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Lesson 1: Approximating Areas with Riemann sums
Topic 6.1: Exploring Accumulations of Change
Let's take a look at the relationship between the graph of a rate ofchange 昀甀nction and the area of the
region between that derivative graph and the x-axis. Consider the tropical cyclone Imelda that brought
more than 40 inches of rainfall over a 72-hour period in Southeast Texas during September 2019.
EX #1: In order to avoid 昀氀ood water entering your home, you rent a pump that can remove water at
a rate of 500 gallons per hour 昀漀r the 昀椀rst 8 hours of pumping. As the storm grew stronge爀Ⰰ
you got a second pump; and, 昀漀r the next 16 hours you could remove 900 gallons per hour
using two pumps.
1A. Draw a graph that models the removal of storm water from your property. Label the axes
correctly.
700
600
嘀⤀
C �
500
'
;
:
'
·!.
'
t··--"�q O .J ⠀㐀. ..� .. ·:· ·:•":"•�·
-+···\
.. ,...... ...,.... , ..... ,.
., . -:- ...... ;...._ ...,. 昀ⴀ 昀℀4450..1u·:11+.
'氀∀:• L. :. ...... u....., ,....:+Iⴀ爀t+
f
� ,: 㔀㨀dt : : u 㨀甀/ ℀㨀: �!�:•: I• I ;•:i:i
��
1?0
'
:
.
:
:
1
1
' . .
::
�
�- �
:
.
i
:
•
'
•
:
:
I I
.
�. -:c·: ;: .. [::�::.�:.� �:·;·_ 16 :: ..�::; �;·:;_;� : :·��:l �
�- , .: .2 .. ::::: :: ·;:;
⸀ꈀ,me, i昀尀 hou爀猀
.
..
1B. When we say "area under a curve"this represents the area between the graph and the
horizontal axis. Find the area under the graph 昀爀om 0 to 8 hours. What does this value
x
opi
o
eld
�
甀퀀
� \,
�
�
)(✀最)
5㬀✀
✀漀
c瘀戀
:•
4000 �I I 5
'
o⸀倀
e ⸀倀rr㔀琀
s琀漀rm w漀⸀J攀爀 was removed dur1A9 �
8 嘀尀ours of pumpin开였 durr✀퀀9 ⸀稀m攀ⴀld䄀瀀.
on
1C. Find the total area under the graph 昀漀r the entire 24-hour period. Explain the meaning in
context to the scenario.
㐀ⴀ00Oⴀ琀 1.41 100 = l81 䌀茀
挀氀ur,nⴀ㤀
攀爀
㰀䄀t
W
rm
s椀瘀
inq
爀渀p
爀Ⰰu
o昀✀
u爀猀
ho
㨀销4
I昀尀
㨀昀er
I me l挀氀漀⸀, a 琀漀ta I of 18/fdO 9　⸀1 I 0㄀㄀s of wa
was r攀ⴀmoved 昀ꄀom 椀栀攀⸀ prop攀爀� b℀䨀 pump.
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Finding Area Using Riemann Sums
Topic 6.2: Approximating Areas With Riemann Sums
What approach can we take if the rate of change 昀甀nction is not constant?
The two big ideas in calculus are the tangent line problem and the area problem. In the ta鸀ent 氀椀ne
problem. you saw how the limit process could be applied to find the slope of the tangent line (i.e. the
derivative) at any point on the curve. A second classic problem in calculus is in 昀椀nding the area under
a curve that is bounded by the graph, the x-axis, and any two x-values.
A basic overview of "areas as limits." In the "limit of rectangles" approach, we take the area under a
curve y = f (x) above the interval [a, b] by approximating a collection of inscribed or circumscribed
rectangles in such a way that, the more rectangles used, the better the approximation. Finall礀Ⰰ the
number of rectangles is increased without limit and, bingo, we get the area! Now known as integration.
Speci昀椀call礀Ⰰ we are interested in 昀椀nding the area A of a region bounded by the x-axis, the graph of a
nonnegative 昀甀nction y = f(x) defined on some interval [a, b].
Note: The requirement that f be nonnegative on [a, b] means that no portion of its graph on the
interval is below the x-axis. By using rectangles, we can see that there are three di昀昀erent ways of
approximating the area for A. This method is commonly known as Riemann Sums.
EX #2: Consider a 30-gallon aquarium is being filled with water at a rate modeled by
r(t) = 6 sin(�t), where tis time in minutes, and r(t) is gallons per minute. Find the volume
of water in the tank a昀琀er 5 minutes by using five rectangles that are 1-minute wide and r(t)
in height.
䨀㨀 r(t)
1 �.�q,
琀椀 4,J䰀⼀3
3 5, 5⸀뀀⸀㌀
4 �lo挀漀
5 5. 5✀⬀3
A攀ꀀ漀⸀ To甀鐀I
�.�9䜀Ⰰ
��--⸀尀-·,-·-·- , -�-.� ⸀尀.⸀⸀똁ᜀ.⸀⸀..⸀⸀ ,
:
,�.
: : :
: :
--- 6
:
1!
··•·ⴀ爀·-··
5
⸀⸀ ⸀⸀⸀尀.
_
rf(I :
'
•
i
;
18. 0 8�
䨀⸀3,⠀儀�5
---· 3
t挀琀nk.
1'
:
-·- -----◄-----
I
�---
A昀⬀爀谀 5 minu琀攀s 琀栀e 漀⸀�u伀䨀wm
✀렀as 㨀销3. �� s 3�1tons 1·n 琀栀e
:
:
---- --- --- ··
㐀✀.539
IJ. 08⤀最
∀✀
:
:
' ···•· :··-···t
: . : l
---· 4
. -____ ) ____ - �- -- --·�--⸀⸀ -_j
.
-.⸀⸀..,..
