One of the points of intersection of the curves y=1+3x-2x^2 and y=(1)/(x) is ((1)/(2),2).

Question

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

Accepted Answer

Solving curves $y=1+3x-2{x}^{2}\&y=\frac{1}{x}$

$\begin{gathered} 2{x}^{3}-3{x}^{2}-x+1=0 \ \Rightarrow \, (2x-1)({x}^{2}-x-1)=0 \ \Rightarrow \, x=\frac{1}{2},x=\frac{1\pm \sqrt{5}}{2} \end{gathered}$

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

$=[x+\frac{3{x}^{2}}{2}-\frac{2x}{3}-\ln x]$

$=\frac{\sqrt{5}+1}{2}+\frac{3}{8}(\sqrt{5}+1{)}^{2}-\frac{1}{12}(\sqrt{5}+1)-\ln (\frac{\sqrt{5}+1}{2})-(\frac{1}{2}+\frac{3}{8}-\frac{1}{12}-\ln \frac{1}{2})$

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

$\begin{gathered} = \frac{1}{24}[12(\sqrt{5}+1)+9(6+2\sqrt{5})- \ 2(5\sqrt{5}+1+3\sqrt{5}(\sqrt{5}+1)-19)]-\ln (\sqrt{5}+1) \ = \frac{1}{24}[14\sqrt{5}+15]-\ln (\sqrt{5}+1) \ 	herefore \, \, l=14,m=15,n=1 \ \, I+m+n=30 \end{gathered}$

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