#f(x) = int(sqrt(x+3) - x)dx#
Split the integral into two separate integrals:
#f(x) = int(sqrt(x+3) )dx - int(x)dx#
u-substitution is your best friend
let #u = x+3#
let #du = 1 dx#
let #dx = du#
find the integral by substituting u into the square root and simplify:
#f(x) = int(u^(1/2))du -int(x)dx#
#f(x) = (u^(1/2 + 1))/(3/2) - x^2 / 2 + C #
#f(x) = u^(3/2) / (3/2) - 1/2 x^2 + C #
#f(x) = 2/3u^(3/2) - 1/2 x^2 + C #
Plug x+3 back in for u:
#f(x) = 2/3(x+3)^(3/2) - 1/2 x^2 + C #
plug in #f(1)=-4# to find the c-value
#-4 = 2/3(1+3)^(3/2) - 1/2 (1)^2 + C #
#-4 = 2/3(4)^(3/2) - 1/2 + C #
# -7/2 = 16/3 +C#
#C=-53/6#
plug in the c-value into your integrated equation
#f(x) = 2/3(x+3)^(3/2) - 1/2 x^2 + C#
#f(x) = 2/3(x+3)^(3/2) - 1/2 x^2 - 53/6 #
