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

Feb 17, 2017

$\frac{2}{3} {\left(x + 3\right)}^{\frac{3}{2}} - {x}^{2} / 2 - \frac{53}{6}$

#### Explanation:

f(x)= $\int \sqrt{x + 3} \mathrm{dx} - \int x \mathrm{dx}$

$f \left(x\right) = \frac{2}{3} {\left(x + 3\right)}^{\frac{3}{2}} - {x}^{2} / 2 + C$

Given $f \left(1\right) = - 4$, we can solve for C

$- 4 = \frac{2}{3} {\left(4\right)}^{\frac{3}{2}} - \frac{1}{2} + C$

$- 4 = \frac{2}{3} {2}^{3} - \frac{1}{2} + C$

$C = - 4 + \frac{1}{2} - \frac{16}{3} = - \frac{53}{6}$

$f \left(x\right) = \frac{2}{3} {\left(x + 3\right)}^{\frac{3}{2}} - {x}^{2} / 2 - \frac{53}{6}$

Feb 17, 2017

$f \left(x\right) = \frac{2}{3} {\left(x + 3\right)}^{\frac{3}{2}} - \frac{1}{2} {x}^{2} - \frac{53}{6}$

#### Explanation:

$f \left(x\right) = \int \left(\sqrt{x + 3} - x\right) \mathrm{dx}$

Split the integral into two separate integrals:

$f \left(x\right) = \int \left(\sqrt{x + 3}\right) \mathrm{dx} - \int \left(x\right) \mathrm{dx}$

let $u = x + 3$
let $\mathrm{du} = 1 \mathrm{dx}$
let $\mathrm{dx} = \mathrm{du}$

find the integral by substituting u into the square root and simplify:
$f \left(x\right) = \int \left({u}^{\frac{1}{2}}\right) \mathrm{du} - \int \left(x\right) \mathrm{dx}$
$f \left(x\right) = \frac{{u}^{\frac{1}{2} + 1}}{\frac{3}{2}} - {x}^{2} / 2 + C$
$f \left(x\right) = {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right) - \frac{1}{2} {x}^{2} + C$
$f \left(x\right) = \frac{2}{3} {u}^{\frac{3}{2}} - \frac{1}{2} {x}^{2} + C$

Plug x+3 back in for u:

$f \left(x\right) = \frac{2}{3} {\left(x + 3\right)}^{\frac{3}{2}} - \frac{1}{2} {x}^{2} + C$

plug in $f \left(1\right) = - 4$ to find the c-value
$- 4 = \frac{2}{3} {\left(1 + 3\right)}^{\frac{3}{2}} - \frac{1}{2} {\left(1\right)}^{2} + C$
$- 4 = \frac{2}{3} {\left(4\right)}^{\frac{3}{2}} - \frac{1}{2} + C$
$- \frac{7}{2} = \frac{16}{3} + C$
$C = - \frac{53}{6}$

plug in the c-value into your integrated equation
$f \left(x\right) = \frac{2}{3} {\left(x + 3\right)}^{\frac{3}{2}} - \frac{1}{2} {x}^{2} + C$
$f \left(x\right) = \frac{2}{3} {\left(x + 3\right)}^{\frac{3}{2}} - \frac{1}{2} {x}^{2} - \frac{53}{6}$