How do you evaluate the definite integral #int sqrt(2/x)dx# from [1,8]?

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Answer 1 · 2016-12-27T12:50:12

#8-2sqrt2#

firstly change the function of #x# from roots to powers.

#int_1^8sqrt(2/x)dx=int_1^8sqrt2(x)^(-1/2)dx#

now use the power rule and evaluate.

#=sqrt2[x^(-1/2+1)/(-1/2+1)]_1^8#

#=sqrt2[2x^(1/2)]_1^8#

#=2sqrt2[sqrt8-sqrt1]#

#2sqrt2[2sqrt2-1]#

#=8-2sqrt2#