## Diyadiya Group Title Evaluate 2 years ago 2 years ago

1. wasiqss

evaluate diya LOl

Equation editor is not working ?

3. wasiqss

it is !

Okay integrate -1 to 3/2 |x sin(pi x)| dx

5. wasiqss

ok i will

tell me how to start

Equation editor is not working :/

8. wasiqss

ok diya do it by parts i did it

but there is modulus

10. wasiqss

ohh dint c the modulus

11. wasiqss

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)

14. wasiqss

plz show us how to remove modulus

full solution is there lol

i just wanna know that starting part

17. razor99

ummm guyzzz

18. Mertsj

f=x, f'=dx g'=sin pix g=-1/picospix

Hm What ? sorry can you explain

20. Mertsj

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)

22. Mertsj

$f(x)=x$ $f'(x)=dx$ $g'(x)=\sin \pi x$ $g(x)=-\frac{1}{\pi}\cos \pi$

23. wasiqss

bt man how to remove modulus

Equation editor isn't working

Can you see the post @wasiqss

26. Mertsj

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 ]

28. wasiqss

diya i can see tho :P

Wait let me refresh

30. wasiqss

mertsj jus remove the modulus first then i will integrate

31. Mertsj

|dw:1330277056196:dw|

32. Mertsj

Is that the problem or not?

33. wasiqss

yes

34. Mertsj

Then integrate it using the substitutions I gave you.

|dw:1330277124894:dw|

36. wasiqss

modulus is thea man

38. Mertsj

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 ?

40. wasiqss

yeh diya thats my question

41. Mertsj

Why would you? Integrate the function.

42. wasiqss

mertsj we have do something with modulus first

43. Mertsj

Why?

This is the solution given in my book

45. wasiqss

diya this is techniqu for removing modulus that u need to learn from someone cox im not able to recall it

46. Mertsj

Right. Since it's absolute value there are actually two integrals to do.

47. Mertsj

Where was the modulus removed in diyadiya's example post?

48. Mertsj

Which step?

49. Zarkon

find where x sin(pi x) is positive and negative

50. wasiqss

as in how

51. Mr.Math

Which part of the solution don't you understand Diya?

the beginning

53. Mr.Math

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

55. Mr.Math

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?

56. Mr.Math

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 ?

58. Mr.Math

Because when x in (-1,0), the angle, πx in (-π,0), right?

yeah

60. Mr.Math

So you know why sin(πx) would be negative in this interval?

okay sine is negetive in 3rd & 4th Quadrant

62. Mr.Math

And then the xsin(πx) is positve in (-1,0).

63. Mr.Math

Now in the interval (0,1) is xsin(πx) positive or negative?

negative ?

65. Mr.Math

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

67. Mr.Math

Alright

x is positive in (0,1)

70. Mr.Math

Positive

72. Mr.Math

Great. So xsin(πx) over all is obviously..?

positive

74. Mr.Math

What is the next interval we should consider now?

(1,3/2) ?

76. Mr.Math

Right, (1,3/2).

77. Mr.Math

And xsin(πx) will be positive or negative in this interval?

positive

79. Mr.Math

Why?

hm x is positive in (1,3/2)

its not ?

82. Mr.Math

positive ?

wait

No its negative

since the angle is b/w (pi ,3pi/2)

87. Mr.Math

Very good. That means that xsin(πx) will be negative in (1,3/2), right?

Yeah

89. Mr.Math

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?

90. Mr.Math

We have found that it's positive in (-1,1) and negative in (1,3/2). You agree?

why (-1,1) ?

92. Mr.Math

Didn't we just find that it's positive in (-1,0) and in (0,1)?

Oh yes !!

94. Mr.Math

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

97. Mr.Math

Awesome!