нужно найти точки пересечения:
sqrt(2x+1) = x-1
2x+1 = x^2 - 2x + 1
x^2 - 4x = 0
x(x - 4) = 0
x = 0 x = 4
![S = \int\limits^4_0 {\sqrt{2x+1}-(x -1)} \, dx = [\frac{1}{3}(2x+1)^{\frac{3}{2}} - \frac{x^2}{2} + x] |_0^4=\\ = [\frac{1}{3}(2*4+1)^{\frac{3}{2}} - \frac{4^2}{2} + 4] - [\frac{1}{3}(2*0+1)^{\frac{3}{2}} - \frac{0^2}{2} + 0] =\\ = 9 - 8 +4 - \frac{1}{3} = 14/3](/tpl/images/0120/7401/a1a08.png)
