What is #f(x) = int 1/(sqrt(x+3) # if #f(2)=1 #?

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Answer 1 · 2016-01-25T10:20:54

#f(x) = 2sqrt(x+3)+1-2sqrt(5)#

First, we evaluate the indefinite integral using #u# substitution.

Let #u = x+3 => du = dx#

Then

#int1/sqrt(x+3)dx = int1/sqrt(u)du#

#=intu^(-1/2)du#

#=u^(1/2)/(1/2)+C#

#=2sqrt(x+3)+C#

Now, we evaluate at #x=2# to find the value of #C#

#f(2) = 2sqrt(2+3)+C = 1#

#=> C = 1-2sqrt(5)#

Thus, we have

#f(x) = 2sqrt(x+3)+1-2sqrt(5)#