ページ3:

WW
Ko
k² =
(971)(24+1)
6
Need to know!
k³ = [(1+1)] ]²
Recall the fig=(-x²
Aren from [0,1]
width = b-a = f
Use RHS Eval.
A=A1+A2+
h
a=
= ½ (f(±)) + ± (f(ñ)) + ·· + ñà († ½³) + ·· + ½ (fα))
= ñ ( f (h)+ f (å) + ·· + · f($) +¬·~ +f(1))
A =
Ak = f()
- (*)
- 21, AR - 2 A [1-()]
= + (2/11 - 12/²²)
= = (n - +
(n+1) (2n+1)
= 1-
(n+1)(2n+1)
6n2

ページ4:

MAT
021B
10/1
Last
/-
time
Continue:
6n2
Check n-3
A= 1- B+V (283+1) = 0/27 20.481
Try nooo
6.32
(Same as before!)
lim A = /am [1- (nt) (2n+1))
noo
642
=
=
-
lim
177
24+35+1
by2
Example
<
P=123,583
+
+
/
2
3
"
"
"
20
x1
22
x₁ = x-x=2-1 = 1
2x2-x₁ = 3-2=1
Ax3= x3x2=5-3=2.
Ax4 = X4-X3 = 8-5=3
24
· |-/m2 273 (1/07)
1- X'm 2+3/1+1/2+
6
|- 240+0= 2/3
=
6
• Tim ²² + lim
4-700
mo
3/
"Norm" of the Partition: || P||
||P || : the width of the largest subinterval
defined by the
partition.
"Partitions" for
an
interval:
X2 X3
行
42
We have subintervals:
We
... XK
Xn-1 Xn
+
12300
Our example:
||P|1=max
[xo, x] [xx] [xx]... [ XK-+, kk), .., [Xm-x, xn]
,
require that
A = X0 < x <x<<<<xnx<xn=b
=
{AXI, OX2, AX7, AX4}
max {1,1,2,3}
- 3°
#
Let's define Cki
For
Xo, x1, x2, ..., XK,...,.
- xn}
A partition is the set:
P = {x
We want Ck such that
CK = [xx+, xx]
E
our example: C₂ could be 5/2
because
x45625x2
Let's define:
4x1 = x₁ - Xo
SX = X2 x1
25553
the choke is not unique!

ページ5:

Riemann Sum
The Riemann Sum Sp for a function f over an interval
[a,b] is given by
Sp = f(x) XK
Note that this Summation depends on the choice of the partition
P and the points Ck where the function is evaluated.
Thus there
are
many
Such summations and we would
expect
that the summation would depend on the details.
著
*2
f(cz)Axe
x
Definite Integral
Let f defined on [a,b].
We say
J is the definite integral of f over [a, b]
and J
=
lim f(c)x
Po
if the following condition satisfies:
S.t.
||P|| <8 and Ck € [ xx, xx] = | §ƒ (4)ox - J | < E
any
choive in the interval
Specifies how small || P || should be
to make the outcome error < E
C₂
f(48)4x8
Note there
f(C8) <0
f(18) 078<0.
| f(C8) sx8|= Area of rectangle.
If the above limit exists, we write the definite
integral as
J =
f(x) dx

ページ6:

