高専の数学解説

第5章 5.40

(1)

F(x)=∫x3x1−t2 dt=[12{t1−t2+sin⁡−1t}]x3xF_{(x)} = \int_x^{\sqrt{3}x} \sqrt{1-t^2}\,dt = \left[ \frac{1}{2}\{t\sqrt{1-t^2} + \sin^{-1}t\} \right]_x^{\sqrt{3}x}

=12{3x1−3x2+sin⁡−13x−x1−x2−sin⁡−1x}= \frac{1}{2}\{\sqrt{3}x\sqrt{1-3x^2} + \sin^{-1}\sqrt{3}x - x\sqrt{1-x^2} - \sin^{-1}x\}

dFdx=12{31−3x2+3x−3x1−3x2+31−3x2−1−x2−x−x1−x2−11−x2}\frac{dF}{dx} = \frac{1}{2}\left\{ \sqrt{3}\sqrt{1-3x^2} + \sqrt{3}x\frac{-3x}{\sqrt{1-3x^2}} + \frac{\sqrt{3}}{\sqrt{1-3x^2}} - \sqrt{1-x^2} - x\frac{-x}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} \right\}

=12{31−3x2+3(1−3x2)1−3x21−3x2−1−x2−(1−x2)1−x21−x2}= \frac{1}{2}\left\{ \sqrt{3}\sqrt{1-3x^2} + \frac{\sqrt{3}(1-3x^2)\sqrt{1-3x^2}}{1-3x^2} - \sqrt{1-x^2} - \frac{(1-x^2)\sqrt{1-x^2}}{1-x^2} \right\}

=31−3x2−1−x2= \sqrt{3}\sqrt{1-3x^2} - \sqrt{1-x^2}

(1)

から　dFdx=31−3x2−1−x2=0\dfrac{dF}{dx} = \sqrt{3}\sqrt{1-3x^2} - \sqrt{1-x^2} = 0　となるとき

31−3x2=1−x23(1−3x2)=1−x28x2=2x=±12\sqrt{3}\sqrt{1-3x^2} = \sqrt{1-x^2} \qquad 3(1-3x^2) = 1-x^2 \qquad 8x^2 = 2 \qquad x = \pm\frac{1}{2}

0≤x≤130 \leq x \leq \dfrac{1}{\sqrt{3}} から x=12x = \dfrac{1}{2}

このとき

F(x)=F(12)=12{324−34+π3−124−14−π6}=π12F_{(x)} = F_{(\frac{1}{2})} = \frac{1}{2}\left\{ \frac{\sqrt{3}}{2}\sqrt{\frac{4-3}{4}} + \frac{\pi}{3} - \frac{1}{2}\sqrt{\frac{4-1}{4}} - \frac{\pi}{6} \right\} = \frac{\pi}{12}

(3)

lim⁡x→0F(x)x=lim⁡x→012{3x1x2−3+1xsin⁡−13x−x1x2−1−1xsin⁡−1x}ロピタルの定理から\lim_{x \to 0}\frac{F_{(x)}}{x} = \lim_{x \to 0}\frac{1}{2}\left\{ \sqrt{3}x\sqrt{\frac{1}{x^2}-3} + \frac{1}{x}\sin^{-1}\sqrt{3}x - x\sqrt{\frac{1}{x^2}-1} - \frac{1}{x}\sin^{-1}x \right\} \quad \text{ロピタルの定理から}

=lim⁡x→012{31−3x2+31−3x2−1−x2−11−x2}=12{23−2}=3−1= \lim_{x \to 0}\frac{1}{2}\left\{ \sqrt{3}\sqrt{1-3x^2} + \frac{\sqrt{3}}{\sqrt{1-3x^2}} - \sqrt{1-x^2} - \frac{1}{\sqrt{1-x^2}} \right\} = \frac{1}{2}\{2\sqrt{3}-2\} = \sqrt{3}-1

解説: r31bn1z