnitz’s notation)
f ( x + ∆x) − f ( x)
d
f ( x) := lim
∆x→0
dx
∆x
Now, consider the reverse problem; given the derivative (slope) at any given location, can
we reverse engineer and get back a function that has the above property. If the function was
continuous to begin with, we can recover back uniquely up to vertical translation, since differentials knock off constant terms.
We thus call F the antiderivative of a (continuous) function f if ∀ x ∈ I, a given interval, we
have F 0 ( x) = f ( x).
1. Find the most general antiderivative of the given functions
(a) f ( x) = x32 − x54
(b) f (θ ) = eθ + sec(θ ) tan(θ )
(c) f ( x) = ( x2 + x + 1)/ x
Solution. The only subtlety here is that the function domain is restricted. We have
(a)

 −3 +
x
F ( x) =
 −3 +
x
5
3x3
5
3x3
+ C1
if x ∈ (−∞, 0)
+ C2
if x ∈ (0, ∞)
as the constant needs not be the same. We take the function −3/ x which has derivative 3/ x2
and correspondingly 53 x−3 which gives −5x−4 .
(b) The antiderivative of eθ is eθ and sec(θ ) for sec(θ ) tan(θ ). The interval of definition is any
1
In of the form In = nπ − π2 , nπ + π2 . We have for any such interval an arbitrary function
Cn .
(c) Dividing each term by x gives f ( x) = x + 1 + 1x which has antiderivative
F ( x) =

 1 x2 + x + log(| x|) + C
2
 1 x2
2
1
+ x + log(| x|) + C2
if x ∈ (−∞, 0)
if x ∈ (0, ∞)
2. Find the function f ( x) where
(a) f 00 ( x) = x2 + 3 cos( x), f (0) = 2, f 0 (0) = 3
√
(b) f 00 ( x) = x + x, f (1) = 1, f 0 (1) = 2
Solution.
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(a) We first have f 0 ( x) = x2 + 3 sin( x) + C1 , ∀ x ∈ R. Setting f 0 (0) = 3, we obtain C1 = 2. We
now anti-differentiate
f 0 ( x) =
x3
+ 3 sin( x) + 3
3
⇒ f ( x) =
x4
− 3 cos( x) + 3x + C2
12
and now the constant term C2 = 2.
(b) We get
3
5
2x 2
x3
4x 2
x2
+
+ C1 f ( x) =
+
+ C1 x + C2
f ( x) =
2
3
6
15
0
and it suffices to identify the constants, C1 =
5
6
4
and C2 = − 15
, as to get
5
4x 2
5
4
x3
f ( x) =
+
+ x−
6
15
6
15
3. Find the graph of the antiderivative of f and justify your answer (see Exercises 43-44, page
274).
Solution. You have to consider when the derivative function f is zero and the sign of the
function. The answers are respectively b and a.
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Summations and sequences
Recall that a sequence is an ordered collection of elements a1 , a2 , . . . , an . The following identities prove useful in time when you need to evaluate sequences.
Proposition 1 (Sum of integer sequences)
n
∑i=
i =0
n
∑ i2 =
i =0
n(n + 1)
2
n(n + 1)(2n − 1)
6
The first expression, S, can be evaluated by taking the sum
S=1+
+ · · · + (n − 1) + n
2
S = n + (n − 1) + · · · +
2
+1
altogether, if we add the terms that aligned, we get n × (n + 1) = 2S.
4. Prove that
!2 n
n
n(n + 1) 2
3
∑i = ∑i =
2
i =1
i =1
Solution. The proof is by induction, a very handy technique for proving statements of the
sort. Clearly, the statement hold for n = 1. Suppose the statement hold for n = k, n ≥ 1. Then
one obtains for k + 1
k(k + 1)
(k + 1) +
2
3
2
k2
= (k + 1) × (k + 1) +
4
2
(k + 2)(k + 1)
=
2
?