-·-· l
r·--
'
'
;'
7
8
'
1
2
�
3
__5, _,6
4
,
By using more rectangles with smaller widths, we could achieve a better approximation. In 昀愀ct, if we
considered taking this to a limit of accumulating an infinite number of rectangles we would arrive at
the exact volume of water in the aquarium a昀琀er 5 minutes. This is called the de昀椀nite integral and is
represented by the notation:
r(t)dt
ls
0
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Three T✀昀pes of Riemann Sums
Le昀琀-Endpoint (Lower sum)
Inscribed Rectangles
Right-Endpoint (Upper Sum)
Circumscribed Rectangles
Mid-Point (Average)
- ---+--+--+--j- - ;-- - !�-
y
l
- -----rI -----rl ----rl ---1I---- -----�--
y
-- ----+---+-'
'
_____ •1 ____ 'l
I
a
X
b
''
I
''
----f---tI r t
1
1
-- ----.---'
_____⸀꬀__ 开氀__
I
b
a
! I ---t--r••--1--:-i ! i !
- ---1-- --- :----f-l - , , --· ! ·-- ! -
----t---- -
:
:
X
a
t :I
----⸀⸀'I -----⸀ⴀ-'
---r--r
X
We can de昀椀ne the area A under the graph on the interval to be as follows.
DEFINM INTEGR䄀䰀 DEFINnION:
Letf be continuous on [a, b] and f(x) 氀琀 0 昀漀r all x in the interval. Let [a, b] be partitioned
into n subintervals of equal length 䨀伀x = b-a
. Then, the definite integral off over [a, b] is
n
given by:
n
b
A = lim � f (ck)䨀伀x = f f (x)dx
n➔ 漀漀 L
a
k=l
General Solution Method for Inscribed or Circumscribed Rectangles
(Lower or Upper Sums):
1. Draw a rough graph of the function over the interval.
2. Use 昀漀rmula 䰀氀x = 頀꼀 to determine each subinterval length.
n
3. Compute x-coordinates of rectangles at either left-end or right-end.
4. Compute the areas of each rectangle (inscribed or circumscribed).
5. Find the summation of the approximated areas of the rectangles.
EX #3: Find the area A under the graph of f(x) = �x
+ 1 on the interval (0, SJ using geometry.
2
l�id A= ±h[b, + bz]
n 㴀ⴀ 5
b, : ⴀ昀(漀⤀㴀㴀 I
A㨀㨀 t(s) [ l + ⴀ℀]
i
騁㴀
A = 椀椀
爀鰀
䰀昀
b 2 : f(5) =
f
...... 5
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Error in Riemann Sums
EX #4: For the function shown at right, draw rectangles 昀漀rming
the Left-Hand Sum shaded in blue.
A. Calculate the left-hand sum. Let 昀椀x = 2.
LH 3 = J[2+7ⴀ琀5]
ove(-�琀椀夀戀a氀鈀s
LH 3 = 40
··· 8
y
'
'
• .6
.•
· 5
•.
. 4
...
.⸀爀 ··--•------ •..
. .. ; ...••• • •···
8. Next, draw the rectangles that represent the Right-Hand
Sum. Shade them in red and calculate the approximate
area under the curve given by RHS.
. ·
•
-�- · ·1···
I1
-⸀嬀-
.. .......
··•·
:
X
I
....
. ·2
..
···'·· . ·, .. ,···• -.-
•. I
7
6
5
4
3
2
1
'刀⌀3 㴀ⴀ J [7 + 5+ 1]
RH3 = ��
C. Which sum over-estimates, which under-estimates?
Left-H漀⸀nd ov攀昀猀�iMa琀攀⸀猀
R19ht- Hand un€Ö�阀洀愀tes
D. Repeat steps A - C 昀漀r the function shown at right. Let 昀椀x = 0.5
䰀⸀ⴀ저 3 = f[ I+ 4+ 琀漀]
·8
⸀氀'
:
:
:
..
.. : ··•·-: ......... ',·· ....
.
' ......
• 7 .. --�-----!·· . . . .
,
•
LH3 = 5.5
· ·6
-- 5
t[ 4� ⠀漀+ 1]
r�픁က ,..
·····. - -
'
'
..• .
I
· ., · ·
·,•• •
,�1
· .,. - j
3
� •
·•!
'
: __ _.,,
:
•
0.5
:
,
!
! ...
·i·- •·:.- ., ·--��-•·••.
;
t·· ..
' -1
··4
···2
8.5
'
: ....•.. .: ...
X
1.0
:
1.5
EX #5: Give a hypothesis for when your will get an under-estimate an or over-estimate, depending
on which type of sum you use and the function's behavior.
Concave
Up
Concave
Down
Decreasing
Increasing
f(x)
Left-Hand
under
Le昀琀-Hand
Right-Hand
DV攀昀
Right-Hand
ov攀爀
甀⸀昀✀der
Le昀琀-Hand
U昀尀der
Left-Hand
0尀⼀椀昀
Right-Hand
over
Right-Hand
under
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EX #6: The table below represents the veloci琀礀, v(t), in 昀攀et per second of a car traveling on a
straight road over a SO-second time interval. We want to estimate the distance the car
traveled over this interval using RHS and LHS along with two different approximation
techniques, the midpoint and trapezoid rules. Use the table along with the graph for your
computations.
t (sec)
0
5
10
15
v(t) (昀琀/sec)
0
14
.2
28
\
o-䤀䤀·
,
- 70
52
1
-
25
20
30
�.
ⰀⰀ
80 j
A. Estimate the total distance traveled by the car by
using the Le昀琀-Hand Sum method, Ls, with 昀椀ve
partitions of equal length. Show your setup and
draw the rectangles on graph #1, in blue.
35
40
45
50
86
77'
)
70
:?0
LH 5 = t0Ⰰ琀[O+J0+5�+8漀⬀11] fⰀ숀
LHs - IO ( �9) +t
�, ☀⸀90ft
B. Now use a Right-Hand Sum with 5 partitions, R 5.
Show your setup and draw your rectangles on
s㨀砀+ 8o+ 11+ 14䨀
RH:�· ;�o+
10 C 303) 昀琀
10.0
R ⴀ퀀s =
+f
�
oJ
+,
8"
➔
0
+1
+�
[14
MPs= 10�
MP㄀㔀 � 10 (J䌀漀✀㠀) fi
� �I䌀漀8o ⴀ䘀t
D. Use the Trapezoidal Sum with 5 partitions Show
your setup and show trapezoids on graph #2.
y
. so ...........
