高専の数学解説

第5章 5.36

G(x)=∫0x2e−tsin⁡t dt=[−e−tsin⁡t]0x2+∫0x2e−tcos⁡t dtG_{(x)} = \int_0^{x^2} e^{-t}\sin t\,dt = \left[-e^{-t}\sin t\right]_0^{x^2} + \int_0^{x^2} e^{-t}\cos t\,dt

=[−e−tsin⁡t]0x2+[−e−tcos⁡t]0x2−∫0x2e−tsin⁡t dt= \left[-e^{-t}\sin t\right]_0^{x^2} + \left[-e^{-t}\cos t\right]_0^{x^2} - \int_0^{x^2} e^{-t}\sin t\,dt

=−e−x2sin⁡x2−e−x2cos⁡x2+1−G(x)= -e^{-x^2}\sin x^2 - e^{-x^2}\cos x^2 + 1 - G_{(x)}

2G(x)=−e−x2sin⁡x2−e−x2cos⁡x2+12G_{(x)} = -e^{-x^2}\sin x^2 - e^{-x^2}\cos x^2 + 1

G(x)=12(−e−x2sin⁡x2−e−x2cos⁡x2+1)G_{(x)} = \frac{1}{2}(-e^{-x^2}\sin x^2 - e^{-x^2}\cos x^2 + 1)

G(x)′=12{2xe−x2(sin⁡x2+cos⁡x2)−e−x2(2xcos⁡x2−2xsin⁡x2)}G'_{(x)} = \frac{1}{2}\left\{2xe^{-x^2}(\sin x^2+\cos x^2) - e^{-x^2}(2x\cos x^2-2x\sin x^2)\right\}

=xe−x2(sin⁡x2+cos⁡x2−cos⁡x2+sin⁡x2)=2xe−x2sin⁡x2= xe^{-x^2}(\sin x^2+\cos x^2-\cos x^2+\sin x^2) = 2xe^{-x^2}\sin x^2

G(x)′=0G'_{(x)}=0 のとき

x=0, ±πx = 0,\ \pm\sqrt{\pi}

x=0x=0 のとき G(x)=0G_{(x)}=0 最小値

x=±πx=\pm\sqrt{\pi} のとき G(x)=12(1+1eπ)G_{(x)}=\dfrac{1}{2}\left(1+\dfrac{1}{e^{\pi}}\right) 最大値

解説: r31bn1z