# HIIII im Varun, can anybody tell me how to do this one : "Find the area enclosed by y=g(x), x=1, and x=15 where g is inverse of f given by f(x) = x^3+3x+1"????

Jitender Singh IIT Delhi
7 years ago
Ans:
$f(x) = x^{3}+3x+1$
$x\rightarrow f^{-1}(x)$
$f(f^{-1}(x)) = (f^{-1}(x))^{3}+3f^{-1}(x)+1$
$(f^{-1}(x))^{3}+3f^{-1}(x)+1-x=0$
On solving this, we have
$f^{-1}(x) = \frac{\sqrt[3]{2}}{\sqrt[3]{\sqrt{x^{2}-2x+5}-x+1}}-\frac{\sqrt[3]{\sqrt{x^{2}-2x+5}-x+1}}{\sqrt[3]{2}}$
Area ’A’:
$A = \int_{1}^{15}(\frac{\sqrt[3]{2}}{\sqrt[3]{\sqrt{x^{2}-2x+5}-x+1}}-\frac{\sqrt[3]{\sqrt{x^{2}-2x+5}-x+1}}{\sqrt[3]{2}})dx$
$A = (\frac{3(\frac{1}{4}(\sqrt{x^{2}-2x+5}-x+1)^{4/3}-\frac{2}{(\sqrt{x^{2}-2x+5}-x+1)^{2/3}})}{2\sqrt[3]{2}}-\frac{3(\sqrt{x^{2}-2x+5}-x+1)^{2/3}}{2.2^{2/3}}+\frac{3}{2^{2/3}.(\sqrt{x^{2}-2x+5}-x+1)^{4/3}})_{1}^{15}$
A = 18
Thanks & Regards
Jitender Singh
IIT Delhi