I. Self-intro
II. Course info
III. What is calculus? The biggest achievement of mankind in the past 1000 years.
III. (Derivatives and) Integrals/antiderivatives
Q. (acceleration 99K ) velocity 99K position?
Sun&Earth
mE
2
d2 E
mS mE
=G
2
dt
jEj2
E
jEj
Solve for E from ddtE2 = jEjE3 ! OR integrate twice for E:
Area problem, volume problem
* arbitrary region, square, disk, part of parabola, ellipse
* football, paraboloid
work, population,
"invisible" problems
Extract to
FUNctions antiderivatives/integrals
Rules + chain rule
+ Applications
Techniques
Conceptually easier (than di¤erential calculus), technically much harder!
Sec4.9 Antiderivatives
Simple elementary functions
f (x)
xn x1 ; ex ; sin x; cos x;
1
0
xn + c; ln jxj + c; ex + c; cos x + c; sin x + c;
F (x) w/ F (x) = f (x) n+1
Other Trigs: tan x; cot x; sec x; csc x? How to reverse?
0
0
1
p 1
Fractions: 1+x
2 = (arctan x + c) ;
2 = (arcsin x + c)
1
x
p
p
Others: 1 x2 ; 1 + x2 ?
5
p
eg2. Find all functionS g such that g 0 (x) = 4 sin x + 2x x x :
1
eg3. Find f if f 0 (t) = et + 20 (1 + t2 ) and f (0) = 2:
eg4’Find h if h00 (t) = 9:8 m=s2 ; f (0) = 0; f (1) = 0:
eg4”. A ball comes back to your hand after 1 second. How high did the ball reach
up from your hand?
Sec5.1 Area and distance
Area of rectangle, triangle
Q. ellipse, arbitrary region?
A. Reduce to canonical situation: say three sides straight, and curve side. Cut [a; b]
dx
evenly into n pieces 4x = b na ; sum rectangles as approximate area. When curved
side nice, say f continuous, as we cut …ner and …ner, say n >> 33
40 billions
(QE3 till mid of 2015 at least). The approximate area would approach the real area.
x0 = a, x1 = a + 4x; x2 = a + 2 4 x;
; xk = a + k 4 x;
; xn = a + n 4 x =
a + n b na = b:
1
P
Rn = f (x1 )4x+f (x2 )4x+ +f (xk )4x+ +f (xn )4x = ( nk=1 f (xk ))4x
+ f (xk 1 ) 4 x +
+ f (xn 1 ) 4 x =
P Ln = f (x0 ) 4 x + f (x1 ) 4 x +
( nk=1 f (xk 1 )) 4 x
P
Sn = f (x1 )4x+f (x2 )4x+ +f (xk )4x+ +f (xn )4x = ( nk=1 f (xk ))4x;
xk 1 xk xk :
Def. For a nice function f, say f continuous, the limits limn!1 Rn = limn!1 Ln =
limn!1 Sn exist. We call the limit the area of the region bounded by y = f (x) ; x-axis,
x = a line, and x = b line.
Distance problem.
x 99K t; f (x) 99K v (t) ; 4x 99K 4t = b na
t0 = a, t1 = a + 4t; t2 = a + 2 4 t;
; tk = a + k 4 t;
; tn = a + n 4 t =
b a
a + n n = b:
distance=speed x time
P
Rn = v (t1 ) 4 t + v (t2 ) 4 t +
+ v (tk ) 4 t +
+ v (tn ) 4 t =P
( nk=1 v (tk )) 4 t
Ln = vP
(t0 )4t+v (t1 )4t+ +v (tk 1 )4t+ +v (tn 1 )4t = ( nk=1 v (tk 1 ))4t
Sn = ( nk=1 v (tk )) 4 t; tk 1 tk tk :
v (t) nice, say continuous
limn!1 Rn = limn!1 Ln = limn!1 Sn =real distance
eg1. Area of triangle f (x) = x, x 2 [0; 1] :
eg2. Area under parabola f (x) = x2 ; x 2 [0; 1] : 12 + 22 + 32 +
+ n2 =
n (n + 1) (2n + 1) =6 on p. A35 of the book.
Sec5.1 #30 or HW1 #? Area of disk by limit.
Cut along angle direction
Step1. Cut disk into n pieces evenly along angle = 2 =n; area of each triangle= 12 r r sin 2n ;
total area=n 21 r2 sin 2n
Step2 Recall
x
sin
!0
! 1: Then An = 12 r2 2
Sec5.2 The de…nite integrals
4x = b na ; xk = a + k 4 x
L = limn!1 [f (x1 ) 4 x + f (x2 ) 4 x +
Recall Rn , xk = xk ; Ln ; xk = xk
xk 1+xk
Midpoint rule
:
2
P Mn , xk =
Stretch out
and 4
upper limit
Z b
sin
2
n
2
n
n!1 1 2
! 2r
+ f (xk ) 4 x +
1 = r2 :
+ f (xn ) 4 x] = limn!1
1
dummy variable
f (x)