50.0
:
60
··40
. ··20
�(77) f 14]
� 5(53�] 㴀㸀 J,㰀漀wft
40.0
t··· ·· •··r·········愀쐀· 娀뤀�Ⰰ贀开开
T5 = 䨀ⴀ(io) [0+J(Jo)+漀㄀(5J)+ �(to)ⴀ昀.
㬀栀
30.0
Graph #1
= 㨀最㌀䐀㌀䨀
昀琀ⴀ㴀
2
C. Next, use the Midpoint Sum with 5 partitions, Ms.
The first rectangle is shown in graph #2. Show
your setup.
7
20.0
20.0
30.0
40.0
50.0
Graph #2
䰀氀夀⸀ = he1·g 栀琀
E. Find the average of Ls and Rs, how does this compare to the trapezoidal sum?
Ls;Rs 㴀ⴀⴀ昀(.uqo+3　㌀o] 㴀㴀 dl☀㸀Ⰰ吀O#_ sam攀⸀ ⸀鴀s 吀猀.
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Transforming the Area Accumulation Function by
Altering Its Limits of Integration
X
f
Let's examine what would happen geometrically to F(x) = J(t) dt if we allow k to take on different
values.
o
f
X
EX #3: Given F(x)= f(t) dt
·
k
A. Complete the table of values 昀漀r the function F.
-
F䌀砀)
o.5
X
5
&
2
�-0
1
1�
\l.25
3
4.5
g
䌀尀
X
0
1
4
F(x)
D
8
\
10
··
11
:-:· :
f(琀⤀
·· ···1····1···:·
:
-----10
..--:·9
. ' '
···:·8
. ' '
.. :--7 - -:'' .⸀⸀. -:. - - .:
----� 6 - �- i·•-- +-s - -,-.'' . - .,.'' . - ',__ - _,_'
····:-4
- '. -; 3
''
:-2
... � · 1
•
o
'
'
r
-- - ⸀⸀
•
····-3
. ··ⴀ㐀
'
I
t
- - _ .. _ t
:-
.. -- '
.
-- - ' '
·• ·'r
''
''
: ···:··· :
.
1
.
l
O
O
I
r
---: -- :··- :-- -�
. .
'
⸀尀...⸀⸀.
.. .,
'
''
0
I
-- ⸀⸀''
.
o
'
: -· · t
:
--�---9/1,0-1,1--
:- -:- ---:-. : •- • I
I
I
'
''
- - .⸀娀.
'
-. .'⸀尀 -
r·• • • . .
I
I
•
I
-··:····�----;•►•-�--,
•
I
. - 昀케-:�-
•
-:
'
I
. -'
I
-- • J. o
.
... -2
'
'
'
'
-' - . - -⸀⸀' -- - .⸀尀' - -.
-
•
. .. .
-----1
'
••r·•••-·•••r••··
1
.
⠀瀀
:
1
••• •••••••••-,⸀ⴀ •••
I
t
- - .⸀⸀. - • •
'
I
'
'
'•·+··:··· ;··
I
----
'
I
,
I
----⸀ⴀ----�---·r···I
I
•
t
t
I
I
I
I
B. What is the domain of F? __Q_�_✀堀_�_f _O_
C. Use the concept of "area accumulation" to complete the graph of 䘀⸀ Plot the points in your
table in blue.
D. Where is the zero of 䘀㼀
F(氀伀)- F(o)
㄀⸀㨀㨀 0
E. What is the "total change" in F? -娀脀⠀漀
伀吀---
F. How does the "total change" in F compare to the "net area" bounded by the graph of the
function/and thex-axis?
t⤀椀-⠀漀 㴀㴀 F(10)- F(O)
猀㨀愀⸀me
ar攀⸀a 琀ꈀ d ,s鐀一 \愀⸀āÝ�
"
1 䰀鴀
H. Where does it occur ?
G. What is the maximum value of F? --蜀簀-I. What is the range of F?
O䔀嬀 W � 1 l
EX #4: Now consider the function
A. Where is the zero of G ?
唀됀 =
X
G(x)= J f(t) dt
⸀㌀
7, 5
[퀀嬀. 5) 1 5]
G. What is the range of G?
1
= ⸀鼀
0
G(o): � f(t)搀琀� -
䨀㨀 昀焀琀⤀dt
3
C. What is the domain of G? 嬀瀀) 10]
E. What is the maximum value of G ?
㄀氀
- 4' 贀尀5 _
B. What is G(0) ? __謀븀㴀娀 ⴀ吀
D. What is the "total change" in G ?
-ⴀ㴀氀攀ⴀ礀_
F. Where will it occur? --�-=��ⴀ樀-H. Sketch the graph of G in green.
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Calculus-Based Justification from the graph of g' = F
In differential calculus we used the derivative f' to discuss properties of a function f. In integral
calculus we need to reverse our thinking. Our new goal is to use derivative functions to talk about
antiderivatives. In the example below, we say that g is an antiderivative off. Let's summarize the
relationships between the properties of a function and its antiderivative.
EX #6: Given the graph off, Let g(x) = f xf(t) dt
漀⸀bo瘀攀 x -ax,·s ⴀ昀 pos(i� 0
攀츀ow x- 漀⸀x,s f n䔀㠀需v€
i
A. f is positive on the interval [O, 8), so g is
⠀⤀ 9 on this interval.
B. f changes sign at ✀堀㨀琀 8
so g must have
\ ⠀⤀℀吀,(鈀帀S i
a relative
耀ꔀ 愀⸀XIꈀ蔀 U夀萀
p9sinv攀⸀
+
+
since f goes
to
·�·
+
2
+
攀愀oJj Ve
C. The graph of g is decreasing where f is
n 攀攀鴀尀tiVe
from
y
+
+
+
+
Ⰰ尀
+
of 3
礀尀
. 昀漀r values of
� ⼀✀
g
__
up
____
嬀㤀, 㐀䨀
. so 9 is 开餀䌀䐀 n ⸀쐀 ve
_ _开Ⰰ. for values x > 䰀尀
E. The graph of g is concave down where r is -娀脀d�e_挀爀_e
Ƭő⸀_,_·琀尀 _9
F. The graph oft has a maximum atx =_4_� sog has an inf leenon roiot
D. The graph oft is increasing on
︁⼀ĂõŘď阀言
ⴀ昀
Reasoning and Justification Summary
The antiderivative
When the derivative function. f is ...
