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Evaluate: `int_(-1)^(3//2)|xsinpix|dx` |
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Answer» `|x sin pi x|={((-x)(-sinpix),"if",-1lexlt0),(xsinpix,"if",0lexle1),(x(-sinpix),"if",1ltxle3//2):}` `:.int_(-1)^(3//2)|x sin pi x|dx` `=int_(-1)^(0)sin pix dx +int_(-1)^(1) x sin dx +int_(1)^(3//2)(-x sin pi x)dx` `=int_(-1)^(1)x sin pi x dx - int_(1)^(3//2), x sin pi x dx` `=2int_(0)^(1)x sin x dx-int_(1)^(3//2)xsin pix dx` `=2[{x((-1)/(pi))cospix}_(0)^(1)-int_(0)^(1)1((-1)/(pi))cospix dx]` `-{x((-1)/(pi))cospix}_(1)^(3//2)+int_(1)^(3//2)1((-1)/(pi))cosx dx` `=(2/(pi))+(2/(pi^(2)))[sin pix]_(0)^(1)+{3/((2pi))}"cos"3/2pi+(1/(pi))` `-1/(pi^(2))[sinpix]_(1)^(3//2)` `=(2//pi)+0+0+(1//pi)+(1//pi^(2))` `=(3pi+1)//pi^(2)` |
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