What is #f(x) = int xsqrt(5-2) dx# if #f(2) = 3 #?

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Answer 1 · 2018-04-29T06:34:00

The function #f(x)# is #(x^2sqrt3+6-4sqrt3)/2#.

First, compute the integral:

#f(x)=intxsqrt(5-2)# #dx#

#color(white)(f(x))=intxsqrt3# #dx#

#color(white)(f(x))=sqrt3intx# #dx#

#color(white)(f(x))=sqrt3*x^2/2+C#

Now, set #f(2)# (which is #3#) and the integral evaluated at #2# equal to each other, then solve for #C#:

#3=sqrt3*2^2/2+C#

#3=2sqrt3+C#

#3-2sqrt3=C#

That's the #C# value, so that means that the function is:

#f(x)=sqrt3*x^2/2+3-2sqrt3#

If you would like, you can rewrite it as:

#f(x)=(x^2sqrt3+6-4sqrt3)/2#

Hope this helped!