Вычисли площадь фигуры, ограниченной линиями: y=1/6x, x–9y+9=0, x=13
S= ∫ ba(f(x)–g(x))dx a=0 b=13 f(x)=(1/9)x–1 g(x)=(1/6)x S= ∫ 130((1/9)(x)–1)–(1/6)xdx= =((1/9)·(x2/2) – x –(1/6)x/ln(1/6))|130= =(1/18)·(132–13–(1/6)13/ln(1/6)–(1/6)0/ln(1/6)= =(169/18)–13+(1/ln6)·(1–(1/6)13)