Am i doing this question correctly.

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Homework Statement



The integral of [x]+ln( (1+x)/(1-x) ) (lower limit -0.5 and upper limit is 0.5.)
here [x] show greater integer function means greatest integer less than equal to x. ex [2.3]=0.3,[1.7]=0.7,[-1.4]=0.6 etc.
//sorry that i don't know how to write it in better way

Homework Equations



millions so i can't write all of them here two of them is
integral of f(x) from a to b =integral of f(a+b-x) from a to b.
[1-x]=1-[x]
rest are in your mind.

The Attempt at a Solution



replace x by lower limit+ upper limit-x.
function will changed into [1-x]+ln( (1-x)/(1+x) )
replace [1-x] with 1-[x]. after little modification it will something like this I=integral with limits(1) -I
here I is the integral whose value i want to know.
finally the answer is 1/2 but this is wrong. correct answer as given in book is -1/2. My question is why my answer is wrong?
thanks for any kind of help.

vikash chandola
suport anna hazare(only for indians)
 
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You have the greatest integer function wrong. example: [2.3]=2
As for the rest of the problem, I'm not sure how to do it.
 
\displaystyle \int_{-0.5}^{0.5}\lfloor x \rfloor dx should be a piece of cake.
For -1 ≤ x < 0 , \displaystyle \lfloor x \rfloor = -1.

For 0 ≤ x < 1 , \displaystyle \lfloor x \rfloor = 0.​

For \displaystyle \ln\left(\frac{1+x}{1-x}\right)\,, use the rule for the log of a quotient. Then break-up the integral accordingly.

Do you know the anti-derivative of ln(x) ?
 
ArcanaNoir said:
You have the greatest integer function wrong. example: [2.3]=2
As for the rest of the problem, I'm not sure how to do it.

Oh man. what a foolish work i did. I understand greater integer function as fraction part function. I also applly wrong property that's why i got wrong answer.After thanks you tel me my mistake.Hey sammy what are you writing. this is seeming lengthy method. see how i solve it. thing i do wrong is understand greatest integer function as fraction part function. So the property [-x]=1-[x] is wrong. Correct one is [-x]=-1-[x]. now i will get correct answer.

anti derivative of ln(x) is xln(x)-x+C
 
The problem has [x]. Why are you working with [-x] ?
 
SammyS said:
The problem has [x]. Why are you working with [-x] ?

oh one more sorry. Actually the place where i write [1-x] that is [-x] =-1-[x].
I do this mistake however my trick is correct. thanks for showing me my mistake. If you still say how yo write [-x] then something is wrong. either i am doing one more mistake or you do something wrong.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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