ДЗ №6 | Застосування первісної

$$F\left(x\right)=-\frac{1}{x}-3x^2+2x+C$$   $$0=-1-3+2+C$$  

$$C=2$$  

$$F\left(x\right)=-\frac{1}{x}-3x^2+2x+2$$  

$$S=\int_1^4\left(x^2+1\right)dx=\left(\frac{x^3}{3}+x\right)\int_1^4=\left(\frac{4}{3}+4\right)-\left(\frac{1}{3}+1\right)=24$$  

$$S=\int_0^3\left(9-x^2\right)dx=2\int_0^3\left(9-x^2\right)dx=36$$  

$$S=\int_{-2}^1\left(x^2+2\right)dx=\left(\frac{x^3}{3}+2x\right)\int_{-2}^1=\frac{1}{3}+2-\left(-\frac{8}{3}-4\right)=9$$  

$$S=S_1+S_2$$  

$$S_1=\int_{-3}^0\left(0,5x^2+2x+3\right)dx=\left(\frac{x^3}{6}+x^2+3x\right)\int_{-3}^0=-\left(-\frac{27}{6}+9-9\right)=\frac{27}{6}=4\ \frac{1}{2}$$   $$S_2=\int_0^2\left(3-x\right)dx=\left(3x-\frac{x^2}{2}\right)\int_0^2=6-2=4$$  

$$S=4,5+4=9,5$$  

$$V=\int_0^1\pi\left(x^2+1\right)^2dx=\pi\int_0^1\left(x^4+2x^2+1\right)dx=\pi\left(\frac{x^5}{5}+\frac{2x^3}{3}+x\right)\int_0^1=\pi\left(\frac{1^5}{5}+\frac{2^5}{3}+1\right)=1\ \frac{13}{15}\pi$$  

$$V=\pi\int_0^1\left(\sqrt[]{x}\right)^2dx=\pi\int_0^1xdx=\pi\left(\frac{x^2}{2}\right)\int_0^1=\frac{1}{2}\pi=\frac{\pi}{2}$$  

$$V=\pi\int_0^1\left(\left(x+3\right)^2-\left(2x\right)^2\right)dx=\pi\int_0^1\left(\left(x+3\right)^2-4x^2\right)dx=\pi\left(\frac{\left(x+3\right)^2}{3}-\frac{4x^3}{3}\right)\int_0^1=\pi\left(\left(\frac{64}{3}-\frac{4}{3}\right)-\left(\frac{27}{3}-0\right)\right)=\pi\left(20-9\right)=11\pi$$  

Шуканий шлях знайдемо за допомогою формули Ньютона-Лейбніца:

$$S=\int_1^3\left(3t^2+4t+1\right)dt=\left(t^3+2t^2+t\right)\int_1^3=\left(27+18+3\right)-\left(1+2+1\right)=48-4=44$$