2
yet it suffices to notice that
(k + 2)2
k2 + 4k + 4
=
22
4
which is what we had inside parenthesis; since each side is equal to the other, and k is an
arbitrary natural, then by induction, the statement holds true for any n ∈ N
3
Telescoping series
A telescoping series is one in which the consecutive terms cancel each other, such that we are
left only with the first terms and the last terms.
Example 1
Evaluate ∑i100
=1 (log (i + 7 ) − log (i + 5 )).
Solution. Note that we do not need to enumerate and sum all 100 terms; there is an obvious
pattern to follow. The first few terms are
100
∑ (log(i + 7) − log(i + 5)) = [log(8) − log(6)] + [log(9) − log(7)] + [log(10) − log(8)] + · · ·
i =1
+ [log(106) − log(104)] + [log(107 − log(105)]
and some rearrangement leads to cancellation of terms, like ± log(8). We are left with
− log(6) − log(7) + log(106) + log(107) = log
using a property of logarithms, that log( ab) = log( a) + log(b).
11342
42
iπ
5. Evaluate ∑i10
=1 sin 2 .
Solution. We use the harmonic behavior of the sine function; sin( π2 ) = 1, sin( 3π
2 ) = − 1 and
5π
π
sin(nπ ) = 0 ∀ n ∈ N. Thus, we have the only non-zero terms are sin( 2 ), sin( 3π
2 ), sin ( 2 ),
9π
sin( 7π
2 ), sin ( 2 ) and by mean of cancellation, we get as final answer 1.
Finite summations of sequence terms enjoy some important properties, notably that of linearity, distributivity. They are summarized in the following:
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Proposition 2
n
n
n
∑ ai ± bi = ∑ ai ± ∑ bi
i =1
i =1
n
n
i =1
c∈R
∑ cai = c ∑ ai ,
i =1
n
i =1
c∈R
∑ c = nc,
i =1
Riemann sums
The idea of Riemann sum is to approximate the area under a given continuous curve f on
I = [ a, b] by taking rectangles of given width ∆x := (b − a)/n and taking limn→∞ . One can
show then that the area is recovered in the limit, by Squeeze theorem. An illustration of the
procedure for arbitrary tagged values xi∗ and variable width of ∆x in the interval1 is given
below for the function f ( x) = x3 /20 − − x2 /5 − x/5 + 3 on I = [−3, 3]:
x0
x∗0
x1 x∗1
x2 x∗2 x3 x∗3 x4
x0
x1
x2
x3
x4
Figure 1: Riemann sum.
Notice that the function is continuous. When f takes negative value, it is understood that we
subtract the area of the curve located under the abscissa. The Riemann integral will converge
if the area is finite.
1 As
long as ∆x → 0 as n → ∞, the integral is the same
5
10
10
5
5
2
4
6
8
10
10
5
5
2
4
6
8
10
10
5
5
2
4
6
8
2
4
6
8
2
4
6
8
2
4
6
8
Figure 2: Refinement of a partition for Riemann sums; n = 5, 10, 50
6. Use the midpoint rule with the given value of n to approximate the integral to four decimal
places
Z π
0
sec
x
3
dx,
Solution. We start by graphing the function
6
n=6
1.05
1.04
1.03
1.02
1.01
1
0
0.2
0.4
0.6
0.8
1
and calculate the midpoint values, which are x ∈ {(2n − 1)π /12}6n=1 , or
and the corresponding values
6
π
× ∑ sec
6
j=1
(2 j − 1)π
36
π π 5π 7π 3π 11π
12 , 4 , 12 , 12 , 4 , 12
≈ 3.9379
7. Express the limit as a definite integral on the given interval
n
lim
n→∞
∑ xi sin(xi )∆x,
[0, π ]
i =1
Solution. This becomes simply 0π x sin( x)dx; here ∆x → dx, the differential operator, and
the function is defined on the domain [0, π ].
