Question

вычислите интеграл, с решением ​

Answer 1

Ответ:

а) $0.425$

б) $\frac{1}{3}$

в) $\frac{\sqrt{3}}{3}$

г) $\frac{40}{3}$

Объяснение:

а)

$\int\limits^2_1 {(\frac{x^4}{4}-\frac{3}{x^3} )} \, dx= \int\limits^2_1 {\frac{x^4}{4}} \, dx-\int\limits^2_1 {\frac{3}{x^3}} \, dx=\frac{1}{4}\cdot \frac{x^5}{5}\bigg|\limits^2_1- (-\frac{3}{2x^2}\bigg|\limits^2_1)=\frac{1}{4}\cdot(\frac{2^5}{5}-\frac{1^5}{5})+(\frac{3}{2\cdot2^2}-\frac{3}{2\cdot1^2})=\frac{31}{20}-\frac{9}{8}=\frac{17}{40}=0.425$

б)

$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}} {\frac{dx}{cos^23x}} =\frac{1}{3}tg3x \bigg|\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{1}{3}(tg(3\cdot \frac{\pi}{3}})-tg(3\cdot \frac{\pi}{4}}))=\frac{1}{3}(tg(\pi)-tg( \frac{3 \pi}{4}}))=\\\\\frac{1}{3}(0-(-1))=\frac{1}{3}$

в)

$\int\limits^{-\frac{\pi}{3}}_{-\frac{\pi}{2}} {\frac{dx}{sin^2x}} =-ctg(x)\bigg|\limits^{-\frac{\pi}{3}}_{-\frac{\pi}{2}}=-(ctg(-\frac{\pi}{3})-ctg(-\frac{\pi}{2}))=-(-\frac{\sqrt{3}}{3}-0)=\frac{\sqrt{3}}{3}$

г)

$\int\limits^{1}_{-1} {5(x^2+1)} \, dx =5\int\limits^{1}_{-1} {(x^2+1)} \, dx=\\5\left[ \int\limits^{1}_{-1} {x^2} \, dx+\int\limits^{1}_{-1} {1} \, dx \right] =5\left[ \frac{x^3}{3}\bigg|\limits^{1}_{-1}+x\bigg|\limits^{1}_{-1} \right]=\frac{40}{3}$