Question

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

Answer 1

`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`

Answer 2

`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`