The area bounded by the parabolas $y^{2} = 5 x + 6$ and $x^{2} = y$

\frac{19}{5}

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\frac{21}{5}

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\frac{23}{5}

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\frac{27}{5}

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Solution

The correct option is B $\frac{27}{5}$
$y^{2} = 5 x + 6$ and $x^{2} = y$
Then
$x^{4} = 5 x + 6$
Or
$x^{4} - 5 x - 6 = 0$
$x = - 1$ is an obvious root.
Then
$x^{4} - 5 x - 6 = 0$
$(x + 1) (x - 2) (x^{2} + x + 3) = 0$
Now $x^{2} + x + 3 = 0$ has no real roots since $D < 0$ .
Hence we have to find the area between the curves for x=-1 to x=2.
Therefore
$A r e a$
$= | \int_{- 1}^{2} x^{2} - \sqrt{5 x + 6} . d x |$
$= | [\frac{x^{3}}{3} - \frac{2}{15} (5 x + 6)^{\frac{3}{2}}]_{- 1}^{2} |$
$= | \frac{8}{3} - \frac{128}{15} - [\frac{- 1}{3} - \frac{2}{15}] |$
$= | \frac{9}{3} - \frac{126}{15} |$
$= \frac{126 - 45}{15}$
$= \frac{81}{15}$
$= \frac{27}{5}$ .

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