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