MAT 021 B
The Definite Integral
Def [f(x)dx =
* F C = *
10/3
lim = f(ck) akk.
UPO 70 K=1
• means "the biggest one"
F
P= {xoxx,...,xn}
with axo and
b=xn
b
Axk = xx-xx-1 and Xx-1=CK EXK
So fbx dx
So flx) dx
S
9 :
:
integral sign
lower limit of integration
b
upper
limit of integration
fix: integrand
dx: differential of x.
Compare
M
CK
f(CK) AXK
to
So foydx.
S
x
XK
dx
Discrete
Continuous
Summation
Summation
The Def. Integral
Equally Spaced Subintervals
DX
△x
9
b.
x1
K-4
Kh
n- subintervals
width Interval is xn-xo=b-a
b-a
AX=
xk:
n
X0=4
x150+4x
×₂ = a+20x.
xk=a+kax.
For RHS
->
evaluation
Ck = x
f(ck) = f(xk)
= f(a+kox)
= f (a+k· bra)
lim f(CK)oxk
"
K
-Im fla+k+) (+)
[Pok
= lim = fla+k ba). (ba)
11-+00 k
True for Equally spaced Subintervals
Example evaluate (2x) x
by definition.
<sol>1° lim & fla+k. b-a) (th) = for fluid.
=
11900 K:1
(a=0, b=b, 6x=b-0)
I'm fo+kb
f()+
-
2bk
= lim 26 k
K
= lim ³ (7+1)
1200 m² &
= lim b²(n+1)
1-3∞
n
= b² lim n+1
noo n
n
can also use
√(b,26)
"
triangle
6x2562
2

ページ7:

Theorem
-
Continuous Functions are Integrable:
If a function f is continuous over the interval [a,b],
f has
at most
definite integral
or
if
finitely many jump discontinuities, then the
f(x)dx exists and
function f is integrable over [a,b].
we
Say
that the
Theorem -
Area Under a Curve:
a
y=
If y = f(x) is
closed interval [a,b], then the area under
the curve
f from a to b:
nonnegative
and integrable over
f(x) over [a,b] is
the integral of
A = Sfodx

ページ8:

MAT 021B 106
8 Rules of definite integral (thank yemetically).
1. S f (x) dx = - 5,ª² flxdx
2. Sa² flxdx=0.
3. Sob k f(x) dx = k )
* Sa foldx (k = const.)
4. So [ fin± gera] dx = So fix dx ± So good
土
5. Forde += sex ([asc] U[C,h] = [a,b])
f(x)dx
b
'fain -fmax
6. fmin' (b-a) ≤ for foxy dx = fman (b-a) (fram af te do
1. ff) dxf goodx
1a. So f(x) dx>0
Dummy Variables
[a,b])
(if fag
(if fox) 20 on [a,b])
Assume equally spaced.
Approx Ave =
[a,b]
KY
=
f(CK).
b-a ky
4x
b-a
f(CK)
AX
Riemann Sum
Let ||Po n∞ for
P
b-a
4x=
n
AX
=
b-a
equally spaced subintervals
E.
lim f(c) x = lim & f(ck) ax
1700k
as n∞ the Approx Ave Exact Value
ave f(x) = 1
[a,b]
b
f(x) dx
b-a
↓
=
think of like changing ex-axes
T
Dummy Index:
K+
k = j = 1+2+3+4 = 10.
j=1
its name
f(x)=1-x²
y
f(x)=1-x
A= 2/3
a
Function Redux:
ave fly = ± S'" (1-x") dx
[0,1]
%
(geometrical interpretation)
The Average Value of
Let
b=xn
4=X
Set
up
9
§
Riemann Sum: XK-1 CK ≤ Xk.
C1 C2
baxm
approximate average
[a,b]
BT =
f(CK)
n

ページ9:

Mean Value Theorem
If f
is
Continuous
on
[a, b]
then there exists at least one value c
s.t.
finca
f(c) = fave
4
fmin
So flx)dx.
Example
af(x)=1-x2
in [a b]
The Fundamental Theorem of Calculus, Part I.
Define F(x) = f(de
Find F'(x) = &* fludi
lim-
h→0
F(x+h)- F(x)
h
t
x+h
So, F(x)= flx)
ho
1
xrh
= lim | f(e) dt.
=
lim
h+o
[f(x+4)]
= Jim f(x+h) = f(x)
h+o
* f (4) dt = f(x)
- FTC I
doesn't matter what
Ck = [x, x+h] we plug in
Since Ck →x
as ho
example. Let y = √," (±²+3+) de
Neel for f(x) = # foods.
c
=
1-0
==
= 1-c².
= 3 c² 1/3
-xx
½ = (negative doesn't count),
-
:
So,
C = √
c = [a,b]
E
Find dy?
dx
(Sol> d = √(t'+36) de FTC-1 f(x) = x²+3x
dx
dx
fts)
Recall FTC-1: √, fedt = fus
example: Find ff fr (1++³) dt =>
<Sol> Let u=x³ → du=2xdx
=
(de)
= 2x (+4³)
:
2× (1+x)

