자연계열 2번

(1)① 두 점 $\displaystyle (a,f(a)),(b,f(b))$를 지나는 직선의 기울기는 $\displaystyle \frac{f(b)-f(a)}{b-a}$이고,② $\displaystyle x=c$인 점에서의 접선의 기울기는 $\displaystyle f^{\prime}(c)=\frac1{2\sqrt c}$이다.③ 따라서 $\displaystyle b-a=h$라 하면$\displaystyle \frac{f(b)-f(a)}{b-a}=\frac{\sqrt{a+h}-\sqrt a}h=f^{\prime}(c)=\frac1{2\sqrt c}$이다.④ $\displaystyle \sqrt c=\frac h{2(\sqrt{a+h}-\sqrt a)}=\frac12(\sqrt{a+h}+\sqrt a)$이므로양변을 제곱하면 $\displaystyle c=\frac14(2a+h+2\sqrt{a^2+ah})$이다.

(2)① $\displaystyle c-a=\frac14(-2a+h+2\sqrt{a^2+ah})$이므로② $\displaystyle \lim_{b\to a}\frac{c-a}{b-a}=\lim_{h\to0}\frac{2\sqrt{a^2+ah}-(2a-h)}{4h}$③ $\displaystyle =\lim_{h\to0}\frac{4(a^2+ah)-(2a-h)^2}{4h(2\sqrt{a^2+ah}+(2a-h))}$④ $\displaystyle =\lim_{h\to0}\frac{8a-h}{4(2\sqrt{a^2+ah}+(2a-h))}=\frac12$

(3)① $\displaystyle c=\frac14(2a+3a+2\sqrt{a^2+3a^2})=\frac14(5a+4a)=\frac94a$② $\displaystyle S=\int_a^{4a}\sqrt x\,dx=\Biggl[\frac23x\sqrt x\Biggr]_a^{4a}=\frac23(8-1)a\sqrt a=\frac{14}3a\sqrt a$③ $\displaystyle \begin{aligned}S_1&=\frac{f(a)+f(c)}2\cdot(c-a)=\frac{\sqrt a+\frac32\sqrt a}2\cdot\left(\frac94a-a\right)\\&=\frac54\sqrt a\cdot\frac54a=\frac{25}{16}a\sqrt a\end{aligned}$④ $\displaystyle \begin{aligned}S_2&=\frac{f(c)+f(b)}2\cdot(b-c)=\frac{\frac32\sqrt a+2\sqrt a}2\cdot\left(4a-\frac94a\right)\\&=\frac74\sqrt a\cdot\frac74a=\frac{49}{16}a\sqrt a\end{aligned}$$\displaystyle \therefore\ S_1+S_2=\left(\frac{25}{16}+\frac{49}{16}\right)a\sqrt a=\frac{74}{16}a\sqrt a$⑤ 따라서 $\displaystyle S:(S_1+S_2)=\frac{14}3:\frac{74}{16}=224:222$ 또는 $\displaystyle 112:111$이다.