please solve this.
Asked by prabhat1993 | 16th Feb, 2010, 07:50: PM
|xsin(πx)| dx, limits, -1 to 3/2
From -1 to 0, x will be negative and sin function will also be negative. Hence their product will be positive.
It will also be positive from 0 to 1, it's only negative from 1 to 3/2.
Hence we can write,
|xsin(πx)| dx, limits, -1 to 3/2 = I1 + I2 + (- I3 ) ............for the three limit parts mentioned above.
Now I1 = I2 = x sin(πx) dx, with limits from 0 to 1
= -xcos(πx) /π + sin(πx) /π2 ........integrating by parts,
Put the limits,
Similarly I3, limits,
I3 = -1/π2 + 1/π
Using these, we find,
|xsin(πx)| dx, limits, -1 to 3/2 = 1/π + 1/π2 = (π + 1)/π2
Answered by | 17th Feb, 2010, 07:44: AM
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