高専の数学解説

第9章 9.5

(1)

∫0π2xsin⁡x dx=[x(−cos⁡x)]0π2+∫0π2cos⁡x dx=0+[sin⁡x]0π2=1\int_0^{\frac{\pi}{2}} x\sin x\,dx = \Big[x(-\cos x)\Big]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos x\,dx = 0 + \Big[\sin x\Big]_0^{\frac{\pi}{2}} = 1

(2)

∫12(x+1)ex dx=[(x+1)ex]12−∫12ex dx=3e2−2e−[ex]12=3e2−2e−e2+e=2e2−e\int_1^2 (x+1)e^x\,dx = \Big[(x+1)e^x\Big]_1^2 - \int_1^2 e^x\,dx = 3e^2 - 2e - \Big[e^x\Big]_1^2 = 3e^2 - 2e - e^2 + e = 2e^2 - e

(3)

∫0πe−xsin⁡x dx=[−e−xsin⁡x]0π+∫0πe−xcos⁡x dx=0+[−e−xcos⁡x]0π−∫0πe−xsin⁡x dx\int_0^{\pi} e^{-x}\sin x\,dx = \Big[-e^{-x}\sin x\Big]_0^{\pi} + \int_0^{\pi} e^{-x}\cos x\,dx = 0 + \Big[-e^{-x}\cos x\Big]_0^{\pi} - \int_0^{\pi} e^{-x}\sin x\,dx

2∫0πe−xsin⁡x dx=e−π+1∫0πe−xsin⁡x dx=12(e−π+1)2\int_0^{\pi} e^{-x}\sin x\,dx = e^{-\pi} + 1 \qquad \int_0^{\pi} e^{-x}\sin x\,dx = \frac{1}{2}(e^{-\pi}+1)

(4)

∫0π2x2sin⁡x dx=[x2(−cos⁡x)]0π2+∫0π22x⋅cos⁡x dx=0+[2x⋅sin⁡x]0π2−2∫0π2sin⁡x dx\int_0^{\frac{\pi}{2}} x^2\sin x\,dx = \Big[x^2(-\cos x)\Big]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} 2x\cdot\cos x\,dx = 0 + \Big[2x\cdot\sin x\Big]_0^{\frac{\pi}{2}} - 2\int_0^{\frac{\pi}{2}} \sin x\,dx

=π−2[−cos⁡x]0π2=π−2= \pi - 2\Big[-\cos x\Big]_0^{\frac{\pi}{2}} = \pi - 2

(5)

∫0π2e−xcos⁡x dx=[−e−xcos⁡x]0π2+∫0π2e−x(−sin⁡x) dx=1−[−e−xsin⁡x]0π2−∫0π2e−xsin⁡x dx\int_0^{\frac{\pi}{2}} e^{-x}\cos x\,dx = \Big[-e^{-x}\cos x\Big]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} e^{-x}(-\sin x)\,dx = 1 - \Big[-e^{-x}\sin x\Big]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} e^{-x}\sin x\,dx

2∫0π2e−x(−sin⁡x) dx=1+e−π2∫0π2e−x(−sin⁡x) dx=12(1+e−π2)2\int_0^{\frac{\pi}{2}} e^{-x}(-\sin x)\,dx = 1 + e^{-\frac{\pi}{2}} \qquad \int_0^{\frac{\pi}{2}} e^{-x}(-\sin x)\,dx = \frac{1}{2}\left(1+e^{-\frac{\pi}{2}}\right)

解説: r31bn1z