Question

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

Answer 1

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

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

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