One of the points of intersection of the curves y = 1 + 3x – 2x2 and Let the area of the region enclosed by these curves be where ℓ, m, n ∈ N. Then ℓ + m + n is equal to

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One of the points of intersection of the curves y = 1 + 3x – 2x² and $y = \frac{1}{x} is (\frac{1}{2}, 2)$ Let the area of the region enclosed by these curves be $\frac{1}{24} (ℓ \sqrt{5} + m) - {nlog}_{e} (1 + \sqrt{5})$ where ℓ, m, n ∈ N. Then ℓ + m + n is equal to

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Solving curves $y = 1 + 3 x - 2 x^{2} and y = \frac{1}{x}$
2x³ – 3x² – x + 1 = 0
(2x–1) (x²–x–1) =

$x = \frac{1}{2}, x = \frac{1 \pm \sqrt{5}}{2}$

$Area = \int_{\frac{1}{2}}^{\frac{\sqrt{5} + 1}{2}} (1 + 3 x - 2 x^{2} - \frac{1}{x}) dx$

$= {(x + \frac{3 x^{2}}{2} - \frac{2 x^{3}}{3} - lnx)|}_{1 / 2}^{\frac{\sqrt{5} + 1}{2}}$

$= \frac{\sqrt{5} + 1}{2} + \frac{3}{8} (\sqrt{5} + 1)^{2} - \frac{1}{12} (\sqrt{5} + 1)^{3} - l n (\frac{\sqrt{5} + 1}{2}) - (\frac{1}{2} + \frac{3}{8} - \frac{1}{12} - l n \frac{1}{2})$ $= \frac{1}{24} [12 (\sqrt{5} + 1) + 9 (\sqrt{5} + 1)^{2} - 2 (\sqrt{5} + 1)^{3} - 12 - 9 + 2] - \ln (\frac{\sqrt{5} + 1}{2} \times 2)$

$= \frac{1}{24} [12 (\sqrt{5} + 1) + 9 (6 + 2 \sqrt{5}) - 2 (5 \sqrt{5} + 1 + 3 \sqrt{5} (\sqrt{5} + 1)) - 19] - \ln (\sqrt{5} + 1)$

$= \frac{1}{24} [14 \sqrt{5} + 15] - \ln (\sqrt{5} + 1)$

Now, ℓ = 14, m = 15, n = 1 then ℓ + m + n = 30

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