The area (in sq. units) bounded by the parabola $$y = x^2 - 1$$, the tangent at the point (2, 3) to it and the $$y$$-axis is:

We have the parabola whose equation is $$y = x^{2}-1$$. To write the equation of the tangent at the point $$(2,3)$$ on this parabola, we first need the slope of the tangent.

For a curve $$y = f(x)$$, the slope of the tangent at any point is given by the derivative $$\dfrac{dy}{dx}$$. Here

$$\dfrac{dy}{dx} = \dfrac{d}{dx}(x^{2}-1) = 2x.$$

At the point $$(2,3)$$ we substitute $$x = 2$$ to get

$$m = 2(2) = 4.$$

The point-slope form of a tangent is $$y - y_{1} = m(x - x_{1})$$. Substituting $$(x_{1},y_{1}) = (2,3)$$ and $$m = 4$$, we obtain

$$y - 3 = 4\,(x - 2).$$

Expanding,

$$y - 3 = 4x - 8$$

$$y = 4x - 5.$$

Thus, the tangent line is $$y = 4x - 5$$. Observe that it cuts the $$y$$-axis (where $$x = 0$$) at the point $$(0,-5)$$. The parabola meets the $$y$$-axis at $$(0,-1)$$. Hence the vertical line $$x = 0$$ together with the curves

$$y = x^{2}-1 \quad\text{and}\quad y = 4x - 5$$

encloses the required region. Between $$x = 0$$ and $$x = 2$$ (the $$x$$-coordinate of the point of tangency) the parabola lies above the tangent, because

$$\bigl(x^{2}-1\bigr) - (4x - 5) = x^{2} - 4x + 4 = (x - 2)^{2} \ge 0.$$

Therefore, the area $$A$$ is obtained by integrating the vertical difference “parabola minus tangent’’ from $$x = 0$$ to $$x = 2$$:

$$A = \int_{0}^{2} \Bigl[(x^{2}-1) - (4x - 5)\Bigr]\,dx = \int_{0}^{2} \bigl(x^{2} - 4x + 4\bigr)\,dx.$$

Now we integrate term-by-term:

$$\int x^{2}\,dx = \frac{x^{3}}{3}, \qquad \int (-4x)\,dx = -2x^{2}, \qquad \int 4\,dx = 4x.$$

Putting the limits $$0$$ to $$2$$, we get

$$A = \left[\frac{x^{3}}{3} - 2x^{2} + 4x\right]_{0}^{2} = \left(\frac{2^{3}}{3} - 2(2)^{2} + 4(2)\right) - \left(\frac{0^{3}}{3} - 2(0)^{2} + 4(0)\right).$$

Simplifying the expression at $$x = 2$$:

$$\frac{2^{3}}{3} = \frac{8}{3}, \quad -2(2)^{2} = -2 \times 4 = -8, \quad 4(2) = 8.$$

Adding these three values,

$$\frac{8}{3} - 8 + 8 = \frac{8}{3} + 0 = \frac{8}{3}.$$

The value at $$x = 0$$ is clearly $$0$$, so

$$A = \frac{8}{3}\ \text{square units}.$$

Hence, the correct answer is Option B.