please solve this.

Asked by prabhat1993 | 16th Feb, 2010, 07:50: PM

Expert Answer:

|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,

= 1/π

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

Regards,

Team,

TopperLearning.

Answered by  | 17th Feb, 2010, 07:44: AM

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