I'm having trouble with a problem and looking for some advice on where to start.
The problem is as follows:
On interval $[0, \frac{\pi}4]$, it is true that $t^2 \leq sin(t) \leq t$
Use this fact to find the upper and lower bounds for:
$\int_{0}^{\frac{\pi}4} \sin(t)dt$
The question is from a chapter on comparison theorem and before the fundamental theorem of calculus.
Can anyone provide some advice on where to begin? I am absolutely lost.
From some other problems, I was assuming it was something to do with limits of the interval, but I wasn't sure why the equality would come into play.
Best Answer
Recall the Taylor series: $\sin t = t - O(t^3)$. On $[0,\pi/4]$ this means $\sin t\leqslant t$. Since $t<1$, $t^2<t$, and so $t^2<\sin t$. It follows that $$ 0.161491\approx\frac{\pi ^3}{192}=\int_0^{\frac\pi4} t^2\ \mathsf dt \leqslant \int_0^{\frac\pi4}\sin t\ \mathsf dt\leqslant \int_0^{\frac\pi4} t\ \mathsf dt=\frac{\pi ^2}{32}\approx 0.3084251. $$