Integral definite #int_0^3(xdx)/sqrt(36-x)# ?

Question

#int_0^3(xdx)/sqrt(36-x)#

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Answer 1 · 2018-04-10T04:13:35

#I=288-50sqrt33#

Here,

#I=int_0^3x/sqrt(36-x)dx#

Let,

#sqrt(36-x)=u,(u>0)=>36-x=u^2=>x=36-u^2#

#=>dx=-2udu#

#:.x=0=>u=6 andx=3=>u=sqrt33#

So,

#I=int_6^sqrt33(36-u^2)/u(-2u)du#

#=-2int_6^sqrt33(36-u^2)du#

#=-2[36u-u^3/3]_6^sqrt33#

#=-2(36sqrt33-(33sqrt33)/3)+2(36xx6-6^3/3)#

#=-2sqrt33(36-11)+2xx216(1-1/3)#

#=-2sqrt33(25)+432xx2/3]#

#=-50sqrt33+288#

#=288-50sqrt33#