Let f be a non-negative function defined on the interval [0,1]. If integral$_{0}^{x} sqrt(1-(f'(t))^{2})$ dt $=$ integral$_{0}^{x} f(t)$ dt$, 0 \le x \le 1,$ and $f(0)=0,$ then $(1) f(\frac{1}{2})<\frac{1}{2}$ and $f(\frac{1}{3})>\frac{1}{3} (2) f(\frac{1}{2})>\frac{1}{2}$ and $f(\frac{1}{3})>\frac{1}{3} (3) f(\frac{1}{2})<\frac{1}{2}$ and $f(\frac{1}{3})<\frac{1}{3} (4) f(\frac{1}{2})>\frac{1}{2}$ and $f(\frac{1}{3})<\frac{1}{3}$

1. A $f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$
2. B $f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$
3. C $f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$
4. D $f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$

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Answer: C. $f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$

Applications of Integrals Let f be a non-negative function defined on the interval [0,1]. If integral$_{0}^{x} sqrt(1-(f'(t))^{2})$ dt $=$ integral$_{0}^{x} f(t)$ dt$, 0 \le x \le 1,$ and $f(0)=0,$ then $(1) f(\frac{1}{2})<\frac{1}{2}$ and $f(\frac{1}{3})>\frac{1}{3} (2) f(\frac{1}{2})>\frac{1}{2}$ and $f(\frac{1}{3})>\frac{1}{3} (3) f(\frac{1}{2})<\frac{1}{2}$ and $f(\frac{1}{3})<\frac{1}{3} (4) f(\frac{1}{2})>\frac{1}{2}$ and $f(\frac{1}{3})<\frac{1}{3}$ Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

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