If $m'(x)=3x^2$ and $m(1)=-1$ what is $m(2)$?

Key concept: This is a basic antiderivative problem: recover $m(x)$ from $m'(x)$, then use the initial condition.

Step by step

We are given

$m'(x)=3x^2$

so integrate both sides:

$m(x)=\int 3x^2\,dx=x^3+C$

Now use the condition $m(1)=-1$:

$1^3+C=-1$ $1+C=-1$ $C=-2$

So

$m(x)=x^3-2$

Finally, evaluate at $x=2$:

$m(2)=2^3-2=8-2=6$

Answer: $6$

Pitfall alert

Do not forget the constant of integration $C$. A common mistake is to write $m(x)=x^3$ and stop, which ignores the initial condition. Also, check the sign carefully when solving for $C$.

Try different conditions

If the initial condition were changed to $m(a)=b$, the same method would work: first find $m(x)=x^3+C$, then substitute $x=a$ to solve for $C$, and finally evaluate the desired point.

Further reading

antiderivative, initial condition, constant of integration