Integration by Special Properties (1 Viewer)

[OLDMAN]

I{a-->b}f(x)dx integral, upper bound b, lower bound a integrate
f(x) w.r.t. x.

Must give Spice Girl's Integration notation above a road test.

Using sin2x=2sinxcosx or otherwise,
find I{pi/2-->0} Ln(sinx)dx

 

[Affinity]

Interesting as well. Is there another method besides using the substitution given?

 

[underthesun]

just wondering if that is 0 --> pi/2 or, like stated above, pi/2 --> 0

 

[OLDMAN]

sinx=2sin(x/2)cos(x/2)
Sorry, can't think of another that is materially different.

 

[underthesun]

While at integration, i would also want to ask this question:

in integration, is there such law that

I{a --> b} f(x) dx = - I{b --> a} f(x) dx ??

i just happened to use it a lot without knowing if it's a law or not. Would be scary if i get reduced in marks for using something that's not a law or so..

 

[aspect]

wouldnt call it a law but its usable enough that its one of the things we were told we could use in uni

guess its like counting the area backwards so its negative

 

[underthesun]

my teacher very³ strict, so I have to be careful³. He marks very³ hard, so i have to take extra³ caution..

 

[Affinity]

under the sun: hmm challenge your teacher

I{a --> b} f(x) dx = - I{b --> a} f(x) dx
is right by definition of the definite integral

 

[underthesun]

yah, will ask that first thing tomorrow morning

gotta show innocence³

 

[OLDMAN]

Other Special Properties:
1. I{0 -->a} f(x) dx = I{0 --> a} f(a-x) dx
2. I{-a -->a} f(x) dx = 2I{0 --> a} f(x) dx when f(x) is even
3. I{-a -->a} f(x) dx =0 when f(x) is odd.

 

[Harimau]

OLDMAN, how would you prove the first property?

 

[wogboy]

Simply use the substitution u = a - x

du/dx = -1
dx = -du

therefore,
I{0 --> a}f(a-x) dx
= I{a --> 0}f(u) -du
= -I{a --> 0} f(u) du
= I{0 --> a} f(u) du

now you can use the substitution u=x (which is basically changing that u to an x) to get:

= I{0 --> a} f(x) dx

Hence
I{0 --> a}f(a-x) dx = = I{0 --> a} f(x) dx

 

[OLDMAN]

Thanks wogboy.

 

[ND]

Originally posted by OLDMAN

Using sin2x=2sinxcosx or otherwise,
find I{pi/2-->0} Ln(sinx)dx

I'm stumped. I had a go this morning, but couldn't do it so i asked my teacher today, he didn't know either.

 

[Affinity]

sin(x) = sin(2x)/[2cos(x)]
substitute and work on from there.
remember your log rules, and sketching some graphs might help you realise somethings, good luck

 

[ND]

I didn't think of that, thanks.

 

[freaking_out]

ru sure that this question is in the course???

edit: if so can someone pls. post the solutions up for us?

 

[Affinity]

I think this is a valid question to be in the papers ... *cries* my HSC is coming up

Using sin2x=2sinxcosx or otherwise,
find I{pi/2-->0} Ln(sinx)dx

just an outline:

let I{pi/2-->0} Ln(sinx)dx = K


K

= I{pi/2-->0} Ln[Sin(2x)/2Cos(x)]dx by the substitution

= I{pi/2-->0} Ln[Sin(2x)] - Ln[Cos(x)] - Ln[2] dx

= I{pi/2-->0} Ln[Sin(2x)] dx - I{pi/2-->0} Ln[Cos(x)] dx - I{pi/2-->0} Ln[2] dx


now substitute u = 2x for the first integral, notice the second integral is equal to K (you might want to prove this by using Cos[x]= Sin[Pi/2 - x] ) and integrate the last integral,

K = (1/2)*I{Pi-->0}Ln[Sin(u)] du - K + Pi*Ln[2]/2

notice the integral in the line above = 2K ( Prove it in an exam)
K = (1/2)*2K - K + Pi*Ln[2]/2

K = Pi*Ln[2]/2

 

[OLDMAN]

Yes!
Perhaps another ND in the making.

 

[OLDMAN]

Thought I typed in "MD". But in this context MD, ND should be equivalent.