Answer
$f(x) = \frac{1}{12}(4x^{2} - 10)^{3} -8$
Work Step by Step
1. Integrate $f'(x)$ by substitution
$\int 2x(4x^{2} - 10)^{2}dx$
Let $u = 4x^{2} - 10$
$u' = 8x$
$\frac{du}{dx} = 8x$
$\frac{du}{4} = 8x$
$\frac{1}{4}\int (du)(u^{2}) dx$
$= \frac{1}{4} \times \frac{u^{3}}{3} + c$
$= \frac{1}{12}(u^{3}) + c$
$= \frac{1}{12}(4x^{2} - 10)^{3} + c$
$f(x) = \frac{1}{12}(4x^{2} - 10)^{3} + c$
2. Substitute the point $(2, 10)$ into $f(x)$ to find for $c$
$10 = \frac{1}{12}(4(2)^{2} - 10)^{3} + c$
$10 = \frac{1}{12}(4(4) - 10)^{3} + c$
$10 = \frac{1}{12}(16 - 10)^{3} + c$
$10 = \frac{1}{12}(6)^{3} + c$
$10 = \frac{1}{12}(216) + c$
$10 = \frac{216}{12} + c$
$10 = 18 + c$
$c = 10- 18$
$c = -8$
3. Rewrite the equation of $f(x)$
$f(x) = \frac{1}{12}(4x^{2} - 10)^{3} -8$
