How do you evaluate #int (1) / (sqrt(1 + x))# for [0, 3]?

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Answer 1 · 2015-10-08T17:49:18

You can rewrite this as:

#= int (1+x)^"-1/2"dx#

and now you can simply use the reverse power rule.

#= [2(1+x)^"1/2"]##|_(0)^(3)#

#= [2sqrt(1+x)]##|_(0)^(3)#

#= [2sqrt(1+3)] - [2sqrt(1+0)]#

#= 4 - 2#

#= color(blue)(2)#