Positive
Negative
+
-
x
g(x) = 昀漀 f(t) dt
is ...
Increasing
Decreasing
Increasing
Concave Up
Decreasing
Concave Down
Changes sign
(crosses the 尀娀axis)
Extremum Point
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Lesson 4= Applying Properties of Definite Integrals
Topic 6.6: Applying Properties of De昀椀nite Integrals
In this section, we move beyond 昀椀nite sums (rectangles) and explore what happens in the limit, as the
terms become in昀椀nitely small and the number of rectangles approaches infinity. We will look at some
geometric functions that may contain negative values (below the x-axis), as well as positive values
(above the x-axis), and zero.
Dealing With Negative Area:
At times we will integrate functions that go below the x-axis (negatives).
Here are a few pointers 昀漀r when that happens:
•
•
Area below the x-axis counts as negative area.
The total area between a and b for some curve is really a net area, where the total area below
the x-axis (and above the curve) is subtracted 昀爀om the total area above the x-axis (and below
the curve).
Definite Integral Notation:
Leibniz introduced the notation 昀漀r the definite integral. The Greek letters of summation become
Roman letters in the limit.
The notation
䨀㨀 f(x)dx is read as:
"me def,㄀㄀,'fe 1nte3漀밀 J 昀爀om a 琀漀b. oF f of�
wi琀栀 respe挀琀 琀漀 x.
11
㬀最
The component parts also have names:
琀销e i 椀鴀攀最rand (㨀萀�ivo⸀ꈀive)
䨀쬀p爀鼀
氀椀夀鬀it
1
뤀蠀b
�
琀销e \Ate8r氀贀 ___, L f(x)dx -•-- i��i愀Ⰰble oⴀ䘀 1n攀�r漀⸀琀椀·dr
蜀찀�-퀀蜀,--�
\ow攀挀 I,mit
琀栀e def,n;+e '"+�r⸀輀l
The de昀椀nite integral can be used to determine:
吀栀e ㄀㄀et 愀爀e⸀鴀 ( displacement) bound攀搀 b� 琀销攀⸀ x-砀鴀is 1
椀栀e 昀甀n挀栀o㄀✀ f 䌀砀) and v攀爀t(였砀 lt椀㄀攀猀 X= a 琀漀 x: b.
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EX #1: Use the graph below to evaluate each of the 昀漀llowing. Note that the graph off (x) consists
entirely of line segments and a semi-circle on the interval -8 � x � 13.
> ·+·
匀
䘀漀rm瘀氀⸀퐀:
......i... ✀⸀ · ⴀ爀· ·! ···r·•· ·i-· r·· �·--. · ·:· ···1·····1·····:···+··-r ····:---···i·· i·y-�f笀Ⰰ砀⤀ r····
:•:
:
r
T ; :· :.:::·.i· ·!· .•...•••.• 昀✀
.℀蔀
···r- 1·4: ---1 r •·-·: --1 ··· �-- 11Jt 1-- g-- - ··· ·(··r···1 ····1· ---1·····1 -··: ---- ⴀ봀 bh
. . :•�·. ✀⸀: er
t: :rJ I :.;:: ;:.:
. : �. . � : ' -� +
+
·
2
• •++l]� ·• :• .I ··J•··1 • ·iI.[:• Hf·L t.:.• E.•
.
'
.
2
----i----
.
.
2
:·
2
:
:
:
6
:
8. ✀⸀
10
'
㄀㈀
X
ⴀ넀h[b,+bJ
--�---··(·· ----1-····:·····最娀·--(··:··-··(··1····r ··:········· ·:-··· �-- ·; ----1-·-··(· (·· ⴀ栀···(·
---- :-----�---·-r· T···:----·r · · :··· ·r- · - r- · · r--·; •---i·---: - --r--··1----: · ·r· :- - -· - '
:
:
:
I
:
:
:
-·--4
:
:
:
:
I
:
:
:
�
:
!
-i--
A.
E.
G.
f:f ( X) 搀砀 =
t(4) [ 3 +1j
f:f(x)dx= -JO
-
f�8 f(x)dx = 4 - 㤀⸀J + �0-㨀䨀
B.
�.5
D.
J:f(x)搀砀= Q
F.
f4 f(x)搀砀= -
/"(6) O氀∀ 5 � 㬀琀� 1 ⸀縀 /(x) 15
⸀猀 昀尀ŎƊ I 5 0
s:
-(8-㄀㈀)
ⴀ昀
攀�J椀ꀀ
= -"
I
嘀
/'(-6) 꼀밀p SO ⴀ琀 /(- 琀鈀)
J.
Average rate of change inf over [O, SJ.
is unde昀椀ned
f(s)-f(o) = ✀琀-o 㴀ⴀ Ѐ
5-D
㈀开
5-o
f/(�)= 0
1
K. 嬀ꄀrFalse
10
H.
- ro.s- �
H.
3
f� f(x)dx= J0-ⴀ봀(3J[昀漀ⴀ琀 ㌀崀
L.
Over which values of xis f (x) not
di昀昀erentiable?
f:f (x)dx+ f:f ( x)搀砀 = f:f (x)搀砀
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Properties of Definite Integrals:
The definition of the definite integral off on the interval [a, b] specifies that a < b. Now, howeve爀Ⰰ it is
convenient to extend the de昀椀nition to cover cases in which a = b or a > b.
