С2. Найдем пределы интегрирования
x^2 + 1 = 3 - x^2
2x^2 = 2
x^2 = 1; x1 = -1; x2 = 1
![S= \int\limits^1_{-1} {[(3-x^2)-(x^2+1) ]} \, dx =\int\limits^1_{-1} {(2-2x^2)}\, dx =(2x-2* \frac{x^3}{3} )|^1_{-1}=](https://tex.z-dn.net/?f=S%3D+%5Cint%5Climits%5E1_%7B-1%7D+%7B%5B%283-x%5E2%29-%28x%5E2%2B1%29+%5D%7D+%5C%2C+dx+%3D%5Cint%5Climits%5E1_%7B-1%7D+%7B%282-2x%5E2%29%7D%5C%2C+dx+%3D%282x-2%2A+%5Cfrac%7Bx%5E3%7D%7B3%7D+%29%7C%5E1_%7B-1%7D%3D "S= \int\limits^1_{-1} {[(3-x^2)-(x^2+1) ]} \, dx =\int\limits^1_{-1} {(2-2x^2)}\, dx =(2x-2* \frac{x^3}{3} )|^1_{-1}=")
C3. f(x) = √x; x0 = 1; f(x0) = f(1) = √1 = 1
f ' (x) = 1/(2√x); f ' (x0) = 1/(2√1) = 1/2
Уравнение касательной в точке
y(x) = f(x0) + f ' (x0)*(x - x0) = 1 + 1/2*(x - 1) = 1 + x/2 - 1/2 = x/2 + 1/2
Ответ: y(x) = 0,5x + 0,5