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

1 Answer
Noah G
Mar 22, 2018

#f(x) = 2/3(x -2)^(3/2) + 4(x- 2)^(1/2) - 17/3#

Explanation:

Let #u = x - 2#. Then #du = dx#, and #x = u + 2#.

#f(x) = int x/sqrt(u)dx#

#f(x) = int (u +2)/sqrt(u) du#

#f(x) = int u/sqrt(u) + 2/sqrt(u) du#

#f(x) = int u^(1/2) + 2u^(-1/2) du#

#f(x) = 2/3u^(3/2) + 4u^(1/2) + C#

#f(x) = 2/3(x - 2)^(3/2) +4(x- 2)^(1/2) + C#

Now we solve for #C#.

#-1 = 2/3(3- 2)^(3/2) + 4(3 -2)^(1/2) + C#

#-1 = 2/3 + 4 + C#

#-17/3 = C#

Therefore the function is

#f(x) = 2/3(x -2)^(3/2) + 4(x- 2)^(1/2) - 17/3#

Hopefully this helps!

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