2
dx
= lim
n!1
integrand
a
lower limit
n
X
k=1
f (xk ) 4 x
Rb
RMK. a f (x) dx algebraic area/net distance
Q. How to evaluate de…nite integrals e¤ectively? (Hint: where are antiderivatives?)
Properties of de…nite integrals
Linear
R b ones:
cdx = c (b a)
a
2
Pn
k=1
f (xk )4
Rb
Rb
Rb
[f (x) g (x)] dx = a f (x) dx
g (x) dx
a
a
Rb
Rb
cf (x) dx = c a f (x) dx
Rab
Rc
Rc
f
(x)
dx
+
f
(x)
dx
=
f (x) dx
a
b
a
Comparison property:
Rb
Rb
f (x) g (x) on [a; b] =) a f (x) dx
g (x) dx
a
Rb
in particular, m f (x) M on [a; b] =) m (b a)
f (x) dx
M (b
a
R1
R1 2
R1
eg 0 xdx; 0 x dx; 0 ( 9:8t + 4:9) dt
R2
R2
eg 0 sin tdt = 0; 0 sin xdx = 0: Recall sin (x + ) = sin x:
R3
eg3. 0 ex dx
R1p
eg4. 0 1 x2 dx
R1
eg6. 0 (4 + 3x2 dx)
Counterexample to: f (x) g (x) on [a; b] =) f 0 (x) g 0 (x) on [a; b] :
f (x) = 2 x x = g (x) on [0; 1] ; but f 0 (x) = 1 1 = g 0 (x) :
a)
Sec5.3 RThe fundamental
R b theorem Rofc calculus
c
Recall a f (x) dx = a f (x) dx + b f (x) dx:
Rb
Vary upper limit b in a f (x) dx 99K F (b) : First
R xF (a) = 0:
Q. What’s the relation between f &F ? F (x) = a f (t) dt:
0
A. Fundamental
(x) = f (x) for nice f (x) say continuous.
RF
R x th’0m of Calculus:
0
0
x
0
Examples. 0 1dt = (x) = 1: 0 tdt = 21 x x = x:
Check.
R x+h
Rx
f (t) dt
f (t) dt
F (x + h) F (x)
0
a
a
F (x) = lim
= lim
h!0
h!0
h
h
R x+h
R x R x+h R x
f (t) dt
+ x
f (x) h
a
= lim x
= lim
= f (x) :
= lim a
h!0
h!0
h!0
h
h
h
Rx
0
That is a f (t) dt x = f (x) :
Rb
Rx
Back to the "fundamental" Question: How to calculate a f (t) dt; a f (t) dt?
Fundamental Theorem of Calculus: For nice f (x) ; say continuous. Let G (x) be
any of its anti-derivatives, i.e. G0 (x) = f (x) ; then
Z
b
f (t) dt = G (b)
a
G (a) = G (t)jba :
Rb
Ra
Rb
b
b
Eg. a tdt = 0 Rtdt
tdt = 12 aa 21 bb = 12 t2 a = 12 t2 + 9876 a :
0
x
Proof. F (x) = a f (t) dt =? (G)?
Rx
Rx
0
f (t) dt x = f (x) ; also G0 (x) = f (x) =) a f (t) dt = G (x) + C
a R
a
0 = R a f (t) dt = G (a) + C =) C = G (a) :
x
Thus a f (t) dt = G (x) G (a) :
Rxp
eg2. g (x) = 0R 1 + t2 dt; g 0 (x) =?
x
eg3’. P (x) = 0 cos (t2 ) dt, P 0 (x) =?
R x4
d
sec tdt:
eg4. dx
1
3
Rx
cos(t2 )dt
#5 Final Spring 2008. Find limx!0 0 x
:
R1 2
R3 x
eg6. 0 x dx: eg5. 1 e dx:
R6
eg7’ 1 1t dt:
eg8. Find the area under the cosine curve y = cos x between 0 and b with 0
=2:
R 6x t2 1
R 6x R 0 R 6x R 0
R 3x
d
Hint: HW2A #9 dx
dt =? : 3x = 3x + 0 ; 3x =
:
3x t2 +1
0
f:
b
Sec5.4R Inde…nite integrals
and netRchange
Rb
x
x
Note a f (t) dt or x f (t) dt =
f (t) dt only represent one anti-derivative of
b
R
R
Notation f (t) dt or f (x) dx represent all anti-derivatives.
Basic Table
Rb
Net Change The integral of a changing rate is the net change: a F 0 (t) dt =
F (b) F (a) ; i.e. FTC.
eg. F 0 (t) = V (t) velocity, displacement
F 0 (t) = A (t) acceleration, velocity change
growth rate of population, net change
F 0 (t) = dn
dt
dC(t)
0
F (t) = dt marginal cost, cost increase
rate change of concentration, concentration change (of the product
F 0 (t) = d[c](t)
dt
of chemical reaction)
R
R
eg0 Rx2 dx eg1 (10x4 + 2 sec2 x) dx
cos
d
eg2. sin
R2 2 3
eg4. 0 2x
6x + x23+1 dx
R 9 2t2 +t2 pt 1
dt
eg5. 1
t2
eg6. Velocity v (t) = t2 t 6 m/s. From t = 1 to 4: a. Find displacement. b.
