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प्रश्न
Evaluate `∫_0^(3/2)|x cosπx|dx`
उत्तर
`int_0^(3/2)|xcospix|dx`
`0<x<1/2`
`0<pix<pi/2rArrcospix>0rArr(xcospix)>0`
`|xcospix|=xcospix`
`1/2<x<3/2`
`pi/2<pix<(3pi)/2rArrcospix<0rArr(xcospix)<0`
`|xcospix|=-xcospix`
`I=int_0^(3/2)|xcospix|dx=int_0^(3/2)xcospix+int_(1/2)^(3/2)-(xcospix)`
`I=int_0^(1/2)xcospix-int_(1/2)^(3/2)xcospix`
`intx(cospix)=x(sinpix)/pi-int(sinpix)/pi`
`=x/pi(sinpix)+(cospix)/pi^2`
`I=[(x/pisinpix)+(cospix)/pi^2]_0^(1/2)-[(x/pisinpix)+(cospix)/pi^2]_(1/2)^(3/2)`
`=[1/pi((1/2)-0)+1/pi^2(0-1)]-[1/pi(3/2(-1)-1/2(1))+1/pi^2(0-0)]`
`=(1/(2pi)-1/pi^2)-((-2)/pi)`
`=(5/(2pi)-1/pi^2)`
`=((5pi-2)/(2pi^2))`
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