What is #f(x) = int xsqrt(x^2-2) dx# if #f(3) = 3 #?

1 Answer
Bdub
Apr 11, 2016

#f(x)=1/3 (x^2-2)^(3/2) +3-(7sqrt7)/3#

Explanation:

#f(x)=intxsqrt(x^2-2)dx=intx(x^2-2)^(1/2)dx#

Note that the x outside of the square root is part of the derivative of the inside so we can just integrate #(x^2-2)^(1/2)#

#f(x)=intx(x^2-2)^(1/2)dx = 2/3 (x^2-2)^(3/2)*1/2#-> since the 2 from the derivative of the inside is missing.

#f(x)=1/3 (x^2-2)^(3/2) +C#
#f(3)=1/3 (9-2)^(3/2)+C=3#
#(7sqrt7)/3 +C =3#
#C=3-(7sqrt7)/3#
#f(x)=1/3 (x^2-2)^(3/2) +3-(7sqrt7)/3#

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