Question

Обчислити площу фігури, обмежену лініями ,y=x(2) і y=4x-+3

Answer 1

$y=x^2 \\ y=4x-3 x^2=4x-3 \\ x^2-4x+3=0 \\ D=b^2-4ac=(-4)^2-4*1*3=16-12=4 \\ \\ x_{1} = \frac{-b+ \sqrt{D} }{2a} = \frac{4+2}{2} = \frac{6}{2} =3 \\ \\ x_{2} = \frac{-b- \sqrt{D} }{2a} = \frac{4-2}{2} = \frac{2}{2} =1$

$\int\limits^{3}_{1} {(-3+4x-x^2)} \, dx =-3x+4* \frac{x^2}{2} - \frac{x^3}{3} |^{3}_{1}= \\ \\ =-3x+2x^2 - \frac{x^3}{3} |^{3}_{1}=(-3*3+2*3^2 - \frac{3^3}{3})-(-3*1+2*1^2 - \frac{1^3}{3})= \\ \\ =(-9+2*9 - \frac{27}{3})-(-3+2*1 - \frac{1}{3})= \\ \\ =(-9+18 -9)-(-3+2 - \frac{1}{3})=(18-18)-(-1 - \frac{1}{3})=-(-\frac{4}{3})= \\ \\ =\frac{4}{3}=1,3333$