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

Solution :

$$\eqalign{ & \because {\text{Curve is given as}}: \cr & y = {x^2} - 1 \cr & \Rightarrow \frac{{dy}}{{dx}} = 2x \cr & \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {2,\,3} \right)}} = 4 \cr & \therefore {\text{Equation of tangent at }}\left( {2,{\text{ }}3} \right) \cr & \left( {y - 3} \right) = {\text{4}}\left( {x - 2} \right) \cr & \Rightarrow y = 4x - 5 \cr & {\text{but }}x = 0 \cr & \Rightarrow y = - 5 \cr} $$
Here the curve cuts $$y$$-axis
$$\eqalign{ & \therefore {\text{Required area:}} \cr & = \frac{1}{4}\int\limits_{ - 5}^3 {\left( {y + 5} \right)dy} - \int\limits_{ - 1}^3 {\sqrt {y + 1} \,dy} \cr & = \frac{1}{4}\left[ {\frac{{{y^2}}}{2} + 5y} \right]_{ - 5}^3\frac{{ - 2}}{3}\left[ {{{\left( {y + 1} \right)}^{\frac{3}{2}}}} \right]_{ - 1}^3 \cr & = \frac{1}{4}\left[ {\frac{9}{2} + 15 - \frac{{25}}{2} + 25} \right] - \frac{2}{3}\left[ {{4^{\frac{3}{2}}} - 0} \right] \cr & = \frac{{32}}{4} - \frac{{16}}{3} \cr & = \frac{8}{3}{\text{ sq}}{\text{. units}} \cr} $$