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+1 vote
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in Definite Integrals by (55 points)
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Find the area bounded by the curve 2y² + 2y -x - 2 = 0 and the line x = 2y.

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1 Answer

+1 vote
by (710 points)

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.

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