Question #c5ddf

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Answer 1 · 2017-03-03T00:21:41

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 · 2017-03-03T00:21:50

Please see below.

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 $int_0^2 f(x) dx$ .

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

$= 4+6 = 10$