начиная со 2,решите пожалуйста. Created by nictooo. algebra-ru

начиная со 2,решите пожалуйста...

Twyla27

Ответ:

Объяснение:

$2)\; \; f(x)=x^{-3}\\\\F(x)=\int x^{-3}\, dx=\frac{x^{-2}}{-2}+C=-\frac{1}{2x^2} +C\; ;\\\\M(-1,0)\; :\; \; F(-1)=0\; \; \to \; \; \frac{1}{2(-1)^2}+C=0\; ,\; \; C=\frac{1}{2}\\\\F(x)=-\frac{1}{2x^2}+\frac{1}{2}\; .$

$3)\; \; f(x)=2-\frac{1}{cos^2x}\; ,\; x\in [0,\frac{\pi}{2})\; ,\\\\F(x)=\int (2-\frac{1}{cos^2x})\, dx=2x-tgx+C\; ;\\\\M(\frac{\pi}{4};\frac{\pi}{2})\; :\; \; F(\frac{\pi}{4})=\frac{\pi}{2}\; \; \to \; \; 2\cdot \frac{\pi}{4}-tg\frac{\pi}{4}+C=\frac{\pi}{2}\; ,\; \; \frac{\pi}{2}-1+C=\frac{\pi}{2}\; ,\; C=1\\\\F(x)=2x-tgx+1\; .$

$4)\; \; f(x)=\frac{2}{sin^2x}+1\; ,\\\\F(x)=\int (\frac{2}{sin^2x}+1)\, dx=-2\cdot ctgx+x+C\; ;\\\\M(\frac{\pi}{4};\frac{\pi}{4})\; :\; \; F(\frac{\pi}{4})=\frac{\pi}{4}\; \; \to \; \; -2\, ctg\frac{\pi}{4}+\frac{\pi}{4}+C=\frac{\pi }{4}\; ,\; \; -2+\frac{\pi}{4}+C=\frac{\pi}{4}\; ,\; C=2\\\\F(x)=-2\, ctgx+x+2$

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