???? = 0,25????^(2) + 2,
5???? − 8???? + 14 = 0, вокруг оси ????????
2x^2-5x+2=0
x_(1)=1/2; x_(2)=2
[m]V=π ∫^{2} _{\frac{1}{2}}(\frac{5x+14}{8})^2dx-π ∫^{2} _{\frac{1}{2}} (0,25x^2+2))^2dx=\frac{8}{5}π∫^{2} _{\frac{1}{2}}(\frac{5x+14}{8})^2d\frac{5x+14}{8}-π ∫^{2} _{\frac{1}{2}} (\frac{1}{16}x^4+x^2+4)dx=[/m]
[m]=\frac{8}{5}π\cdot (\frac{(\frac{5x+14}{8})^3}{3}|^{2} _{\frac{1}{2}}-π\cdot (\frac{1}{16}\cdot \frac{x^5}{5}+\frac{x^3}{3}+4x)|^{2} _{\frac{1}{2}}=\frac{4977}{512}π-\frac{23103}{2560}π=\frac{4977\cdot 5-23103}{2560}π[/m]