1. If / is defined at x = a, then
f:f (x) 搀砀 = 0
2. If / is integrable on [a, b], then
f:f (x)搀砀 = - 䨀㨀 f(x)搀砀
Additive Interval Proper琀礀:
3. If/ is integrable on the three closed intervals determined by a, b, and c, then
f :f(x)搀砀 = 䨀㨀 f(x)搀砀 + 䨀㨀 f(x)dx
Properties of De昀椀nite Integrals:
If/ andg are integrable on [a, b] and k is a constant, then
f:k 搀砀 = k (b - a)
s.
EX #2: Use the properties of integrals to evaluate each of the following, given that:
A.
c.
E.
f
t
J: g( X) 搀砀 = - g ⠀砀) 搀⸀x
= -(-3) � 3
f)㄀�= 5(b-a)
s:
B.
5
h(x)搀砀
=b
a
2
s:2•f(x)dx = J
s: +
(x) 㬀�
�J(�)�攀褀
D.
J: [ 5 + g( x)]搀砀 = 匀㬀 5dx + 缀尀 9䌀砀搁ᜀ
b
ⴀ㴀 S(b-愀⤀- 3
[f(x)-h(x)]搀砀=
s: +C砀⤀dx-J:hw J:
=8-(¾)- ⴀ─
1 �.
h(x)搀砀 +
J: h(x)搀砀 =
b
5嘀唀 h砀儀dx = ㈀ⴀ
需阀
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Lesson 5: Antiderivatives and Indefinite Integrals
Basic Rules and Notation
Topic 6.8: Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation
VOCABULARY AND DEFINITIONS:
1. A function F(x) is an antideriva琀椀ve of a function f (x) if F' (x) = f(x) 昀漀r all x in the domain
off. The process of finding an antiderivative is antidi昀昀erentiation.
2. The 昀愀mily of all antiderivatives of a function f (x) is the inde昀椀nite integral off with
respect to x and is denoted by ∫ 𝑓( 𝑥) 𝑑𝑥.
3. If Fis any function such that F'(x) = f(x), then f f(x)dx = F(x) + C is called the general
(an arbitrary constant).
solution and C is called the 挀漀nstant o⠀椀nteqration
,
EX #1: We know that if a function f has an anti-derivative,
F(x), then there is a "fami氀礀 of昀甀nctions" which share
the same anti-derivatives; and, they will differ only by
a constant, C.
A. State three 昀甀nctions F1 , F2, F3 whose derivatives
are f(x) = 3x 2•
昀琀= X 8 + 5
F� = �㨀㌀3- �
.Ⰰ最
昀㌀ 㨀㨀 ∀堀 + ⸀鄀
B. Explain the meaning of "+C" on your functions above.
gr愀瀀h 1s traosl愀⸀⸀琀ed +c or
-挀⸀ J up o爀ⴀ down
EX #2: Applying Basic Rules
Given each derivative, below, find the general anti-derivative by Guess and Check.
A.
B. f'(t) = t 2
g'(x) = 2x
3㘀㰀) = X ⴀ琀 C
㨀开
D.
f(t)ⴀ㴀
搀昀 =cos0
d0
+Ce)= s,o e +⸀輀
E.
tt 3 + ⠀㨀
2 3
F' =-x4
3
F⠀砀) 㴀㴀 椀尀X
✀⼀4
1
I
y=­2
x
C.
�㬀㨀 -ⴀ笀+⸀輀
+C
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f
The expression f(x) 搀砀 is read as the antiderivative offwith respect to x. So, the di昀昀erential
dx serves to identi昀礀 x as the variable of integration. The term "inde昀椀nite integral" is a synonym
昀漀r antiderivative.
Notation 昀漀r antiderivatives:
The operation of finding all solutions of this
equation is called antidifferentiation or
indefinite integration denoted by sign.
dy = f(x)
搀砀
dy= f(x)搀砀
A GENERAL ANTI-DERIVATIV E is denoted by:
愀愀⸀㄀㜀1e of
! Ⰰ눀ns+a渀琀 of in�ꄀ툀on
1ot攀⸀9c搀찀漀漀l
y =f f(x)搀砀 = F(x)+C
t ⸀�ⴀ昀id攀爀,v넀�ve
j
,nt�爀∀a爀搀
EX #3: When integrating, think be昀漀re you work! Practice will help you "discover" many tricks and
techniques that make the integration rules "昀椀t your problems".
J +
Se t
TRICK •I
REWRITE
TRICK •2
MULTIPLE or
DISTRIBUTE
1
cos 2 t
J (3x -1) 2 dx
J
TRICK •3
SE倀䄀RATE
FRACTIONS
x 2 + 4x+ 5
X5
J 嘀堀
(x - 3) 2
TRICK •4
MULTIPLE
O PTIONS
dt
= J 5攀Ⰰt + see t 椀휀
- 5et + 踀t +�
2
- j(9x� - wx +I) dx
-
dx
3x㨀㌀ - 3砀䨀 + x + 焀尀
= j (x- 3 + 4x- 4 +5x- 5 )dx
- �x㨀䨀
I -㌀ⴀ3 - 5 +t
3x
嘀娀x氀昀
dx
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EX #4: Match each function with its antiderivative, where C is a constant.
Function
C
끫뢦 끫룊 = 6끫룊 2 + 6끫룊 − 5
A
끫뢦 끫룊 = 3끫룊 2 − 12끫룊 + 4
D
끫뢦 끫룊 = 4끫룊 3 + 6끫룊 2 − 2끫룊
B
끫뢦 끫룊 = 6끫룊 2 − 8끫룊 − 2
Antiderivative
#4B: For 끫뤲 > 끫뾢, match the integral to its antiderivative.
Integral
G
1
끫뢢끫룊
끫룊
E. 끫롲 끫룊 = ln 2끫룊 − 6 + 끫롬
1
끫뢢끫룊
끫룊 − 3
F. 끫롲 끫룊 = ln(2끫룊 3 − 6) + 끫롬
�
E
�
H
�
F
�
Antiderivative
2끫룊
끫룊 2 − 2
끫뢢끫룊
3끫룊 2
끫뢢끫룊
끫룊 3 − 3
G. 끫롲 끫룊 = ln(2끫룊) + 끫롬
H. 끫롲 끫룊 = ln 3끫룊 2 − 6 + 끫롬
#4C: Match the antiderivative to its integral.