ページ10:

MAT 021B 18
Fundamental Theorem of Calculus, Part 2 (Evaluation Theorem)
If is continuous over [a, b]
and F is
any
then
L
"a
antiderivative of f on [a,b],
foxdx=
<proof>
= F(b) - F(a)
ľ
preparation:
We know that.
Assume: F(x) = f(x)
m slope
=
AX
→> k = (MAX)
2°
K+
k+1
yp
F(b)
F(a)
4y = max
a
b
F(x)
zoom
4yk
Partition P= [5, 2, 3}
XK-1
m=
= f(ck)
F(x)
ex
XK
Axk
F(b) - F(a) = 4yk
particular
[f()]
= lim [f(ce) Axe]
=
of
=S+
f(x) dx
rb
doesn't matter now.
(P0)
4YK = m 4XK
[def of definite inc.]
F (6) - F(a) = f(x) dx
Ja
=
f(CK) AXK
exists Mean Value Theorem
of Differentiation
Suppose y=fus is continuous over a
closed interval [a,b]
and differentiable on the interval's intenter (a,b).
Then there is at least one point c in (a,b)
at which
f(c) =
f(b)-f(a)
b- a

ページ11:

MAT OB 1
Antiderivative
def
F is antideriv. of f on I
Example
is
an
if f'(x) = f(x) for all x in I.
Show that F(x)=&e²* + + sin(2x) + 7
3x
antiderivative of f(x) = " + cos (2x)
F(x) = 14+ & sm (2x) + 7 ]
=
dx
3x
} e³. 3 + & cos (2x). 2
=
'+ = cos (2x)
Q.E.D.
Indefinite Integral
Def. The collection of all antiderivatives of f is called
the indefinite
s f(x)dx
integral off with respect to x and is written as
= F(x) + C
Here:
S: integral sign
a function.
feg:
integrand
dx: differential of x
Fe antiderivative of foxx
C:
Constant of integration
on interval I
I
on
is
F(x)+C
Read this as
S[J] 2x
means
Theorem if F is an antiderivative of f
general antiderivative of f.
the most
C is
an
arbitrary
Const.
F(x) is not unique.
F(x)=2x+8
e.g.
fox = 2
F(x) =2x+]
Recall, there were mathematical notations
for finding derivatives
Let y = for
"Find the derivative of f(x)" is
+ [f(x)] = * = f'(x) = y'
"Find the derivative of [...]
" find the [most genel] antiderivative of [...] with
The
respect
+%.
process of finding untiderivatives is often called integration
Thus we can say:
+ Fox) = for SAD dx = FO+C
examples
=>
Jx tunx) = seex secx, dx = tanx +C.
* [Sec" (x)] = "/dx = sei", +C
.
d
dx
[secx] =
= 2 secx secx toone
= √2 secx tanx dx+C.
= 2 Sec³x tanx
Rules
Formulas
1. Sx"dx =
n+l
2. S sinxdx = -205x+C.
9.S=11x+c (×40)
1. const. mile: Skdx= kx+C.
+C.
2.
conse, multiple nde:
kfdx = k [ fexdx.
dx
10. x = sin x+c
(x^<1)
3. sum/dif rule:
[fest gey] dx = ffos dx ± √ gosdx
4.
power
rule: Sx"dx =
[n+1
+C. (n-1)
Example
sinx +c
3. Scosxdx=s
4. S sex dx =
= tanx +L.
5.fcsc²x dx=-cotx+C.
6. Scccxcotx=- cscx+C.
7. S sexx tanxdx = secx+c.
8. Se* dx=e*+c.
11. S dx = tan x+c.
1+x2
dx
12. √1x14x4 = sec³x+c.
(x>1)
13. √a^dx = en na + C. (a>o, a#1)
Sdx.= √(x + # ) dx.
=
S ( x + 2 x ) dx.
x ½ +1
X+
3/2
×
+23
1/2+1
-½+1
2 + 4 x ½ + C
(C₁+(2).
+ C

