Given curve and tangent

The parabola is $y = x^2 - 1.$

We need the tangent to this parabola at the point $(2,3)$.

Equation of the tangent

Differentiate: $\frac{dy}{dx} = 2x.$

At $x=2$, slope is $m = 2(2)=4.$

So the tangent at $(2,3)$ is $y-3 = 4(x-2).$ Simplifying, $y = 4x - 5.$

Region bounded by the three curves/lines

The boundaries are:

• parabola: $y=x^2-1$
• tangent: $y=4x-5$
• y-axis: $x=0$

The tangent touches the parabola at $x=2$.

On the interval $0 \le x \le 2$, compare the two: $(x^2-1) - (4x-5) = x^2-4x+4 = (x-2)^2 \ge 0.$

Hence, for $0 \le x \le 2$, the parabola lies above the tangent.

Area between the curves from $x=0$ to $x=2$

Therefore area is $A = \int_0^2 \Big[(x^2-1) - (4x-5)\Big]dx.$

Simplify the integrand: $A = \int_0^2 (x^2 -4x +4)dx = \int_0^2 (x-2)^2 dx.$

Evaluate the integral

$A = \int_0^2 (x^2-4x+4)dx$ $= \left[\frac{x^3}{3} - 2x^2 + 4x\right]_0^2$ $= \left(\frac{8}{3} - 8 + 8\right) - 0$ $= \frac{8}{3}.$

Match with options

$\boxed{\frac{8}{3}}$

So the correct option is C.