If $n'(x)=2\sin(2x)$ and $n(0)=0$ what is the absolute maximum of $n(x)$ on $[0,2\pi]$?

Expert intro: Find $n(x)$ from its derivative, then compare critical points and endpoints on the closed interval.

Detailed walkthrough

We start with

$n'(x)=2\sin(2x)$

Integrate:

$n(x)=\int 2\sin(2x)\,dx=-\cos(2x)+C$

Use the initial condition $n(0)=0$:

$-\cos(0)+C=0$ $-1+C=0$ $C=1$

So

$n(x)=1-\cos(2x)$

Now find the absolute maximum on $[0,2\pi]$. Since

$-1\le \cos(2x)\le 1,$

we get

$0\le 1-\cos(2x)\le 2$

The maximum value is

$\boxed{2}$

This happens when $\cos(2x)=-1$, for example at

$x=\frac{\pi}{2},\ \frac{3\pi}{2}$

on the interval $[0,2\pi]$.

💡 Pitfall guide

A common error is to stop after finding critical points and forget to check the endpoints of the interval. For absolute extrema on a closed interval, always evaluate endpoints too. Another mistake is to integrate $2\sin(2x)$ incorrectly; the antiderivative is $-\cos(2x)$, not $\cos(2x)$.

🔄 Real-world variant

If the interval were different, the maximum value might still be $2$, but you would need to verify whether the points where $\cos(2x)=-1$ lie inside the new interval. If the initial condition changed, the graph would shift vertically, and the absolute maximum would shift by the same amount.

🔍 Related terms

absolute maximum, critical point, definite interval