Antiderivative
Integral
L
−6 cos 3끫룊 + 4 + 끫롬
J.
� −18sin(3끫룊 + 1) 끫뢢끫룊
J
6 cos 3끫룊 + 1 + 끫롬
K.
� −18 cos(3끫룊 + 1) 끫뢢끫룊
M
6 sin 3끫룊 + 4 + 끫롬
L.
� 18 sin(3끫룊 + 4) 끫뢢끫룊
K
−6 sin 3끫룊 + 1 + 끫롬
M.
� 18 cos(3끫룊 + 4) 끫뢢끫룊
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BA SIC INTEGRATION RULES
Integration is the "inverse" of di昀昀erentiation. Di昀昀erentiation is the "inverse" of integration.
Differentiation
J
J
J
J
d
-[C] = 0
dx
d
-[kx] = k
dx
d
,
[k f(x)] = k · f (x)
dx
,
d
,
[f(x) ± g(x)] = f (x) ± g (x)
dx
Integration
Odx=C
k dx = kx + C
k f(x) dx = k
J
f(x)dx
[f(x) ± g(x)]dx =
J
f(x)dx ±
J
g(x)dx
GENERAL POWER RULE
I
d
-[x n ] = nxn -1
dx
J
xn+l
xn dx = -- + C · n ꄀ最 -1
n+l
TRIGONOMETRY RULES
d
[sin x] = cos x
dx
d
[cosx] = -sinx
dx
d
dx·[tan x] = sec 2 x
d
[cscx] = -cscxcotx
dx
d
[sec x] = sec x tan x
dx
d
[cotx] = - csc 2 x
dx
J
J
J
cos x dx = sin x +C
sin x dx = - cos x + C
sec 2 x dx = tan x + C
J cscxcotx dx = - cscx+C
J
J
sec x tan x dx = sec x + C
csc 2 x dx = -cotx+ C
TRANSCENDENTAL FU NCTION RULES
J
J
J� =
d
-[e x ]= e x
dx
e x dx=e x + C
ax
a x dx=-+C
ln a
d
-[a x ]= a x Ina
dx
d
1
dX [In x] = -X x > 0
dx
lnlxl + C
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SO䰀嘀ING DIFFERENTIAL EQUATIONS
In Example 1,we explored a simple di昀昀erential equation (DEQ), dy =3x2.
dx
So, a di昀昀erential equation is just a statement about the derivative ofy (with respect to x). We
昀漀und a 昀愀mily of functions that had the same derivative. This antiderivative is the general
solution. Now,our goal will be to solve 昀漀r y with a speci昀椀c solution 昀漀r the constant of integration,
+C, based on an initial condition that can be used to 昀椀nd C.
EX #5: Solve the differential equations subject to given conditions.
A. f'(x) = 2x-1 , f(Z) = 5
�= 爀䨀x-1
J d� 㨀㨀S⸀䨀(�x-1) dx
� 㨀㨀 X - Xⴀ⬀ ⸀焀
5= 㬀Ⰰl-J�䌀Ⰰ
挀开㨀㨀 ⸀㌀
f (x) = X 䨀⸀ - X + 㨀㌀
tG) =-3
5 ay =SJ 攀漀s x 挀昀 x
y = d s,'r, x + �
- 3 㨀琀 J s i" (�) � ⸀焀
-3= �紀蔀
B. f'(x) =Zcosx,
琀Ⰰ㨀㨀-5
[-�-(x)-= -�-⸀猀,.-A x---5 1
ⴀⰀ
D. f'(x)=3x2 -4x+2,f(-1)=-4
S 漀⸀y = J(3xJ - 氀礀x+�) dx
y = x3 - Jx㨀氀 +伀鄀x + 开輀
-⸀뀀 = (-1⤀㌀- J(-1)㬀㈀+ �(-1) +⸀漀
䌀Ⰰ㴀㨀 椀⸀
+䌀砀)= tx3+ 넀ⴀxi - 1x+fⰀ䨀
E. g'(t) = 鐀ꈀ, g(9) = 6
J d y = S t ㄀⼀⸀㄀ dt
y � +t 3⼀㈀3 + 䌀Ⰰ
䌀漀= 䨀娀(9) 㐀怀+ ⸀錀
�㬀㬀 18 + 开漀
t㴀娀 - l J
[䌀鐀x)= ⤀⠀ 3 -� 怀栀 +Jx+ 椀䨀
F. f'(0) = sin 0, f(O) = 2
Sdd � S s,⠀⤀e de
yⴀ㴀 - ⸀蜀S 8 + G
J = 䌀䐀SO + �
䌀Ⰰ = 3
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Lesson 6: The Fundamental Theorem of Calculus
and Definite Integrals
Topic 6.7: The Fundamental Theorem of Calculus and De昀椀nite Integrals
We have now seen the two major branches of calculus:
1) differential (tangent line problem)
2) integral (area problem)
Leibniz and Newton independently discovered a connection between these two branches, stated
in昀漀rmally, that di昀昀erentiation and (de昀椀nite) integration are inverse operations. In an earlier lesson,
we used the Fundamental Theorem of Calculus, Part I to define new functions, represent
accumulation 昀甀nctions, and provide information about the antiderivative function.
THE FUNDAMENTAL THEOREM OF CALCULUS - PART I
If f(x) is continuous on [a, b] then, g(x) = f:J(t)dt is continuous on [a, b]
and it is di昀昀erentiable on (a, b) also g'(x) = f(x).
An alternate notation can be seen by taking the derivative of both sides of F(x):
dF d x
-=- f f(t)dt=f(x)
dx dx a
In words we can say: The Fundamental Theorem Part I says that if we de昀椀ne a function as the
definite integral of another function, f, then the new function is an antiderivative off.
Now we want to turn our attention to the FTC-Part II which will give us a tool 昀漀r computing definite
integrals in a quick and efficient manner.
