please solve this.

|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.