Find distance.
Sec5.5.
The Rsubstitution rule:
anti-chain
R
R
R rule
Q. e2x dx; sin ( x) dx; 2xe2x dx; (x + 1)1000 dx
A. "Simplifying notation" Anti-chain rule
Z
Z
0
f (g (x)) g (x) dx = f (u) du
where u = g (x) ; du = g 0 (x) dx:R
Compare to chain rule. Let f (u) du = F (u) : Then
d
d
d
R = dx
F
(u) = F 0 (u) dx
u = f (g (x)) g 0 (x)
dx
R
d
d
L = dx
f (g (x)) g 0 (x) dx = f (g (x)) g 0 (x) :
dx
For de…nite integral
Z b
Z g(b)
u=g(x)
0
f (g (x)) g (x) dx =
f (u) du:
a
eg0.
R
2xe2x dx;
R
xe2x dx;
g(a)
R
e2x dx;
R
sin ( x) dx;
4
R
(x + 1)1000 dx
R 3
4
Rx
p cos (x + 2) dx
+ 1dx
R 2x
p x
dx
R p1 4x2 5
2
R 1+x x
tan xdx
R4p
eg7. 0 2x + 1dx
R2
eg8. 1 (3 dx5x)2
Re
eg9. 1 lnxx dx
2
2
2
Appl. Area enclosed by ellipse x2 + ay2 = 1; then xa2 + yb2 = 1:
q
p
R2 p
R2
R2
2
A = 2 b 2 1 xa2 dx = b 2 2 a1 a2 x2 dx = b a1 2 2 a2
ab:
p
R2
#73. 2 (x + 3) 4 x2 dx by geometric consideration.
Integral
by symmetry
Ra
Ra
R0
e
(x)
dx
=
2
e
(x)
dx
=
2
e (x) dx:
0
a
R aa
o (x) dx = 0:
a
eg1.
eg2.
eg3.
eg5.
eg6.
x2 dx = b a1 a2 =
Sec6.1 Area between curves
Rb
I slice along horizontal direction a T (x) B (x) dx
Rd
II slice along vertical direction c R (y) L (y) dy
Rb
Rd
I’/II’geometric area a jT (x) B (x)j dx; c jR (y) L (y)j dy:
eg1. Find the area of region bounded above by y = ex ; below by y = x; and
bounded on two sides by x = 0 and x = 1:
eg2. Find the area of the region enclosed y the parabola y = x2 and y = 2x x2 :
eg5. Find the area of the region bounded by the curves y = sin x; y = cos x; x = 0;
x = =2:
eg6. Find the area enclosed by the line y = x = 1 and the parabola y 2 = 2x + 6:
Sec6.2 Volume
Regular shape: cube, cylinder
Vol=base height
Irregular shape: ball, ellipsoid, cone, paraboloid, hyperboloid
Z b
2
2
vol =
Rout
Rinner
durotation
a
The dummy rules!
eg1. Volume of ball with radius r:
eg2. volume
of solid obtained by rotating about the x-axis the region under the
p
curve y = x from 0 to 1:
eg3. Find the volume of the solid obtained by rotating the region bounded by
y = x3 , y = 8; and x = 0 about the y-axis.
eg4. Volume of the solid obtained by rotating about x-axis the region, which is
bound by y = x and y = x2 :
5
eg6. Volume of the solid by rotating the region in eg4 about x = 1:
#67 Find the volume common to two spheres, each with radius r, if the center of
each sphere lies on the surface of the other sphere.
HW3B#5 Three regions from top to bottom
in the rectangle w/ vertices (0; 0),
p
A(1; 0) ; B (1; 3) ; and C (0; 3) ; cut by y = 3 4 x and y = 3x : R3 ; R2 ; R1: Volume R3
about AB; volume R2 about OA; volume R1 about OC:
Sec6.3 Volume by cylindrical shell
eg1. Volume of solid by rotating about y-axis the region enclosed by y = 2x2
and y = 0:
Washer method: slice along the rotating direction
Z ymax
x2L dy
x2R
vol =
x3
0
Q. ymax =? xR =? (y) ; xL =? (y)? Need to solve cubic equation y = 2x2 x3 for x:
Hard. Or impossible in other situations like y = 2x2 x7 ; y = x 2 sin x:
Abandon.
Shell method: slice away from rotating axis.
volume of “thin”shell=base height =base y = 2 xdx y =y 2 xdx
Note. base area= (x + dx)2
x2 = 2xdx + (dx)2
2 xdx: Another way
base area=base length width=2 x dx; or 2 (x + dx) dx 2 xdx:
vol =
Z
b
y (x) 2 xdx:
a
eg2. Find the volume of the solid obtained by rotating about y-axis the region
between y = x and y = x2 : (w.s.)
#63 #14 Find the volume of the solid obtained by rotating about x-axis the region
enclosed by x + y = 3 and x = 4 (y 1)2 . (s.)
HW3C.#10 Volume of solid torus (donut-shaped solid) with radii r and R.
