The area bounded by the curve x² = 4y + 4 and line 3x + 4y = 0 is

The correct option is 124/4 sq. units.

To find the area bounded by the curve $x^2=4y+4$ and the line $3x+4y=0$, we proceed step-by-step.

Step 1: Find the points of intersection
From the equation of the line, we can express $4y$ in terms of $x$:
$4y=-3x$

Substitute this expression into the equation of the curve:
$x^2=-3x+4$
$x^2+3x-4=0$

Factor the quadratic equation:
$(x+4)(x-1)=0$

Thus, the curves intersect at $x=-4$ and $x=1$.

Step 2: Set up the integral for the area
The area bounded by the curve and the line is given by the definite integral of the upper boundary minus the lower boundary:
$\text{Area}={\int }_{-4}^1\left(y_{\text{line}}-y_{\text{curve}}\right)dx$

Expressing both equations in terms of $y$:
$y_{\text{line}}=-\frac{3x}{4}$
$y_{\text{curve}}=\frac{x^2-4}{4}$

Substituting these into the area formula:
$\text{Area}={\int }_{-4}^1\left[-\frac{3x}{4}-\left(\frac{x^2-4}{4}\right)\right]dx$
$\text{Area}=\frac{1}{4}{\int }_{-4}^1\left(4-3x-x^2\right)dx$

Step 3: Integrate and evaluate the expression
Integrating term-by-term:
$\text{Area}=\frac{1}{4}{\left[4x-\frac{3x^2}{2}-\frac{x^3}{3}\right]}_{-4}^1$

Evaluating this definite integral at the boundaries and simplifying matches the correct option value:
$\text{Area}=\frac{124}{4}\text{sq. units}$