The area bounded by the curve y = x² – 1 and the straight line x + y = 3 is

The correct option is 17√17/6 sq. units.

To find the area bounded by the curve $y=x^2-1$ and the straight line $x+y=3$, we first need to determine their points of intersection by setting their equations equal to each other.

From the equation of the line, we can express $y$ in terms of $x$:
$y=3-x$

Now, equating the two expressions for $y$:
$x^2-1=3-x$

Rearranging this into a standard quadratic equation form:
$x^2+x-4=0$

We can find the roots of this quadratic equation using the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, where $a=1$, $b=1$, and $c=-4$:
$x=\frac{-1\pm \sqrt{1^2-4\left(1\right)(-4)}}{2\left(1\right)}=\frac{-1\pm \sqrt{1+16}}{2}=\frac{-1\pm \sqrt{17}}{2}$

Let the two intersection points have $x$-coordinates:
$\alpha =\frac{-1-\sqrt{17}}{2}$ and $\beta =\frac{-1+\sqrt{17}}{2}$

The bounded area is located between these limits. In this interval, the straight line $y=3-x$ lies above the parabola $y=x^2-1$. Therefore, the area $A$ is given by the definite integral:
$A={\int }_{\alpha }^{\beta }\left[(3-x)-(x^2-1)\right]dx={\int }_{\alpha }^{\beta }(4-x-x^2)dx$

Integrating the terms:
$A={\left[4x-\frac{x^2}{2}-\frac{x^3}{3}\right]}_{\alpha }^{\beta }$

Evaluating this expression yields:
$A=4(\beta -\alpha )-\frac{{\beta }^2-{\alpha }^2}{2}-\frac{{\beta }^3-{\alpha }^3}{3}$

We can simplify this using algebraic relations of the roots $\alpha$ and $\beta$ of the equation $x^2+x-4=0$:
$\alpha +\beta =-1$
$\beta -\alpha =\sqrt{17}$

Let's find the values of the components:
${\beta }^2-{\alpha }^2=(\beta -\alpha )(\beta +\alpha )=\left(\sqrt{17}\right)(-1)=-\sqrt{17}$

For the cubic difference term:
${\beta }^3-{\alpha }^3=(\beta -\alpha )({\alpha }^2+\alpha \beta +{\beta }^2)=(\beta -\alpha )\left[{(\alpha +\beta )}^2-\alpha \beta \right]$

Since the product of the roots is $\alpha \beta =-4$:
${\beta }^3-{\alpha }^3=\sqrt{17}\left[{(-1)}^2-(-4)\right]=\sqrt{17}(1+4)=5\sqrt{17}$

Substitute these values back into the area equation:
$A=4\sqrt{17}-\frac{-\sqrt{17}}{2}-\frac{5\sqrt{17}}{3}$

Now, factor out $\sqrt{17}$ and find a common denominator of 6:
$A=\sqrt{17}\left(4+\frac{1}{2}-\frac{5}{3}\right)=\sqrt{17}\left(\frac{24+3-10}{6}\right)=\frac{17\sqrt{17}}{6}$

Thus, the area bounded by the curves is $\frac{17\sqrt{17}}{6}$ square units.