Sec6.4 Work
work=force distance =F d (Nm) J Joul
Constant force
eg1. How much work is done in lifting 1:2kg box up to 0.7 m high from ground?
g 9:8m=s2 :
Varying force
eg2. A force of x2 + 2x lbs (weight unit now, not the mass unit anymore) acts on
the particle when it is x feet from the origin, How much work done in moving it from
x = 1 to x = 3?
eg3. A force of 40N is required to hold a spring that has been stretched from its
natural length 10cm to 15cm. How much work is done in stretching the spring from
15cm to 18cm?
eg4. A 200-lb cable is 100 ft long and hangs vertically from the top of a tall
building. How much work is required to pull the cable to the top of the building?
6
1st way. Constant force/varying distance 2nd way. Varying force/“constant”
distance
eg5. A tank has the shape of an inverted circular cone with height 10m and base
radius 4m: It is …lled with water a height of 8m. Find the work required to empty
the tank by pumping all of the water that from the top of the tank. Density of the
water is 1000kg/m3 :
Midterm I review
Integrals: area, volume, distance, work, ...
theorem of calculus
RFundamental
b
* a f (t) dt = F (b) F (a) ; F 0 (t) = f (t) :
Rx
Rb
d
d
f
(t)
dt
=
f
(x)
;
f (t) dt = f (x) :
* dx
dx x
a
Anti + - chain rule, +, -, u-substitution
Other properties: comparison, symmetry
Applications.
Rb
Rb
area= a jf (s) g (s)j ds; net area= a [f (s) g (s)] ds
Rb
2
2
volume= a ( Rout
Rin
) durotation washer way,
Rb
volume= a height 2 Radius dthickness shell way
Rt
Rt
distance= t12 jV (t)j dt; net distance= t12 V (t) dt:
Sec5.3 HW2A#9, sec5.4 eg6, sec5.5 eg6/9, sec6.1 eg5, sec6.2 eg4, sec6.3 eg2.
Sec6.5 Average of a function
Physics: average velocity
R t2
V (t) dt
(net) distance
= t1
:
Vave =
time
t2 t1
Geometry. average height
have
(net) area
=
=
base
Abstract.
R x2
x1
h (x) dx
x2
x1
:
Rb
f (x) dx
:
b a
Mean value theorem for integrals: If f is continuous on [a; b] ; then there exists
Rb
c 2 [a; b] s.t. f (c) = fave or a f (x) dx = f (c) (b a) :
Rb
Analytic proof. m (b a)
f (x) dx M (b a) ; then
a
fave =
min
Rb
a
a
f (x) dx
b a
max
=fave
As f is continuous, then there exists c s.t. f (c) = fave 2 [min; max] :
eg0. V (t) = 2t2 t3 ; t 2 [0; 2] ; …nd average velocity.
eg1. Find the average of f (x) = 1 + x2 on [ 1; 2] :
7
Sec7.1 Integration by parts
d
dx
R
[f (x) g (x)] = f (x) g (x) + f (x) g (x) 99K f g =
Z
Z
0
f g dx = f g = f 0 gdx
0
0
R
f 0 gdx +
R
f g 0 dx or
that is partilaly integrate g 0 :
Another notation: f = u; g = v; df = f 0 dx; dg = g 0 dx
Z
Z
udv = uv
vdu:
The hard part is to write the integral
as a product of one function (f) and the derivative of another (g), such that
the product of the derivative of the …rst (f) and the second (g) is easier to
integrate.R
R
eg1. R xex dx; eg2 x sin xdx
eg3. t2 et dt
R
R1
eg4. R ln xdx; eg5 0 arctan xdx
eg6. R ex sin xdxR
eg7. sin2 xdx; sin3 xdx
D
+ t2
2t
+ 2
0
t2 et
D
+ t2
2t
+ 2
0
I
et
et
et
et
2tet + 2et t2 ( cos t)
I
sin t
cos t
sin t
cos t
2t ( sin t) + 2 cos t
Sec7.2
Trigonometric integrals
R
m
sin x cosn xdx
split odd powers
Z
Z
Z
k
m
2k+1
m
2
sin cos
xdx = sin cos x cos xdx = sinm 1
Z
k
u=sin x
=
um 1 u2 du
sin2 x
k
cos xdx
du=cos xdx
half even powers
Z
Z
1
2
sin xdx =
Z
Z
cos 2x
dx;
2
Z
2
cos xdx =
Z
1 + cos 2x
dx
2
Z
sec x (sec x + tan x)
1
sec xdx =
dx =
d (sec x + tan x) = ln jsec x + tan xj + C
sec x + tan x
sec x + tan x
Z
Z
Z
csc x (csc x + cot x)
1
csc xdx =
dx =
d (csc x + cot x) = ln jcsc x + cot xj + C
csc x + cot x
csc x + cot x
8
Z
Z
1
[sin (Ax Bx) + sin (Ax + Bx)] dx
2
Z
Z
1
[cos (Ax Bx) cos (Ax + Bx)] dx
sin Ax sin Bxdx =
2
Z
Z
1
cos Ax cos Bxdx =
[cos (Ax Bx) + cos (Ax + Bx)] dx
2
R
R
I eg1.R sin3 xdx; eg2. sin5 x cos2 xdx
2
eg3. sin
R xdx R
R
2
II eg4.