ページ12:

MAT 021B 10/13
Suppose / Ex²-3, you =0,
DE
-
IC
6x3
Sdy of( 6x = 3) olx → y = x² - 3x + C.
dx
Apply I.C.
u-substitutions:
| y=2x² - 3x + c general sol.
0 = 2 (1)²= 3(1) +C
c = |
| y=2x³-3x+1 particular solution
method 1.
Let's evaluate √(x+1)*xdx = [(x*x) xdx =
example
=S
S tanx ds =
Sinx
cos x
-f-du
dx
= - In|u| + C .
lulose c
+ C
= |In | Secx|+C
example
√
secx dx
=
Ssecx
Secx + tanx
secx+tanx
Let u = cosx
→ du = - Sinx dx.
dx
= f seix + sectare dx.
u ·→du= (sec²x + sexx canx) dx.
S(x²+2x²+ x)dx.
=
=
+
method 2: Let u=x+1 →
-=2x→ du= 2xdx.
f(x²+1) = x² du += f +
=
=
(171) + C
3.2
+C
+C
example
77
=
兽++++C
Example (√x71) xdx.
Let u= x²+1
→ du= 2xdx. →
= √ 4½ du
= In \ tanx + secx/+c
Sex dx
= √ 1+ex - e* dx.
ex+1
= Sdx - ST
=
Su
x-bu+c
Try u=e+1 →du = "dx
adu.
x- · en le* +1| +C.
"
=
www same conse.
* equivalent answer
Example Se
24
+ C =
32.2
+C
3/2
Let u=x' du=3x+dx.
= √ udu = u² + c =
2.3
know: [smx] = cosx
d
dex [cosx] == s
->
++C
·S.
cosxex smx+C
E - sinx — sinxdx =
-C
->>>
The Integrals of six trig functions:
1. S cosxdx.
= sinx + c
2. S sinx clx= - cosx+c
3. S tanxdx= In | secx/+c
4. S secx = en tanx+secx/+C
5. Scotx dx = In/sinx |+c
sex. Seat funt.
6. Sesexdx = -lalesextox t

ページ13:

MAT
021B 10/15
Substitution for Definice Integrals
Let
x-b
S f (g(x) g'(x) dx.
•x=a
g(x) = u
4-965)=1
cu=g'(x)dx.
Sf(u) du = F(B)-F(A).
u = g(a)=A
example
5/44
cosxsinx
dx
COSX + Sin
Let us cosxsinx
XE O
=
431
1√2
= en lull,
= ln JΣ - In |
=
== ln 2
du= (cosx - sinx)dx.
when x=0+4= |
| when x = 1/4 + 1/1 + 1/12 = √2 |
Area Interpretation of
fos<o for integrals
Az
b
>x
A
A₁ = - Sa² fessex >0
A₂ = S₂ fexdx
>0
→ Sea Food x = Son Ax dx + So festex
= - A₁ +A₂
-
Area
(between fix) and
x-axis
Def of Even&Odd Functions
over Symmetric Intervals [-a, a]
1. A function is even
of X E
[-a, a]
Satisfy
on
L-a, a] if for all values
f(x) = f(x).
2. A function is odd on [-a, a] if for all values
of x-a, a] satisfy:
1.
f(x) = f(x)
f(-a)=4(a)
f(-a) = f(a)
fla)
2.
f(-a)=-f(a)
Z
f(-a)
All About Even and Odd Functions
1.
2.
common even functions: (E)
a) constants: 1, 3, √2,...
b) even powers : x², x, ¨·‚׳‚x*,...
c) these trig
cosx & secx
Common odd functions (0)
a) odd
powers: Xxx, ..., xx³-
b) these trig: tanx, Sinx, csex, cotx
3.
How to combine by (±)
0 ± 0 = 0
E ± E =E
E = 0 = X
0 ± E=X
0xE=0
4.
0*0=E
EXE =E
E¼o= 0

