The area enclosed by the parabola x² = 6y and the line x - 6y +2 = 0 is:

The correct option is 3/4.

To find the area enclosed by the parabola $x^2=6y$ and the line $x-6y+2=0$, we first determine their points of intersection.

From the equation of the line, we can express $6y$ in terms of $x$:
$6y=x+2$

Substituting $6y=x^2$ from the parabola's equation into this relation gives:
$x^2=x+2$
$x^2-x-2=0$

Solving this quadratic equation by factoring:
$(x-2)(x+1)=0$

This gives the x-coordinates of the intersection points as:
$x=-1$ and $x=2$

The area $A$ enclosed between the line and the parabola is given by the integral of the upper curve (the line) minus the lower curve (the parabola) from $x=-1$ to $x=2$:
$A={\int }_{-1}^2\left(y_{\text{line}}-y_{\text{parabola}}\right)dx$

Expressing $y$ in terms of $x$ for both curves:
$y_{\text{line}}=\frac{x+2}{6}$
$y_{\text{parabola}}=\frac{x^2}{6}$

Setting up the integral:
$A={\int }_{-1}^2\left(\frac{x+2}{6}-\frac{x^2}{6}\right)dx$
$A=\frac{1}{6}{\int }_{-1}^2\left(2+x-x^2\right)dx$

Integrating the terms:
$A=\frac{1}{6}{\left[2x+\frac{x^2}{2}-\frac{x^3}{3}\right]}_{-1}^2$

Evaluating the expression at the upper limit $x=2$:
$V_{\text{upper}}=2\left(2\right)+\frac{2^2}{2}-\frac{2^3}{3}=4+2-\frac{8}{3}=6-\frac{8}{3}=\frac{10}{3}$

Evaluating the expression at the lower limit $x=-1$:
$V_{\text{lower}}=2(-1)+\frac{{(-1)}^2}{2}-\frac{{(-1)}^3}{3}=-2+\frac{1}{2}+\frac{1}{3}=\frac{-12+3+2}{6}=-\frac{7}{6}$

Subtracting the lower limit value from the upper limit value:
$A=\frac{1}{6}\left(\frac{10}{3}-\left(-\frac{7}{6}\right)\right)$
$A=\frac{1}{6}\left(\frac{10}{3}+\frac{7}{6}\right)$
$A=\frac{1}{6}\left(\frac{20}{6}+\frac{7}{6}\right)$
$A=\frac{1}{6}\left(\frac{27}{6}\right)$
$A=\frac{27}{36}$

Simplifying the fraction by dividing the numerator and denominator by 9:
$A=\frac{3}{4}$

Thus, the area enclosed by the parabola and the line is 3/4 square units.