tan
xdx;
tan
xdx;
tan3 xdx
R
R
R
eg5. sec xdx; sec2 xdx; sec3 xdx
sec2 x 1
R
R
R
0
3
sec
xdx
=
sec
x
(tan
x)
dx
=
sec
x
tan
x
sec
x
tan
xdx = sec x tan x
R
R
R x tan
3
sec3 xdx
+
sec
xdx
=
sec
x
tan
x
+
ln
jsec
x
+
tan
xj
sec
xdx:
R
#27R tan5 x sec4 xdx
III sin 3x cos 5xdx
sin Ax cos Bxdx =
Sec7.3
R p Trig substitutions— beating square roots
1 x2 dx area under
R circle
R
x = sin ; dx = cos d ; cos cos d = 12 (1 + cos 2 ) d = 12 + 41 sin 2 + C =
p
1
arcsin x + 12 x 1 x2 + C:
2
R Rp
2
x2 dx; x = a sin
Rpa
1 + x2 dx area under hyperbola
R
R
x = tan ; dx = sec2 d ; sec sec2 d = sec3 d =
= 21 sec tan +
p
p
1
ln jsec + tan j + C = 21 1 + x2 x + ln 1 + x2 + x + C:
2
Rp
a2 + x2 dx; x = a tan
Rp
x2 1dx area under vertical
R hyperbola
R
x = sec ; dx = sec tan d ; tan sec tan d = p
(sec3
sec ) dp =
=
1
1
1
1
1
2
2
sec tan + 2 ln jsec + tan j 2 ln jsec + tan j+C = 2 1 + x x 2 ln 1 + x + x +
2
C: R p
x2 a2 dx; x = a sec
p
R6
Ra
R
x
HW5c #2 3p2 t3 p1t2 9 dt; #3 0 5x2 a2 x2 dx; #4 px2 +x+2
dx
Sec7.4 Partial fractions
Fundamental Theorem of Algebra
Q (x) = xn + an 1 xn 1 +
= (x r1 ) (x r2 )
(x
where (x
s1 )2 + t21 = x2
s1 )2 + t21
(x
2s1 x + s21 + t21 = x2
9
+ a1 x + a0
s2 )2 + t22
2s1 x + t~21 ; irreducible.
Consequence
P (x)
bn 1 x n 1 +
+ b1 x + b0
= n
n
1
Q (x)
x + an 1 x
+
+ a1 x + a0
A2
A1
+
+
=
x r1 x r2
B1 x + C1
B2 x + C2
+
+
+
2
2
(x s1 ) + t1 (x s2 )2 + t22
eg.
x3 x + 1
A B
D
E
C
= + 2+
+
+
3
2
x x
x 1 (x 1)
x2 (x 1)
(x 1)3
x4
x3 + 2x 1
A Bx + C
Dx + E
= + 2
+
2
2
x
x +1
x (x + 1)
(x2 + 1)2
Recall
R 1
R 1
dx
=
ln
jxj
+
c;
dx = ln jx 1j + c
x 1R
R x1
R 1 1
1
dx =
dx
=
arctan
x
+
c;
dx
=
2
2
2
x +1
x +a
a2 ( x )2 +1
a
c:
R
eg0. x21 1 dx
R 3
R x3 +x+1
eg1. xx +x
dx;
eg1’
dx
x2 1
R x21+2x 1
eg2. 2x3 +3x2 2x dx
R
eg4.
R
eg5.
R
eg9.
x4 2x2 +4x+1
dx
x3 x2 x+1
2
2x x+4
dx; eg6.
3
px +4x
x+4
dx
x
R
1
a
R
1
x 2
+1
a
( )
4x2 3x+2
dx
4x2 4x+3
Sec7.5 Strategy for integration
Basic table
"
Rp
p
x (1 + x) Rdx
1. Simplify eg.
R
2. "Obvious" substitution eg. x2x+1 dx; x (x + 1)2008 dx
3.
R
R
3
Trig fcn eg. R sin
xdx;
sin 5x
8x dx
p
p
R cos
n
2
2 dx;
a
x
ax
+ bdx
Radicals eg.
R x+5
RationalRfcn eg. R x2 +x 2 dx R
IBP eg. xex dx x sin xdx; ln xdx
4. Try again.
Sec7.7 Approximation of integrals
*R Most functions have no explicit anti-derivatives
x2
(probability,
statistic)
R e dx
R
R
p
2
2
sin (x ) dx; cos (x ) dx (compare w/ sin ( x) dx doable)
p
R
1 + sin2 d (elliptic integrals, length of ellipse)
10
d xa =
1
a
arctan xa +
* Integrand only has sample data in experiments, reality
What to do?
Rx
Rx
Give a new name, eg. 1 1t dt = ln x; eg. 2 ln1 t dt = Li (x)
Approximate
Linear approximation: left endpoint rule, right endpoint rule, midpoint rule,
trapzoidal rule.
