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Integral(0,pi,cos(2x),dx) = ???

Integral(0,pi,cos(2x),dx) = 0

hint: [ sin(2x)/2 ]' = cos(2x)

hint: [ sin(2x)/2 ]' = cos(2x)

Integral(0,pi,sin(x)sin(x),dx) = ???

Integral(0,pi,sin(x)sin(x),dx) = pi/2

There are at least two ways:

(1) integrate by parts: page 87 in OLDTIMER

(2) use: 2sin(x)sin(x) = 1-cos(2x), and then Integral(0,pi,cos(2x),dx)=0

There are at least two ways:

(1) integrate by parts: page 87 in OLDTIMER

(2) use: 2sin(x)sin(x) = 1-cos(2x), and then Integral(0,pi,cos(2x),dx)=0

Integral(-1,1,sqrt(1-x^2),dx) = ???

Integral(-1,1,sqrt(1-x^2),dx) = pi/2

hint: x=cos(t), later use: Integral(0,pi,sin(x)sin(x),dx) = pi/2

Notice the significance of this integral.

It calculates half the area of the unit circle, which is pi.

hint: x=cos(t), later use: Integral(0,pi,sin(x)sin(x),dx) = pi/2

Notice the significance of this integral.

It calculates half the area of the unit circle, which is pi.

Integral(0,pi,cos(x)cos(x),dx) = ???

Integral(0,pi,cos(x)cos(x),dx) = pi/2

Use:

(1) cos(x)cos(x) = 1 - sin(x)sin(x)

(2) Integral(0,pi,sin(x)sin(x),dx) = pi/2

Or you can integrate by parts: page 87 in OLDTIMER

Use:

(1) cos(x)cos(x) = 1 - sin(x)sin(x)

(2) Integral(0,pi,sin(x)sin(x),dx) = pi/2

Or you can integrate by parts: page 87 in OLDTIMER

Let F be an algerba in X.

Let f : X -> |R\{oo,-oo} be F-measurable.

What can we conclude about function [f] ?

([x] denotes the integer part of real number x)

Let f : X -> |R\{oo,-oo} be F-measurable.

What can we conclude about function [f] ?

([x] denotes the integer part of real number x)

g:|R->|R, g(x)=[x] is an increasing function,

hence gof:X->|R is measurable.

page 91,90 in 2nd measure

hence gof:X->|R is measurable.

page 91,90 in 2nd measure

Let (X,d) be a metric space.

Let A be an uncountable subset of X satisfying:

(1) \/(a>0) /\(x,y:-A) x!=y => d(x,y)>=a

What can we conclude?

Let A be an uncountable subset of X satisfying:

(1) \/(a>0) /\(x,y:-A) x!=y => d(x,y)>=a

What can we conclude?

X does not contain a countable dense subset.

Every dense subset of X is uncountable.

Every countable subset of X is not dense.

page 193 in OLDTIMER

Every dense subset of X is uncountable.

Every countable subset of X is not dense.

page 193 in OLDTIMER

(1) f : |R x |R -> |R

(2) /\x,y:-|R f(y,x) = f(x,y)

(3) a,b :- |R

(4) /\x,y:-|R f(x,y) <= f(a,b)

Can we conclude that a=b ?

(2) /\x,y:-|R f(y,x) = f(x,y)

(3) a,b :- |R

(4) /\x,y:-|R f(x,y) <= f(a,b)

Can we conclude that a=b ?

NO.

Let f(1,0)=f(0,1)=7 and otherwise: f(x,y)=0.

Let a=1, b=0.

Let f(1,0)=f(0,1)=7 and otherwise: f(x,y)=0.

Let a=1, b=0.

(1) f : |R x |R -> |R

(2) /\x,y:-|R f(y,x) = f(x,y)

(3) a,b :- |R

(4) /\x,y:-|R (x,y)!=(a,b) ==> f(x,y) < f(a,b)

Can we conclude that a=b ?

(2) /\x,y:-|R f(y,x) = f(x,y)

(3) a,b :- |R

(4) /\x,y:-|R (x,y)!=(a,b) ==> f(x,y) < f(a,b)

Can we conclude that a=b ?

YES.

Transform (4) into (5) /\x,y:-|R f(x,y) >= f(a,b) ==> (x,y)=(a,b).

Then by (2), f(b,a)=f(a,b), and f(b,a)>=f(a,b).

By (5), (b,a)=(a,b). Hence a=b.

Transform (4) into (5) /\x,y:-|R f(x,y) >= f(a,b) ==> (x,y)=(a,b).

Then by (2), f(b,a)=f(a,b), and f(b,a)>=f(a,b).

By (5), (b,a)=(a,b). Hence a=b.

Draw the set { (x,y):-|R^2 : x^2 + y^2 <= x and x^2 + y^2 <=
y }.

It is the intersection of two closed disks.

(1) center (0 , 1/2) radius 1/2

(2) center (1/2 , 0) radius 1/2

(1) center (0 , 1/2) radius 1/2

(2) center (1/2 , 0) radius 1/2

Let (x,y):-|R^2 satisfy x^2 + y^2 = y and x>=0.

How can we express the distance of point between (x,y) and (0,0)?

How can we express the distance of point between (x,y) and (0,0)?

There are two ways:

1) sqrt(y)

2) sin(a), where a = arctan(y/x) = sin( arg(x+iy) )

page 91 in OLDTIMER

1) sqrt(y)

2) sin(a), where a = arctan(y/x) = sin( arg(x+iy) )

page 91 in OLDTIMER