The area bounded by the curve $x^{2} + 2 x + y - 3 = 0,$ the $x -$ axis and the tangent at the point, where it meets the $y -$ axis is

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Solution

The equation of the curve is $x^{2} + 2 x + y - 3 = 0$
or $x^{2} + 2 x + 1 + y - 3 - 1 = 0$ (add and subtract $1$ )
$\Rightarrow y - 4 = - {(x + 1)}^{2}$
The curve meets the $y -$ axis at $(0, 3)$ and the $x -$ axis at $(- 3, 0)$ and $(1, 0)$
The tangent at $(0, 3)$ is $y - 3 = - 2 (x - 0)$
The required area $=$ area of $△ O A B -$ the shaded area $(O C B)$
$= \frac{1}{2} . \frac{3}{2} .3 - \int_{0}^{1} (4 - {(x + 1)}^{2}) d x$
$= \frac{9}{4} - {[4 x - \frac{{(x + 1)}^{3}}{3}]}_{0}^{1}$
$= \frac{9}{4} - [(4 - \frac{8}{3}) + \frac{1}{3}]$
$= \frac{9}{4} - 4 + \frac{7}{3} = \frac{7}{12}$ .sq.unit

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