Nonlinear approximation: Simpson’s rule
Left endpoint rule: 4xf (xi 1 ) 4x [f (x0 ) + f (x1 ) +
+ f (xn 1 )]
Right endpoint rule: 4xf (xi ) 4 x [f (x1 ) + f (x2 ) +
+ f (xn )]
x1 +x2
1
+
f
+
+ f xn 12+xn i
Midpoint rule: 4xf xi 12+xi 4 x hf x0 +x
2
2
4x
2
Trapezoidal rule: 4x f (xi
[f (x0 ) + 2f (x1 ) +
1 )+f (xi )
2
f (x0 )+f (x1 )
2
4x
+
f (x1 )+f (x2 )
2
+
+
f (xn
1 )+f (xn )
2
+ 2f (xn 1 ) + f (xn )]
Simpson’s rule-parabolic approximation
Approximating parabola y = At2 + Bt + C through three points, ( h; y0 ) ; (0; y1 ) ;
(h; y2 ) ; y0 = f (x0 ) ; y1 = f (x1 ) ; y2 = f (x2 )
To determine A; B; C:
At2 + Bt + C
h;0;h
= y0 ; y1 ; y2
8
< Ah2 Bh + C = y0 (1)
C = y1
(2)
Solve for A=B=C in terms of y0 ; y1 ; y2 :
:
2
Ah + Bh + C = y2 (3)
Rh
h3 + 2Ch = h3 (2Ah2 + 6C) =
Approximate area= h (At2 + Bt + C) dt = 2A
3
h
(y0 + 4y1 + y2 ) ; since
3
(1)+(3) ) 2Ah2 + 2C = y0 + y2
4 (2) ) 4C = 4y1
) 2Ah2 + 6C = y0 + 4y1 + y2 :
OR solve for A directly, and plug in.
x0 ; x1 ; x2 h3 (y0 + 4y1 + y2 )
x2 ; x3 ; x4 h3 (y2 + 4y3 + y4 )
x4 ; x5 ; x6 h3 (y4 + 4y5 + y6 )
x2m 2 ; x2m 1 ; x2m
h
3
h
3
(y2m
2
+ 4y2m
1
+ y2m )
(y0 + 4y1 + 2y2 + 4y3 + 2y4 + 4y5 + 2y6 +
+ 2y2m
2
+ 4y2m
1
+ y2m )
R1 2
eg7. Use Simpson’s rule to approximate 0 ex dx w/ n = 10:
HW6C#4
The width (in meters) of a kidney-shaped swimming pool were measured at 2
meter intervals as indicated in the …gure. Use Simpson’s rule to estimate the area of
the pool.
Sec7.8 Improper integrals
R1
Q. Area under the whole hyperbola? 0 x1 dx
11
=
A problem:
integrand unbounded II
integral interval unbounded
I
R1 1
R1
R1
2
1
More examples: 1 x2 dx; 1 1+x
e x dx
2 dx;
1
Improper integrals
R1 Rb
R1
I Integral intervals unbounded: a ;
;
1
1
R1
R1
II Integrand unbounded: 0 x1 dx; 0 p1x dx:
Formal de…nition
Type I
Rb
R1
Rb
If limb!1 a f (x) dx exists, then a f (x) dx = limb!1 a f (x) dx and we say
the improper integral convergent.
Otherwise, the
R b improper integral is divergent.
Rb
Rb
If lima!1 1 f (x) dx exists, then 1 f (x) dx = lima! 1 a f (x) dx and we
say the improper integral convergent.
Otherwise,
the improper
Rc
R 1 integral is divergent.
R1
Rc
If 1 f (x) dx and c f (x) dx are both convergent, then 1 f (x) dx = 1 f (x) dx+
R1
f (x) dx; and we say the improper integral convergent.
c
Otherwise, the improper integral is divergent.
Type II Improper integrals
Rb
Suppose f (x) continuous on (a; b] and f (x) unbounded near a: Then a f (x) dx =
Rb
limt!a+ t f (x) dx; if the limit exists, as a …nite number. We say the improper integral convergent. If the limit doesn’t exist, the improper integral divergent.
Rb
Suppose f (x) continuous on [a; b) and f (x) unbounded near b: Then a f (x) dx =
Rt
limt!b a f (x) dx; if the limit exists, as a …nite number. We say the improper integral
convergent. If the limit doesn’t exist, the improper integral divergent.
Suppose f (x) continuous on [a; b] except c and f (x) unbounded near c: If
Rc
Rb
Rb
both a f (x) dx and c f (x) dx are convergent, we say a f (x) dx convergent and
Rb Rc Rb
Rb
= a+
a
R 1c 1: Otherwise, a f (x) dx is divergent.
eg1. 1 x dx convergent or divergent
R0
eg2. Evaluate 1 xex dx
R1 1
eg3. Evaluate 1 1+x
2 dx
R1
eg4. For what value p the integral 1 x1p dx convergent?
