Maharashtra State BoardHSC Arts 12th Board Exam
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Evaluate ∫0(3/2) |x cosπx| dx - Mathematics and Statistics

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Evaluate `∫_0^(3/2)|x cosπx|dx`

 
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Solution

`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))`

 

Concept: Evaluation of Definite Integrals by Substitution
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