Question #c5ddf

2 Answers

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#

Mar 3, 2017

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#