The slope of the tangent to a curve y = f(x) at [x, f(x)] is 2x + 1. If the curve passes through the point (1, 2), then area bounded by the curve, x-axis and the line x = 1 is

$\frac{5}{6}$

$\frac{d y}{d x}=2 x+1 \Rightarrow y=x^2+x+c$

It passes through (1, 2) ∴ c = 0

∴  $y+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2$

It is a parabola with vertex $\left(-\frac{1}{2},-\frac{1}{4}\right)$

∴ Required area

$=\int\limits_0^1 y d x=\int\limits_0^1\left(x^2+x\right) d x=\frac{5}{6}$.

Hence (1) is the correct answer.