Question

Question #c5ddf

Answer 1

Hence we have that

`int f'(x)dx=int (3x^2)dx`

`f(x)=x^3+c`

Because `f(-1)=2` we have that

`f(-1)=(-1)^3+c`

`2=-1+c`

`c=3`

Finally `f(x)=x^3+3`

Now we can calculate the integral

`int_0^2 f(x)dx=int_0^2 (x^3+3)dx=[x^4/4+3x]_0^2=10`

Answer 2

Please see below.

Explanation:

From `f'(x) = 3x^2`, we conclude `f(x) = x^3+C` for some constant `C`.

Since we are told that `f(-1) = 2`, we see that `(-1)^3 +C = 2`, and we conclude that `C = 3`

To finish, we are asked to evaluate `int_0^2 f(x) dx`.

`int_0^2 (x^3+3) dx = {: x^4/4+3x]_0^2`

`= 4+6 = 10`