R
2. Antiderivatives
We can view the integral as being the “inverse” of the differential, in a sense that will come
clear soon. For example, if you have the function f ( x) = x3 , than its antiderivative corresponds to F ( x) = 14 x4 , since the differential F 0 ( x) gives f ( x). Note however that F ( x) + c,
where c ∈ R is any constant, would have given back the same differential.
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3. Riemann sums, definite and indefinite integrals
We will define the definite integral as a limit of a partition, for now on a closed interval I
on which the function f ( x) is continuous. Let I = ( a, b) and ∆( x) = (b − a)/n with { xi∗ }in=1
points in each subinterval of the form [ xi−1 , xi ]. Then, the definite integral is defined to be
Z b
a
n
∑ f (xi∗ )∆x
n→∞
f ( x)dx = lim
i =1
A definite integral is a functional, that is it gives a numerical value once calculated. This
is different from the indefinite integral, which gives back a function (the antiderivative, or
primitive) which describes the area. One can then evaluate this function between the points
( a, b) to get the definite integral.
4. The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is the link between the inverse operations of differentiation and integration; it will provide us with our first known (in)definite integral and is a
fundamental result for this course.
Theorem 3
Given a function f continuous on a given interval I = [ a, b], the function F defined by the
relationship
F ( x) =
Z x
a
f (t)dt,
∀a≤x≤b
is continuous on I and differentiable in the open set ( a, b). Moreover, F 0 ( x) = f ( x), that is to
say F is an antiderivative of f ( x) and f ( x) is the derivative of g( x). This gives us
d
dx
Z x
a
f (t)dt
Note that we can reverse the integral bounds by considering the negative of the integral; if
the expression appearing in the upper bound is not x, but rather some h( x) differentiable, we
can use the chain rule and make a proper substitution.
The second part of the FTC assess that for any antiderivative F of f continuous on I, then
Z b
a
f ( x)dx =
Z b
a
F 0 ( X )dx = F (b) − F ( a)
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(constants of integration are irrelevant, as they cancel.)
8. Use the Fundamental Theorem of Calculus to find the derivative of the function
(a)
(b)
(c)
(d)
(e)
R √
g( x) = 0x 1 + 2tdt
R
F ( x) = x10 tan(θ )dθ
R cos( x)
y= 1
(t + sin(t))dt
R1
3
y = 1−3x 1+u u2 du
R
y = e0x sin3 (t)dt
Solution.
√
(a) 1 + 2x
R
R
(b) − tan( x) since xa f ( x)dx = ax f ( x)dx for a ∈ R.
(c) Change of variable and chain rule time. We take u = cos( x), du = − sin( x)dx. By the
chain rule,
du
dy du
dy
d
u
=
=
∈ t1 (t + sin(t))dt ·
dx
du dx
du
dx
= (u + sin(u)) · (− sin( x)) = − sin( x) · (cos( x) + sin(cos( x))
(d) Let s = 1 − 3x, ds = −3dx; then by the Fundamental theorem of Calculus and the chain
rule, one obtains
3(1 − 3x)3
dy
s3
=
= (−3) · −
dx
1 + s2
1 + (1 − 3x)2
(e) −e x sin3 (e x )
9. Use the Fundamental Theorem of Calculus to evaluate the integral, or explain why it does
not exist.
(a) π2π cos(θ )dθ
R
(b) ππ/4 sec2 (θ )dθ
R
(c) 19 2t dt
R
(d) −−ee2 3x dx
R
9
Solution.
(a) sin(2π ) − sin(π ) = 0
(b) Does not exist because we have an vertical asymptote at π /2 ∈ [π /4, π ].
(c) The antiderivative is 2t / log(2); we get log1(2) (29 − 28 ).
(d) The integral exists since zero is the only point at which the function is undefined; the
antiderivative is 3 log(| x|); this gives
3 log(e) − 3 log(e2 ) = 3 − 6 = −3.
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