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evaluate diya LOl
Equation editor is not working ?
it is !

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Other answers:

Okay integrate -1 to 3/2 |x sin(pi x)| dx
ok i will
tell me how to start
Equation editor is not working :/
ok diya do it by parts i did it
but there is modulus
ohh dint c the modulus
w8 tellling u
okay
http://ncertbooks.prashanthellina.com/class_12.Mathematics.MathematicsPartII/Integral%20ch_7%2006.11.06.pdf Page number 351 (its written on the top right corner)
plz show us how to remove modulus
full solution is there lol
i just wanna know that starting part
ummm guyzzz
f=x, f'=dx g'=sin pix g=-1/picospix
Hm What ? sorry can you explain
Do you use u and v or do you use f(x) and g(x) when you integrate by parts.?
for parts f(x) and g(x)
\[f(x)=x\] \[f'(x)=dx\] \[g'(x)=\sin \pi x\] \[g(x)=-\frac{1}{\pi}\cos \pi \]
bt man how to remove modulus
Equation editor isn't working
Can you see the post @wasiqss
I thought you wanted to integrate the function. Didn't know you were trying to remove the modulus.
Mertsj i cant see what you posted ,it just shows [maths processing error ]
diya i can see tho :P
Wait let me refresh
mertsj jus remove the modulus first then i will integrate
|dw:1330277056196:dw|
Is that the problem or not?
yes
Then integrate it using the substitutions I gave you.
|dw:1330277124894:dw|
modulus is thea man
1 Attachment
f(x)=x f'(x)=dx g'(x)=sin pi x g(x)= -1/pi cos pi x
So we dont have to remove modulus first ?
yeh diya thats my question
Why would you? Integrate the function.
mertsj we have do something with modulus first
Why?
This is the solution given in my book
1 Attachment
diya this is techniqu for removing modulus that u need to learn from someone cox im not able to recall it
Right. Since it's absolute value there are actually two integrals to do.
Where was the modulus removed in diyadiya's example post?
Which step?
find where x sin(pi x) is positive and negative
as in how
Which part of the solution don't you understand Diya?
the beginning
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.
MrMath i cant see Latex
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?
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.
Why is it in the 3rd or 4th Quadrant ?
Because when x in (-1,0), the angle, πx in (-π,0), right?
yeah
So you know why sin(πx) would be negative in this interval?
okay sine is negetive in 3rd & 4th Quadrant
And then the xsin(πx) is positve in (-1,0).
Now in the interval (0,1) is xsin(πx) positive or negative?
negative ?
Lets cut it into two pieces. Is x positive or negative in (0,1)?
wait a minute , i'm trying to understand the previous thing :D 1min
Take your time :-)
Alright
x is positive in (0,1)
Good. What about sin(πx)?
Positive
Great. So xsin(πx) over all is obviously..?
positive
What is the next interval we should consider now?
(1,3/2) ?
Right, (1,3/2).
And xsin(πx) will be positive or negative in this interval?
positive
Why?
hm x is positive in (1,3/2)
its not ?
Correct. But what about sin(πx)?
positive ?
wait
No its negative
since the angle is b/w (pi ,3pi/2)
Very good. That means that xsin(πx) will be negative in (1,3/2), right?
Yeah
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?
We have found that it's positive in (-1,1) and negative in (1,3/2). You agree?
why (-1,1) ?
Didn't we just find that it's positive in (-1,0) and in (0,1)?
Oh yes !!
Okay, the next step is to write it as it's written in your book. :-)
Yeah Got it :D
Thank You soo much :D that was a great explaination :D
Awesome!
Now i can do the rest :-) Thanks again :)
You're welcome!

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