Question

Determine a função z= z(x), x ∈ R, tal que (dz)/(dx) = 1/2x + 3 e z(- 1) = 0

Answer 1

$\displaystyle \sf \frac{dz}{dx} = \frac{1}{2}\cdot x + 3 \\\\ dz = \frac{1}{2}\cdot x\ dx + 3 \ dx \\\\ \int dz = \int \frac{1}{2}\cdot x\ dx + \int 3 \ dx \\\\\\ z = \frac{1}{2}\cdot \frac{x^2}{2}+3x + C\\\\\\ z(x) = \frac{1}{2}\cdot \frac{x^2}{2}+3x + C \\\\ z(-1) = 0\ \therefore \ \frac{1}{2}\cdot \frac{(-1)^2}{2} + 3 \cdot (-1) + C = 0$

$\displaystyle \sf C + \frac{1}{4}-3= 0 \\\\ C = 3-\frac{1}{4} \\\\ C = \frac{12-1}{4} \to C = \frac{11 }{4} \\\\ Da{\'i}} : \\\\ z(x) = \frac{1}{2}\cdot\frac{x^2}{2}+3x+\frac{11}{4} \\\\ \huge\boxed{\sf \ z(x) = \frac{x^2}{4}+3x+\frac{11}{4}\ } \checkmark$