Given that/represents the RATE at which the function F(x) is changing,that isf(x) = F'(x), we
b
know that f J(x)dx represents the TOTAL (NET) CHANGE IN Fon the interval [a, b].
a
THE FUNDAMENTAL THEOREM OF CALCULUS- 倀䄀RT II
If a 昀甀nction/ is continuous on a closed interval [a, b] and F(x) is an antiderivative off on the
interval [a,b],then
f f(x)搀砀= F(b)-F(a)
a
b
If you like,you can also write this as: f g'(t) dt= g(b)-g(a)
a
In words we can say: The de昀椀nite integral of a function over [a, b] is equal to the NET change,
or displacement, in the antiderivative off over the same interval. This is commonly known as the
Evaluation Theorem.
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Using Proper Notation
When applying the Fundamental Theorem of Calculus, we will 昀漀llow the notation below:
fab
f(x)dx = F(x) b
a
= F(b) - F(a)
EX #1: Find the value of the de昀椀nite integrals by using the Evaluation Theorem stated above.
B.
L ( 2x - I)' = J II � X ✀圀 - 4 X + 䤀⤀ dx
Ⰰ鼀
1x 3 -Jx ⸀㄀ +x氀椀
(f-a+1)- c-i-漀㈀-1)
I椀鐀
3
E.
f,'䨀⬀ 搀砀 �
3[f X 琀甀J�
愀氀 (� 漀⼀㬀㄀ - ' 3⼀㐀)
I 氀昀
ⴀⴀ
c.
f3F 搀砀
f 5 x- dx
D.
Jof\sin x+ l)搀砀
夀匀
- cos 堀匀 + ✀堀 J O
(- 䌀⸀o⸀猀 ㄀昀 + ㄀吀) - (- cos O + O)
(I +㄀爀) - (−\+O)
㄀㄀+ 2
㄀⼀㨀ꠀ
⸀먀 嬀开 J X Y⸀㼀J :
⠀氀嬀䨀 [ 爀椀-䨀吀]
- 8-�㔀销
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The Fundamental Theorem of Calculus - Special Cases
So 昀愀r, we have seen thatwhen givenF(x) =ff(t)dt , f (t) has always been the derivative of F (x).
In the next exampleswewillwant to explore some special caseswhere f (t) mayor maynot be the
derivative of F(x). Wewant to find a general rule 昀漀r evaluating expressions such as:
d [g[x]
f(t) dt
dx [
l
l
d [g(x]
In otherwords,whycan'twe saythe result isALWAYS - J f(t)dt =f(g(x))?
dx
a
EX #3: Find the derivative in two di昀昀erent ways. Will your result always be f(x) = F' (x)?
CASE
#1:
Upper bound:
Function of x
Lower bound:
constant
J: cost dt ✀堀
䘀⠀x) = s,/, t I 㬀爀
Upper bound:
function of x
Lower Bound:
constant
F (x) - s, -昀尀 x - S 椀∀ n ㄀Ⰰ
F (x) 㨀琀 Sin x
Upper bound:
Function of x
Lower Bound:
Function of x
娀 搀琀
� ( I +t ) 3/㬀㼀
F(x)㨀㨀 3
J
Ftx) = cos x
F(xJ=t(1+x3)¾-f
Fcx)= (\ ⬀ⴀX 3) ㄀⼀J • 3xJ
F(x)= f(iⴀ琀 t㌀鈀搀⸀_ j(1+0/P
搀夀iŠŬ ćğ
of X⸀㌀
3
琀ⴀ (x) = ;(I+ x 3) � �ⴀ℀
f(x) 焀꜀ F꜀儀
0
⤀⠀3
-
䘀琀 x) = 琀ꌀ5 x i - 漀혀�s 砀⤀
s,爀椀 t ⸀� Ftx)= -sm x"1. � x +
F(砀⤀ = J 挀㼀 sint 搀琀ⴀ⬀
X 2
�
l�x
Ⰰ⸀
+ s i昀尀 (eos x)⠀ⴀSr琀椀x)
s,nt搀琀 1
F(x)㨀㨀-j　⸀ s,nt 搀琀 +
F (⸀砀)㨀㨀 -Jx S1爀∀x�F(x): COS (✀堀 㬀Ⰰ✀⤀ - ℀吀OS (COS ✀堀)
s,n x (si爀椀(cosx)∀⤀
F(x) =
#3
F (x); S ,·" X
F(x) =
F(x)-f:
#2
Find F'(x)
Find antiderivative F(x)
fcosr
X
2
sin t dt
⤀嘀
l
cos⤀⠀ •
眀舀
昀最㘀㰀)㨀琀 r fx)
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Integrating a Piecewise Function
12 p ⸀瘀
EX #7: Consider the piecewise defined function, whose
graph is shown at right.
10
{3x2 x 猀砀 I
f(x) =
5 x>I
8
6
4
Evaluate each of the 昀漀llowing integrals.
2
-2
A.
氀✀ 3ⴀ㼀
= x3 1_� + 5x I�
f f(x)搀砀
㨀㨀 s:
B.
5 䨀�
I!
ⴀꀀJ(x)搀砀
=
J
-2
X
2
鐀숀 + � � 5 휀븀
- 嬀㄀3-(-昀꜀ + 5(4-1)
= 5x
= 5( 8-3)
= �
, 4 ⴀ⬀ z + 5(3)
㨀�
=
Integration with a Calculator
EX #8: If f is an antiderivative of
I
S0
1
cos( x2 )
挀漀�/ ⴀ㴀 fC 1)- fC >
e
0
x
such that f(l) = �, find f(0)
o. 59oosoqq = t- f (o)
f (o) = ⴀ紀- o. 59005oqq
f (o)z-o.J5"7'
EX #9: Unleaded gasoline is being pumped into an empty underground storage tank 昀漀r the opening
of a new convenience store at the rate of 600崀ꈀ gallons per hour. How many gallons of
gasoline can the supply truck deliver if the driver pumps 昀漀r 2 hours?
� = 漀鈀o䨀ꈀ
5 dV ∀✀5 㬀崀w昀琀 搀넀
V = 5/ boo 䨀吀 搀琀
V= 800@ 8Ųƌ � 'l31. 3108 9�
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EX #1: Integrate by u-substitution.
