高専の数学解説

第5章 5.16

(1)

∫0axa2−x2 dx=[−12(a2−x2)3 23]0a=13a3\int_0^a x\sqrt{a^2-x^2}\,dx=\left[-\frac{1}{2}\sqrt{(a^2-x^2)^3}\,\frac{2}{3}\right]_0^a=\frac{1}{3}a^3

(2)

∫02a2ax−x2 dx=∫02aa2−(x−a)2 dx=[12{(x−a)a2−(x−a)2+a2sin⁡−1x−aa}]02a\int_0^{2a}\sqrt{2ax-x^2}\,dx=\int_0^{2a}\sqrt{a^2-(x-a)^2}\,dx=\left[\frac{1}{2}\left\{(x-a)\sqrt{a^2-(x-a)^2}+a^2\sin^{-1}\frac{x-a}{a}\right\}\right]_0^{2a}

=12{a2π2−a2(−π2)}=πa22=\frac{1}{2}\left\{a^2\frac{\pi}{2}-a^2\left(-\frac{\pi}{2}\right)\right\}=\frac{\pi a^2}{2}

(3)

∫−11dx(1+x2)3=[14{x(x2+1)2+3(12(xx2+1+tan⁡−1x))}]−11\int_{-1}^{1}\frac{dx}{(1+x^2)^3}=\left[\frac{1}{4}\left\{\frac{x}{(x^2+1)^2}+3\left(\frac{1}{2}\left(\frac{x}{x^2+1}+\tan^{-1}x\right)\right)\right\}\right]_{-1}^{1}

=14{14+32(12+π4)+14−32(−12−π4)}=12+316π=\frac{1}{4}\left\{\frac{1}{4}+\frac{3}{2}\left(\frac{1}{2}+\frac{\pi}{4}\right)+\frac{1}{4}-\frac{3}{2}\left(-\frac{1}{2}-\frac{\pi}{4}\right)\right\}=\frac{1}{2}+\frac{3}{16}\pi

(4)

∫0ax2(a2+x2)2 dx=[12(−xx2+a2+1atan⁡−1xa)]0a=12{−12a+1a⋅π4}=π−28a\int_0^a \frac{x^2}{(a^2+x^2)^2}\,dx=\left[\frac{1}{2}\left(-\frac{x}{x^2+a^2}+\frac{1}{a}\tan^{-1}\frac{x}{a}\right)\right]_0^a=\frac{1}{2}\left\{-\frac{1}{2a}+\frac{1}{a}\cdot\frac{\pi}{4}\right\}=\frac{\pi-2}{8a}

解説: r31bn1z