The gradient function of a curve is given by the expression 2x + 1. If the curve passes through the point (3, 10), (a) Find: (i) The equation of the curve (ii) The values of x, at which the curve cuts the x-axis (b) Determine the area enclosed by the curve and the x-axis.

$$\frac{dy}{dx} = 2x + 1$$ $$y = \int (2x + 1) dx$$ $$y = \frac{2x^2}{2} + x + C$$ $$y = x^2 + x + C$$

Badilisha $x=1$ na $y=4$ katika equation:

$$4 = (1)^2 + (1) + C$$ $$4 = 1 + 1 + C$$ $$4 = 2 + C$$ $$C = 4 - 2$$ $$C = 2$$

Equation ya curve ni $\boxed{y = x^2 + x + 2}$.

a) ii) Tafuta thamani za $x$, ambapo curve inakata x-axis.

Weka $y=0$ katika equation ya curve:

$$x^2 + x + 2 = 0$$

Tumia discriminant $\Delta = b^2 - 4ac$. Kwa equation hii, $a=1$, $b=1$, $c=2$.

$$\Delta = (1)^2 - 4(1)(2)$$ $$\Delta = 1 - 8$$ $$\Delta = -7$$

Kwa kuwa discriminant ($\Delta$) ni hasi ($\Delta < 0$), equation haina mizizi halisi. Hii inamaanisha curve $\boxed{\text{haikati x-axis}}$.

b) Bainisha eneo lililofungwa na curve na x-axis.

Kutokana na matokeo ya sehemu (a) (ii), curve $y = x^2 + x + 2$ haikati x-axis. Kwa kuwa coefficient ya $x^2$ ni chanya ($1 > 0$), parabola inafunguka juu na iko juu ya x-axis kabisa.
Kwa hivyo, $\boxed{\text{hakuna eneo lililofungwa na curve na x-axis}}$.