ページ14:

1. If fis even
on
[-a, a]
then foot = 2 fra forde.
2.
then
odd
So flx) dx = 0x
even
"
-easy!!
2.
easy!!!
old
f(x)=-f(x) old
f(x) = f(x) even
example
example
a) x(x+2) = odd →
m m
ochd
even
=even→
even
b) x²(x²+2)
m
even
m
dx
• L'x = (x+2)
=25'x²(x+2)dx
Godd
S sinxx dx. = f (odd) dx =
x²+1
Seven
-
0

ページ15:

MAT 021B 17
f(ck)
fix)
Area between curves
Assume fix gox) on [a, b] g(ck) -
Ak= (ht) (width)
ges
example
X=D
x=sing.
y=sinx.
-y=%2
14
=
[flck)-g(ck)] [oxk]
XK1XK
Xo
xn.
n
Totul Area Ak = [f(ck)_g(ck)] oxk
Take ||pl|0
Ka
=> Total Area = lim & [f(ck) – gcck)] oxk .
IPII to Kel
=
=
S
[fox-g(x)] dx.
~
other perspective:
x-(x²-x)
Fox)=x
i.e.
:
A-a
=△A.
Sites de "goods
"[40-g(x)]
a
example area between fox, g(x)?
1° To find the intersection, set
f(x)=g(x)
x=xxx²-2x=0
x(x-2)=0
x=0 or x=2
2° Area = S₁₂ [ fixy - gas] dx = f (-x²+2x) dx = (- 3ת³ +
=-
+4==告
Draw
a Picture!
Let's integrate along the
Area lim[f(k)-g()) <Yk
PUK=1
y-axis
C
syk
+
gly)
Fly)
Integrate along x-axis
y=sin x
A = S [- sin x] dx.
(Hard!
x = siny. (sin(sin x) asing).
Integrate along the y-axis:
T42
x
A = (siny - 0) dy = (-cosys |
= [-cos³] - [- cos 0 ] = 0-(-1) = 1
18.
=
Se² [f(y) - gcy)] dy

ページ16:

CH6. Volumes of Solids of Revolution
Y
example
y=4-x²
x=4-y
(x=QN4-y.).
the right/lefe half
x
For single disk:
0
V=πr² 6x = Tur² oxk
rk=f(ck)
Vtotal = VK =
K=1
n
[fa] ox
Vtotal = lim π [f(CK)] Axk
=
UPUT TO
6
So Tv [fo)] "dx
.
DISK METHOD (assumption: fix)=0 over [a,b]>
example volume of
(h,r)
cone
BY
For rotation around x:
For
V= []"dx
rotation about y:
V = x ( [fly]]" day.
"
==π√ √ ² (4-y) dy
=
TV
(4y-1)|
=π(16-8)
=
8π
f(x)=y.
h(y)=x
Washer Method
yp
fas
geo
line:
y = mx +b
b=0
rise
m= run
y=
×
2-o = 1/
h-o
=
= (1) ***
=
πC [f(x) dx = x
rh
✓✓
3
Hollow solid of Revolution
Assume fzg. and bath continuous and nonnegative on [a, b]
VfV of S. of R. from fix) about x-axil
Vg =
"
"
g(x)..
→ V = V4 - Ug = √(fax)] x - x 240)'de
"
=
TV
dx
Not: TUS" [40) - gw³)] "dx ~ geometric interpret ??

ページ17:

example 10.5
y.
y=x.
y=1
V = πU Sa³ { [Pics]² = [g(x)]" } dx.
== π 5,² (x²=-1) dx.
9
= ≈ [ ¸x³ − ×], = x[ ( §-2) – (5'−1)]
-
=