Найдите интеграл от 0 до pi/2 (1/2 sin x/2+1/3 cos x/3) dx и интеграл от 1 до 4 (x^(2)+x корень из x+x/корень из x) dx

Ответ:

zoga2003

23.05.2020 20:55

$\\\int \limits_0^{\frac{\pi}{2}}{\frac{1}{2}\sin \frac{x}{2} + \frac{1}{3}\cos \frac{x}{3}}\, dx=\\ \frac{1}{2}\int \limits_0^{\frac{\pi}{2}}{\sin \frac{x}{2}}\, dx+\frac{1}{3}\int \limits_0^{\frac{\pi}{2}}{\cos \frac{x}{3}}\, dx=(*)\\ t=\frac{x}{2},u=\frac{x}{3}\\ dt=\frac{1}{2}\,dx,du=\frac{1}{3}\,dx\\ \frac{1}{2}\int \limits_0^{\frac{\pi}{2}}{2\sin t}\, dt+\frac{1}{3}\int \limits_0^{\frac{\pi}{2}}{3\cos u}\, du=\\ \int \limits_0^{\frac{\pi}{2}}{\sin t}\, dt+\int \limits_0^{\frac{\pi}{2}}{\cos u}\, du=\\ \Big[-\cos t\Big]_0^{\frac{\pi}{2}}+\Big[\sin u\Big]_0^{\frac{\pi}{2}}=\\ (*)=\Big[-\cos \frac{x}{2}\Big]_0^{\frac{\pi}{2}}+\Big[\sin \frac{x}{3}\Big]_0^{\frac{\pi}{2}}=\\ -\cos \frac{\pi}{4}-(-\cos 0)+\sin \frac{\pi}{6}-\sin 0=\\ -\frac{\sqrt2}{2}+1+\frac{1}{2}=\\ -\frac{\sqrt2}{2}+\frac{3}{2}$

$\\\int \limits_1^4 {x^2+\sqrt x+\frac{x}{\sqrt x}}\, dx=\\ \int \limits_1^4 {x^2+\sqrt x+\sqrt x}\, dx=\\ \int \limits_1^4 {x^2+2\sqrt x}\, dx=\\ \Big[\frac{x^3}{3}+2\cdot\frac{2x^{\frac{3}{2}}}{3}\Big]_1^4=\\ \Big[\frac{x^3}{3}+\frac{4x^{\frac{3}{2}}}{3}\Big]_1^4=\\ \frac{4^3}{3}+\frac{4\cdot4^{\frac{3}{2}}}{3}-(\frac{1^3}{3}+\frac{4\cdot1^{\frac{3}{2}}}{3})=\\ \frac{64}{3}+\frac{4^{\frac{5}{2}}}{3}-(\frac{1}{3}+\frac{4}{3})=\\ \frac{64}{3}+\frac{2^{\frac{10}{2}}}{3}-\frac{5}{3}=\\ \frac{59}{3}+\frac{32}{3}=\\ \frac{91}{3}$