Question

найти площадь фигуры ограниченной линиями y+x=1;y+3x=1;x=y;x=2y

Answer 1

$y+x=1\; ,\; \; y+3x=1\; ,\; \; x=y\; ,\; \; x=2y\\\\y=1-x\; ,\; \; y=1-3x\; ,\; \; y=x\; ,\; \; y=\frac{x}{2}\\\\Tochki\; \; peresecheniya:\; \; \left \{ {{y=1-x} \atop {y=x}} \right. \; \left \{ {{x=1-x} \atop {y=x}} \right. \; \left \{ {{x=\frac{1}{2}} \atop {y=\frac{1}{2}}} \right. \\\\\left \{ {{y=1-x} \atop {y=\frac{x}{2}}} \right.\; \left \{ {{\frac{x}{2}=1-x} \atop {y=\frac{x}{2}}} \right. \; \left \{ {{x=\frac{2}{3}} \atop {y=\frac{1}{3}}} \right.$

$\left \{ {{y=1-3x} \atop {y=x}} \right.\; \left \{ {{4x=1} \atop {y=x}} \right.\; \left \{ {{x=\frac{1}{4}} \atop {y=\frac{1}{4}}} \right. \\\\\left \{ {{y=1-3x} \atop {y=\frac{x}{2}}} \right.\; \left \{ {{x=\frac{2}{7}} \atop {y=\frac{1}{7}}} \right.\\\\S=\int\limits^{1/2}_{1/4}\, (x-\frac{x}{2})\, dx-\int\limits_{1/4}^{2/7}\, (1-3x-\frac{x}{2})\, dx+\int\limits^{2/3}_{1/2}\, (1-x-\frac{x}{2})\, dx=\\\\=\int\limits^{1/2}_{1/4}\, \frac{x}{2}\, dx-\int\limits^{2/7}_{1/4}\, (1-\frac{7x}{2})\, dx+\int\limits^{1/2}_{2/3}\, (1-\frac{3x}{2})\, dx=$

$=\frac{x^2}{4}\Big |_{1/4}^{1/2}+\frac{2}{7}\cdot \frac{(1-\frac{7x}{2})^2}{2}\Big |_{1/4}^{2/7}-\frac{2}{3}\cdot \frac{(1-\frac{3x}{2})^2}{2}\Big |_{1/2}^{2/3}=\\\\=\frac{1}{4}\cdot (\frac{1}{4}-\frac{1}{16})+\frac{2}{7}\cdot (0-\frac{1}{128})-\frac{2}{3}\cdot (0-\frac{1}{32})=\frac{11}{168}$