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`int_(0)^(2)xsqrt(x+2)dx (x+2=t^(2)" रखिए ")`

Answer» माना `x+2=t^(2)" "rArr" "dx=2t dt`
`x=0` पर `t=sqrt(0+2)=sqrt2`
`x=2" पर "t=sqrt(2+2)=2`
`therefore " "int_(0)^(2)x sqrt(x+2)dx`
`=int_(sqrt2)^(2)(t^(2)-2)sqrt(t^(2)).2t dt`
`=2int_(sqrt2)^(2)(t^(2)-2t^(2))dt`
`=2[(t^(5))/(5)-(2t^(3))/(3)]_(sqrt2)^(2)`
`=(2)/(15)[3t^(5)-10t^(3)]_(sqrt2)^(2)`
`=(2)/(15)[(96-80)-(12sqrt2-20sqrt2)]`
`=(2)/(15)(16+8sqrt3)=(16)/(15)(2+sqrt2)`


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