A.
J +
(x 2
1/(2x) dx
㬀紀
B.
唀⸀= 堀✀ + \
J
搀甀� �⤀⠀ 琀�
S✀1 u + 焀尀
St笀⤀ 㐀ⴀ X + ⸀焀
3
u
t +G
f( 㨀琀 + 1)
C.
J
3
+ 攀Ⰰ
Bx⸀䨀4x 2 + 1 dx
D.
氀䄀 � 4X�+ I
d䰀䰀 = ox dx
3
x (x 2 + 2) dx
搀甀 � 㤀ꀀx 鈀尀x
℀娀 鈀븀 㨀琀 ⤀⠀ d x
3 u
u
¾S d
3
t [ 椀錀 u ꬀뀀 ⬀ⴀ 䌀Ⰰ
f( 4x +I) + L
E.
J
U= X J + 㬀㈀
s u ㄀⼀� �騀
�u ¾+⸀輀
¾
㬀椀
U=4x
愀甀= 㐀ⴀd�
J' 挀⸀os 䰀琀 밀騀
s·u�椀�
㨀氀
4cos4x dx
f (4xz x+ 3)6 dx
㨀需 + J) 4 + 挀⸀
x
f(
F.
u: 堀娀x�+ 3
x
䨀ꔀ� 8砀✀ 挀琀
ⴀ琀椀� = � dX
ⴀ⬀5 u-✀ⴀ du
t[t u- J + C
5
� ( 4�+3r s +⸀鄀
f ㌀䨀 (7 - 2x x dx
3)4
2
䰀氀㨀㬀 7-�x 3
du= -㐀㸀x J dx
-椀ⴀ挀琀u = X�d 㼀䌀
-f s u ✀氀⼀㌀椀�3
⬀ⴀ C
Y
J
u
ⴀ
-琀娀
3 ¼+ L
)
x
J
� (7
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Definite Integrals.
EX#4:
EX#S:
J x 3 Inx dx
1
-e-x
u=
X
v=
du=
IM
dv=
u=
e-x관찀
- xcx 1; + 㬀瘀Ⰰ⸀ e-x 䄀윀
-砀攀-x+ (-e-t)/:
- 4e-� - e-∀✀-(o-e 0)
du=
昀渀x
1 밀需
v = 笀ⴀ
dv= X
砀夀
✀䤀
3 ⠀唀
e
3
开䰀昀砀 x琀㐀爀� X le - ㄀贀5
x�
✀琀
I
I
tx 4蜀x - ¾ ·ⴀ㐀 x�J �
4 -(o-i;
最븀
ⴀ䰀e
I琀漀
䰀㐀
- 5e- 4 ti
3
4
䨀븀
J氀倀 € ⴀ⬀ I琀漀
Using Integration by parts and u-substitution to evaluate.
EX#6:
u=
du=
EX#7:
J arctan x dx
ian- 'x
I
l+X�
v=
dv=
X
i 瀀�
= X ijŔt - J, :x椀最 ☀鄀
㨀㨀 A ěDz t -錁匀 I X 㨀氀+j ) 氀ⴀ䌀⸀
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u=
v=
du=
dv=
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Long Division
Sometimes we need to integrate an improper rational expression with a higher degree in the
numerator or the same degree in both the numerator and denominator. Long division can help reduce
the integrand to a polynomial plus a proper 昀爀action.
J
EX#S:
3x 3 + 8x 2 - lOx + 15
dx
x+ 4
EX#6:
-耀萀 3
� -10 15
I䜀Ⰰ -J4
� -J�
3 - ✀椀
☀Ⰰ (- q
S(a⸀砀 㬀椀 - ⴀ笀x +" + ;:䤀昀 ) dx
⠀℀3-�x 㬀需 + bx- 9 � /✀开+41 + 挀䨀
J
EX#7:
x3 +x
dx
x- 1
䨀樀 1 0
� 1
i
-1
1
� �
J
+ x - llx + 20
-----dx
x2 + 2x - 3
2x3
2
�x-3
- ⸀㌀⤀⠀㴀㈀ - 5x +漀椀o
tx + tl +� x +J �lx-1l+t
1
䨀开尀✀3+⸀稀/ +Jx+ �(x-0 + 挀开
-
ꄀ툀 - 琀椀 攀鬀,�- ii+ ⸀鄀
㨀䰀
+ 䌀⸀
1)
(X夡ᨀ
fX -
x +�x-3 )漀氀x3+x� - 11ⴀ堀ⴀ⬀䨀⸀o
-䨀ꀀ㨀㌀ + ✀⬀x✀ⴀ - "x)
�
3
�
S (ⴀ봀 + �ci-1� dx
d
s �2 + x + 䨀开 + 㬀⸀,) dx
3⼀尀
Y漀琀
�x-氀鄀) X -5
-(X - I)
- 騀砀
EX#8:
1 O
J
x- S
dx
2x- 2
-(-3x� -�x+ q)
x+"
s·�⤀⠀-3+ X+II
dx*
X㬀琀+�⤀⠀-3
S Jx-3-
�
X+3
+ⴀ帀 dx
X-I
= ⸀꜀㄀栀_ 3x-J�IX+3l+3�lx-il+(�
挀䰀 +怀褀 椀鈀arti愀䨀
(X琀娀3)䌀砀-t) xt3 x-, 1
X + II
㨀㨀
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kn�
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Theorem:
(　1
jl xP
EX#5:
is convergent when
Sketch the 昀甀nction f(x) = ✓x�
5
i
p
>I
and divergent when
p 㴀㨀 ⸀椀
over [O, 5].
A. Why would f2 ✓x�z dx be considered
improper?
伀琀 x� � ⴀ爀e⤀⠀) h⸀퐀 M
i昀딀oi攀ꈀ dis 琀脀nti椀尀ui� .
.
C. Now 昀椀nd Lim S(t)
tⴀ⬀2+
\i爀渀 (㨀䨀昀挀-琀氀Jt-�)ⴀ㴀
琧鐀JT
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