Verified answer

Resposta:

Opção 3

Explicação passo a passo:

[tex]\frac{dy}{dx} = 2x^3+3x^2+4\\\\dy = (2x^3+3x^2+4)dx\\\\\int\limits^a_b {} \, dy =\int\limits^a_b {(2x^3+3x^2+4)} \, dx \\\\y = \int\limits^a_b {2x^3} \, dx +\int\limits^a_b {3x^2} \, dx +\int\limits^a_b {4} \, dx \\\\y = \frac{x^4}{2} +x^3+4x + C[/tex]

[tex]f(x) = \frac{x^4}{2} +x^3+4x + C[/tex]

f(0) = -2

[tex]\frac{0^4}{2}+0^3+4*0+C=-2\\\\ C = -2[/tex]

A função é

[tex]f(x) = \frac{x^4}{2} +x^3+4x -2[/tex]

f(-1) = ?

[tex]f(-1) = \frac{(-1)^4}{2} +(-1)^3+4 * (-1)x -2\\\\f(-1) = \frac{1}{2} -1-4-2\\\\f(-1) = \frac{1}{2} -7\\\\f(-1) = -\frac{13}{2}[/tex]

Opção 3