What is $f (x) = \int {(x + 3)}^{2} + 3 x d x$ $f(x) = int (x+3)^2+3x dx$ if $f (5) = 2$ $f(5)=2$ ?

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Answer 1 · 2017-05-20T16:25:11

$f (x) = \frac{1}{3} x^{3} + \frac{9}{2} x^{2} + 9 x - \frac{1183}{6}$ $f(x) = 1/3 x^3 + 9/2 x^2 + 9x - 1183/6$

$f (x) = \int ({(x + 3)}^{2} + 3 x) d x$ $f(x) = int ((x+3)^2+3x)dx$

This integral could be solved by doing a $u$ $u$ substitution where $u = x + 3$ $u=x+3$ , OR by expanding out the binomial in the parentheses.

I'll expand out the value in the parentheses for simplicity (and to use less extraneous variables).

$f (x) = \int (x^{2} + 6 x + 9 + 3 x) d x$ $f(x) = int(x^2+6x+9+3x)dx$

$f (x) = \int (x^{2} + 9 x + 9) d x$ $f(x) = int(x^2+9x+9)dx$

$f (x) = \frac{1}{3} x^{3} + \frac{9}{2} x^{2} + 9 x + C$ $f(x) = 1/3 x^3 + 9/2 x^2 + 9x +C$

Plug in the initial condition and solve for $C$ $C$ (an arbitrary constant):
$f (5) = 2$ $f(5) = 2$

$2 = \frac{1}{3} {(5)}^{3} + \frac{9}{2} {(5)}^{2} + 9 (5) + C$ $2 = 1/3 (5)^3 + 9/2 (5)^2 + 9(5) +C$

I used a calculator in this step (not necessary)
$C = \frac{- 1183}{6} \approx - 197.1667$ $C = (-1183)/6 approx -197.1667$

Using this value for $C$ $C$ in the equation for $f (x)$ $f(x)$ :

$f (x) = \frac{1}{3} x^{3} + \frac{9}{2} x^{2} + 9 x - \frac{1183}{6}$ $f(x) = 1/3 x^3 + 9/2 x^2 + 9x - 1183/6$

What is f(x) = int (x+3)^2+3x dxf(x)=∫(x+3)2+3xdxf(x) = int (x+3)^2+3x dx if f(5)=2 f(5)=2f(5)=2 ?