Verified answer

Объяснение:

№2.

[tex]\displaystyle\\f(x)=2+4x-3x^2\\\\F(x)=\int {(2+4x-3x^2)} \, dx =\int2dx+\int4xdx-\int3x^2dx=\\\\=2x+4\frac{x^2}{2} -3\frac{x^3}{3} +C=2x+2x^2-x^3+C.\\\\A(2;4) \ \ \ \ \Rightarrow\\4=2*2+2*2^2-2^3+C\\\\4=4+8-8+C\\\\4=4+C\\\\C=0\ \ \ \ \Rightarrow\\\\F(x)=2x+2x^2-x^3.[/tex]

№3.

[tex]\displaystyle\\f(x)=2x+2\\\\F(x)=\int(2x+2)dx=\int2xdx+\int2dx=2\frac{x^2}{2} +2x+C=x^2+2x+C.\\\\(1;4)\ \ \ \ \Rightarrow\\\\4=1^2+2*1+C\\\\4=1+2+C\\\\4=3+C\\\\C=1\ \ \ \ \Rightarrow\\\\F(x)=x^2+2x+1\\\\[/tex]

Ответ: Б. F(x)=x²+2x+1.

№4. (а)

[tex]\displaystyle\\f(x)=1+x^2\\\\F(x)=\int(1+x^2)dx=\int1dx+\int x^2dx=x+\frac{x^3}{3} +C\\\\(-3;9)\ \ \ \ \Rightarrow\\\\9=-3+\frac{(-3)^2}{3} +C\\\\9=-3+\frac{9}{3} +C\\\\9=-3+3+C\\\\C=9\ \ \ \ \Rightarrow\\\\F(x)=\frac{x^3}{3}+x+9.[/tex]