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wasiqss
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evaluate diya LOl
Diyadiya
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Equation editor is not working ?
wasiqss
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it is !
Diyadiya
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Okay integrate -1 to 3/2 |x sin(pi x)| dx
wasiqss
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ok i will
Diyadiya
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tell me how to start
Diyadiya
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Equation editor is not working :/
wasiqss
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ok diya do it by parts i did it
Diyadiya
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3
but there is modulus
wasiqss
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ohh dint c the modulus
wasiqss
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w8 tellling u
Diyadiya
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3
okay
wasiqss
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plz show us how to remove modulus
Diyadiya
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3
full solution is there lol
Diyadiya
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i just wanna know that starting part
razor99
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ummm guyzzz
Mertsj
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f=x, f'=dx g'=sin pix
g=-1/picospix
Diyadiya
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Hm What ? sorry can you explain
Mertsj
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Do you use u and v or do you use f(x) and g(x) when you integrate by parts.?
Diyadiya
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for parts f(x) and g(x)
Mertsj
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\[f(x)=x\]
\[f'(x)=dx\]
\[g'(x)=\sin \pi x\]
\[g(x)=-\frac{1}{\pi}\cos \pi \]
wasiqss
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bt man how to remove modulus
Diyadiya
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Equation editor isn't working
Diyadiya
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Can you see the post @wasiqss
Mertsj
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I thought you wanted to integrate the function. Didn't know you were trying to remove the modulus.
Diyadiya
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Mertsj i cant see what you posted ,it just shows [maths processing error ]
wasiqss
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diya i can see tho :P
Diyadiya
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Wait let me refresh
wasiqss
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mertsj jus remove the modulus first then i will integrate
Mertsj
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|dw:1330277056196:dw|
Mertsj
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Is that the problem or not?
wasiqss
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yes
Mertsj
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Then integrate it using the substitutions I gave you.
Diyadiya
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|dw:1330277124894:dw|
wasiqss
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modulus is thea man
Diyadiya
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Mertsj
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f(x)=x
f'(x)=dx
g'(x)=sin pi x
g(x)= -1/pi cos pi x
Diyadiya
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So we dont have to remove modulus first ?
wasiqss
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yeh diya thats my question
Mertsj
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Why would you? Integrate the function.
wasiqss
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mertsj we have do something with modulus first
Mertsj
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Why?
Diyadiya
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This is the solution given in my book
wasiqss
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diya this is techniqu for removing modulus that u need to learn from someone cox im not able to recall it
Mertsj
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Right. Since it's absolute value there are actually two integrals to do.
Mertsj
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Where was the modulus removed in diyadiya's example post?
Mertsj
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Which step?
Zarkon
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find where x sin(pi x) is positive and negative
wasiqss
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as in how
Mr.Math
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Which part of the solution don't you understand Diya?
Diyadiya
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the beginning
Mr.Math
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Alright. You must know that the definition of absolute value, that's |x|=x, if x>=0 and -x if x<0. So we have to find when \(x\sin(\pi x)\) is positive and when it's negative.
Diyadiya
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MrMath i cant see Latex
Mr.Math
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Lets first consider the interval from (-1,0). x is obviously negative in this interval, and so is sin(pi*x) because sin(pi x) would lie either in the third or fourth quadrant for x in (-1,0). And hence xsin(pi*x) is the product of two negative number which is positve in (-1,0).
Following so far?
Mr.Math
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Alright. You must know that the definition of absolute value, that's |x|=x, if x>=0 and -x if x<0. So we have to find when xsin(πx) is positive and when it's negative.
Diyadiya
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Why is it in the 3rd or 4th Quadrant ?
Mr.Math
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Because when x in (-1,0), the angle, πx in (-π,0), right?
Diyadiya
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yeah
Mr.Math
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So you know why sin(πx) would be negative in this interval?
Diyadiya
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okay sine is negetive in 3rd & 4th Quadrant
Mr.Math
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And then the xsin(πx) is positve in (-1,0).
Mr.Math
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Now in the interval (0,1) is xsin(πx) positive or negative?
Diyadiya
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negative ?
Mr.Math
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Lets cut it into two pieces. Is x positive or negative in (0,1)?
Diyadiya
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wait a minute , i'm trying to understand the previous thing :D
1min
Mr.Math
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Take your time :-)
Diyadiya
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Alright
Diyadiya
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x is positive in (0,1)
Mr.Math
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Good. What about sin(πx)?
Diyadiya
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Positive
Mr.Math
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Great. So xsin(πx) over all is obviously..?
Diyadiya
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positive
Mr.Math
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What is the next interval we should consider now?
Diyadiya
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(1,3/2) ?
Mr.Math
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Right, (1,3/2).
Mr.Math
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And xsin(πx) will be positive or negative in this interval?
Diyadiya
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positive
Mr.Math
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Why?
Diyadiya
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hm x is positive in (1,3/2)
Diyadiya
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its not ?
Mr.Math
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Correct. But what about sin(πx)?
Diyadiya
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positive ?
Diyadiya
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wait
Diyadiya
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No its negative
Diyadiya
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since the angle is b/w (pi ,3pi/2)
Mr.Math
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Very good. That means that xsin(πx) will be negative in (1,3/2), right?
Diyadiya
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Yeah
Mr.Math
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So now since you know where our expression is positive and negative you can rewrite it using the definition of absolute value. Can you do that?
Mr.Math
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We have found that it's positive in (-1,1) and negative in (1,3/2). You agree?
Diyadiya
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why (-1,1) ?
Mr.Math
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Didn't we just find that it's positive in (-1,0) and in (0,1)?
Diyadiya
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Oh yes !!
Mr.Math
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Okay, the next step is to write it as it's written in your book. :-)
Diyadiya
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Yeah Got it :D
Diyadiya
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Thank You soo much :D that was a great explaination :D
Mr.Math
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Awesome!
Diyadiya
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Now i can do the rest :-)
Thanks again :)
Mr.Math
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You're welcome!