What is #f(x) = int sqrt(x+3) -x dx# if #f(1)=-4 #?

2 Answers
Feb 17, 2017

#2/3 (x+3)^(3/2) - x^2 /2 -53/6#

Explanation:

f(x)= #int sqrt (x+3)dx - int x dx #

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

Given #f(1) = -4#, we can solve for C

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

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

#C= -4 +1/2 -16/3 = -53/6#

#f(x)=2/3 (x+3)^(3/2) - x^2 /2 -53/6#

Feb 17, 2017

#f(x) = 2/3(x+3)^(3/2) - 1/2 x^2 - 53/6 #

Explanation:

#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 #