# Question #c5ddf

Hence we have that

$\int f ' \left(x\right) \mathrm{dx} = \int \left(3 {x}^{2}\right) \mathrm{dx}$

$f \left(x\right) = {x}^{3} + c$

Because $f \left(- 1\right) = 2$ we have that

$f \left(- 1\right) = {\left(- 1\right)}^{3} + c$

$2 = - 1 + c$

$c = 3$

Finally $f \left(x\right) = {x}^{3} + 3$

Now we can calculate the integral

${\int}_{0}^{2} f \left(x\right) \mathrm{dx} = {\int}_{0}^{2} \left({x}^{3} + 3\right) \mathrm{dx} = {\left[{x}^{4} / 4 + 3 x\right]}_{0}^{2} = 10$

Mar 3, 2017

#### Explanation:

From $f ' \left(x\right) = 3 {x}^{2}$, we conclude $f \left(x\right) = {x}^{3} + C$ for some constant $C$.

Since we are told that $f \left(- 1\right) = 2$, we see that ${\left(- 1\right)}^{3} + C = 2$, and we conclude that $C = 3$

To finish, we are asked to evaluate ${\int}_{0}^{2} f \left(x\right) \mathrm{dx}$.

${\int}_{0}^{2} \left({x}^{3} + 3\right) \mathrm{dx} = {\left.{x}^{4} / 4 + 3 x\right]}_{0}^{2}$

$= 4 + 6 = 10$