The area of the region bounded by the curve x² = 4y and the straight line x = 4y – 2 is

The correct option is 9/8 sq. units.

To find the area of the region bounded by the curve $x^2=4y$ and the straight line $x=4y-2$, we first need to determine their points of intersection.

From the equation of the line, we can express $4y$ in terms of $x$:
$4y=x+2$

Substitute this expression for $4y$ into the equation of the parabola $x^2=4y$:
$x^2=x+2$
$x^2-x-2=0$

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

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

The region is bounded between $x=-1$ and $x=2$. In this interval, the line lies above the parabola. Thus, the area $A$ is given by the integral:
$A={\int }_{-1}^2(y_{\text{line}}-y_{\text{parabola}})dx$

Expressing both curves in terms of $x$:
$y_{\text{line}}=\frac{x+2}{4}$
$y_{\text{parabola}}=\frac{x^2}{4}$

Set up the integral:
$A={\int }_{-1}^2(\frac{x+2}{4}-\frac{x^2}{4})dx$
$A=\frac{1}{4}{\int }_{-1}^2(2+x-x^2)dx$

Evaluate the integral:
$A=\frac{1}{4}{[2x+\frac{x^2}{2}-\frac{x^3}{3}]}_{-1}^2$

Substitute the upper limit ($x=2$):
${\text{Value}}_{\text{upper}}=2\left(2\right)+\frac{2^2}{2}-\frac{2^3}{3}=4+2-\frac{8}{3}=6-\frac{8}{3}=\frac{10}{3}$

Substitute the lower limit ($x=-1$):
${\text{Value}}_{\text{lower}}=2(-1)+\frac{{(-1)}^2}{2}-\frac{{(-1)}^3}{3}=-2+\frac{1}{2}+\frac{1}{3}=\frac{-12+3+2}{6}=-\frac{7}{6}$

Subtract the lower limit value from the upper limit value:
${[2x+\frac{x^2}{2}-\frac{x^3}{3}]}_{-1}^2=\frac{10}{3}-(-\frac{7}{6})=\frac{20}{6}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}$

Now, multiply by the factor $\frac{1}{4}$ outside the integral:
$A=\frac{1}{4}\times \frac{9}{2}=\frac{9}{8}\text{ sq. units}$