R5
eg5. Find 2 px1 2 dx
R =2
eg6. 0 sec xdx convergent or divergent?
R3
eg7. 0 x 1 1 dx; evaluate if possible.
R1
eg8. 0 ln xdx
1
R1
R1
R1
4x
2
2
eg9. Show that 0 e x dx is convergent. eg9’ 4 e x dx < 4 e 4x dx = e4
=
4
1
e 16
4
< 0:0000001
R1
2
Q. 1 e x dx convergent or divergent? Compare w/ model, easier things.
12
RMK.
e
r2
1
0
R1
e
1
x2
= : Thus
2
dx
=
RR
R2
e
Z
x2 y 2
1
e
dxdy =
x2
dx =
R2 R1
p
0
0
e
r2
rdrd = 2
R1
0
e
r2
rdr =
:
1
Comparison
Test: Suppose 0 g (x)
R1
R 1 f (x) for a x < 1; then
If Ra f (x) dx convergent, thenR a g (x) dx convergent.
1
1
If a g (x) dx divergent, then a f (x) dx divergent.
Sec 8.1 Arclength
Q. Circumference of circle, ellipse, arclength of other curves?
A. 1st way, straighten out
B. Cut into pieces, each approximate each piece by a straight segment, add, then
take limit.
eg. circumscribed polygons for circle.
Calculus
opp
dx
opp
adj
q
q
(dx)2 + [f 0 (x) dx]2 = 1 + (f 0 )2 dx
= tan = f 0 (x) ) opp = f 0 (x) dx
Rbp
Length of curve = a 1 + f 02 dx
Rb
RMK. Compare w/ area a f (x) dx; 0th order derivative f nice, eg. cont. Now
Rbp
arclength a 1 + f 02 dx; 1st order derivative f 0 nice, eg. f 0 cont.
Other notation/way
s
s
Z b
Z d
Z bq
2
2
dx
dy
2
2
L=
1+
dx =
(dx) + (dy) dx =
+ 1dy:
dx
dy
a
c
a
=
eg0. Arclength of y = x for x 2 [0; 1] :
eg0.5 Cirumference of x2 + y 2 = 1:
5
1
x 2:
eg1. Sec8.1 #9 arclength of y = x6 + 10x
3; 1
eg2. Find the arclength of parabola y 2 = x from (0; 0) to (1; 1) :
Review for Midterm II
More applications: work, average, arclength
(after the midterm, moments, center of mass ...)
Techniques for integrations
IBP
Trig integrals
Trig sub
Partial fractions
Rationalizing sub
13
Approximation
linear: left, right, midpoint, trapezoid
nonlinear: Simpson’s rule
R 1 (parabolic)
Improper integrals, eg 0 x1p dx:
Sec8.3 Center of mass
Q. Where to sit on a seesaw?
You, fulcrum, your younger sister
Math/physics
1-d: m1 x1 + m2 x2 = 0
moment=m1 x1 + m2 x2 + m3 x3
2 x2 +m3 x3
=x
Average of moment=center of mass= m1 xm1 +m
1 +m2 +m3
2-d:
x-moment m1 x1 + m2 x2 + m3 x3 = Mx (My in book)
y-moment m1 y1 + m2 y2 + m3 y3 = My (Mx in book)
centroid (x; y)
Mx
m1 x1 + m2 x2 + m3 x3
=
m1 + m2 + m3
m1 + m2 + m3
My
m1 y1 + m2 y2 + m3 y3
=
y=
m1 + m2 + m3
m1 + m2 + m3
x=
3-d?
Continuous version
B
eg. straight vertical bar x = x; y = yT +y
2
mass= (yT yB )
Ry
x-moment Mx = x mass = x (yT yB ) = yBT x dy
Ry
B
y-moment My = y mass = yT +y
(yT yB ) = 21 (yT2 yB2 ) = yBT y dy
2
R
eg’straight horizontal bar x = xL +x
;y=y
2
mass= (xT xB )
R xR
1
2
2
L
(x
x
)
=
(x
x
)
=
x dx
x-moment Mx = x mass = xR +x
R
L
L
2
xL
R xR 2 R
y-moment My = y mass = y (xR xL ) = xL y dx
Centroid of plate with mass density kg/m2
Rb
Rb
m = a (yT yB ) dx = a [f (x) g (x)] dx
Rb
Rb
Horizontal moment Mx = a x (yT yB ) dx = a x [f (x) g (x)] dx
Rb
Rb 1
B
Vertical moment My = a yT +y
(y
y
)
dx
=
[f 2 (x) g 2 (x)] dx
T
B
2
a 2
Rb
x
Mx
x=
= Ra b
m
a
Rb 1
My
y=
= aR2b
m
[f (x)
[f (x)
[f 2 (x)
g (x)] dx
g (x)] dx
g 2 (x)] dx
g (x)] dx
Rc
RMK. Natural way for My = 0 y (xR
to …nd xR and xL in terms of y:
a
[f (x)
14
Rb
x [f (x) g (x)] dx
Rb
[f (x) g (x)] dx
a
Rb 1 2
g 2 (x)] dx
constant a 2 [f (x)
=
Rb
[f (x) g (x)] dx
a
constant
=
a
xL ) dy: But in general, could be hard
Rb
Rb
RMK. 2 x Aear = 2 x a [f (x) g (x)] dx = a 2 x [f (x)
of A around y-axis. This is Theorem of Pappus in 400sAD.
