Given curve is
2y2 + 2y – x – 2 = 0
⇒ 2(y2 + y) = x + 2
⇒ (y2 + y + \(\frac{1}{4}\) – \(\frac{1}{4}\)) \(=\frac{\mathrm x+2}{2}\)
⇒ \(\left(y+\frac{1}{2}\right)^2\) \(=\frac{\mathrm x+2}{2}+\frac{1}{4}\)
⇒ \(\left(y+\frac{1}{2}\right)^2\) \(=\frac{2\mathrm x+5}{4}\)
Hence, given curve is parabola whose vertex is \(\left(\frac{-5}{2}, \frac{-1}{2}\right)\).
⇒ \(y=-\frac{1}{2}+\frac{1}{2}\sqrt{2\mathrm x+5}\)
For intersection point of given curve and line x = 2y, y = \(\frac{\mathrm x}{2}\)
we have 2y2 + x – x –2 = 0
⇒ 2y2 – 2 = 0
⇒ y2 = 1
⇒ y = \(\pm\)1
Hence, intersection point are (–2, –1) & (2, 1).
The intersection point of curve and x-axis is (–2, 0).
For the intersection point of curve and y-axis, we put x = 0, then we get
2y2 + 2y – 2 = 0
⇒ y2 + y – 1 = 0
⇒ \(y=\frac{-1\pm \sqrt{1+4}}{2}\)
\(=\frac{1\pm \sqrt 5}{2}\)
Required area
\(=\displaystyle\int^{-2}_{\frac{-5}{2}}\)\(\Big[\left(\frac{-1}{2}+\frac{1}{2}\sqrt{2\mathrm x+5}\right)\)\(-\left(\frac{-1}{2}-\frac{1}{2}\sqrt{2\mathrm x+5}\right)\Big]d\mathrm x\) \(+\displaystyle\int ^2_{-2}\)\(\left(\frac{-1}{2}+\frac{1}{2}\sqrt{2\mathrm x+5}-\frac{\mathrm x}{2}\right)d\mathrm x\)
\(=\displaystyle\int ^{-2}_{\frac{-5}{2}}\)\(\sqrt{2\mathrm x+5}\,d\mathrm x\) \(+\left[\frac{-\mathrm x}{2}-\frac{\mathrm x^2}{4}+\frac{1}{4}\times \frac{2}{3}(2\mathrm x+5)^{\frac{3}{2}}\right]^2_{-2}\) \(\left(\because \int \sqrt{a\mathrm x}\,d\mathrm x=\frac{1}{a}\frac{2}{3}\mathrm x^{\frac{3}{2}}\right)\)
\(=\frac{1}{2}\times \frac{2}{3}\left[(2\mathrm x+5)^{\frac{3}{2}}\right]^{-2}_{-\frac{5}{2}}\) \(-\frac{1}{2}\left(2+2\right)\) \(-\frac{1}{4}(4-4)+\frac{1}{6}(27-1)\)
\(=\frac{1}{3}(1-0)-2-0+\frac{1}{6}\times 26\)
\(=\frac{1}{3}-2+\frac{13}{3}\)
\(=\frac{14}{3}-2\)
\(=\frac{14-6}{3}\)
\(=\frac{8}{3}\)
Hence, area bounded by given curve and line is \(\frac{8}{3}\) square unit.