g (x)] dx =volume
eg4. Find the center of semicircle plate of radius r:
Sec8.3 #29? Find the centroid of the region bounded by y = x2 and x = y 2 :
1
1
; 0
x < 1: Assume
HW8 #7 A plate bounded by y = 1+x
2 and y =
1+x2
density is 1. Find the mass. Where is the centroid?
or
Sec 9.1 Modeling with di¤erential equations
Nature
Newton’s second law
d2 E
E
mE 2 = GmE mS
dt
jEj3
d2 E
E
=
2
dt
jEj3
it is hard but possible to solve.
Society
Population P (t)
dP
P (t)
dt
dP
= kP
dt
P (0) known
Solve P (t) = Cekt
P (0) = Cek 0 = C: So P (t) = P (0) ekt :
Eg. P (2012) = 10million e0:02 2012 > 1025 = 10 1m 1b 1b; impossible.
> 0; P % initially, then & after certain threshold population K, eg. due
Flaw dP
dt
to limited resources.
dP
kP (K P ) ; (+; ; ; )
dt
dP
P
~ 1 P :
= kP (K P ) = kKP 1 K
= kP
dt
K
We’re able to solve it and will do it. The graph of solutions are:
K equilibrium
Di¤erential equations
E 00 = E= jEj3 2nd order di¤erential equ
P 0 = kP 1st order di¤erential equation
Compare x2 3x + 2 = quadratic, algebraic equ
eg1. Given a …rst order di¤erential equation with initial condition y 0 + y tan x =
cos2 x; y (0) = 1: Verify y = sin x cos x cos x is a solution.
#3. For what values r does the function y = erx satisfy the di¤erential equation
00
2y + y 0 y = 0?
#14. Co¤ee at 950 C in a room of 200 C:
a) when the co¤ee cools most quickly? What happens to rate of cooling at time
goes by?
b) Newton’s law’s of cooling: rate of cooling is proportional to the temperature
di¤erence. Write down a di¤erential equ for the temperature of co¤ee. What is the
initial temperature?
15
c) Make a rough sketch of the graph of solution to b):
Sec 9.3 Separable equations & Sec 9.4 Models for population growth
P
= 0:08P 1 1000
eg0’(HW9A #6, Sec9.4 eg1) dP
dt
2
dy
eg1. Solve dx
= xy2 : Find the solution with y (0) = 2:
= 2 + 2u + t + tu
#9 du
dt
eg0’Find P (40) ; P (80) ; given P (0) = 100: (Sec9.4 eg2.) At what time does the
population reach 900?
#41 A tank contains 1000L of brine w/ 15kg of dissolved salt. Pure water enters
the tank at rate of 10L/min. The solution is kept thoroughly mixed and drains from
the tank at the same rate. How much salt is in the tank a) after t minutes; b) after
20 minutes.
HW9B #4
du
dt
=
2t+sec2 t
;
2u
u (0) =
5:
Sec3.8 #13. A roasted turkey is taken from an oven when its temperature has
reached 1850 F and is placed on a table in a room where the temperature is 750 F: a)
If the temperature of the turkey is 1500 F after half an hour, what is the temperature
after 45minutes. b) When will the turkey have cooled to 1000 F ?
Review for Final Saturday Dec 8 1:30–4:20 KNE 120
Integrals: area, volume, length, distance, work,
Theorem of Calculus
RFundamental
b
* a f (t) dt = F (b) F (a)
Rb
Rb
d
d
0
f
(t)
dt
=
f
(b)
=
F
(b)
;
f (t) dt = f (a) =
* db
da a
a
Rules: anti + - (No ), & chain
Techniques
* u-substitution
R x
* IBP
xe dxR
2
* Trig integrals
R p sin d , area of ellipse
x2 a2 dx
* Trig sub
R x
* P. F, x2 1 dx
R pt+1
* Rationalizing sub
dt
t
F 0 (a) :
** Approximation eg. trapezoidal rule, Simpson’s rule
** Improper integrals
Applications:
* area, volume length, distance, work, center of mass, average
* di¤erential equations, population, temperature, concentration,
Preview for M126 Calculus III
one variable
+ n1 +
series 1 + 21 + 13 + 14 +
(= 1)
16
1 + 212 + 312 + 412 +
+ n12 +
(= 2 =6)
3
4
n
2
+ xn! +
ex = 1 + x + x2 + x3! + x4! +
arctan x = x 13 x3 + 15 x5 71 x7 +
; eg.
multiple variables
algebra of vectors
vector functions
partial derivatives
multiple integrals
4
=1
1
3
+
1
5
1
7
+
E
eg. Solve E• = jEj
3 ; E = (x; y; z) orbit of the Earth around the Sun.
R1
p
x2
eg. 1 e dx =
:
17