Area enclosed by $f^{-1}(x)$ given $f(x)$

Question

Find the area enclosed by $g(x),x=-3,x=5$ and the $x$ axis where $g(x)$ is the inverse of $f(x)=x^3+3x+1$.

Clearly, there's no easy way to directly find the inverse of $f(x)$.So, how to go about this? I am supposed to do this using techniques of finding area by definite integrals.

Thanks for any help!

Answer 1

Hint: Starting from the integral $$\int_{-3}^5f^{-1}(x)dx$$ do the substitution $f^{-1}(x)=y$ using the derivative of inverse formula: $$(f^{-1}(x))'=\frac1{f'(f^{-1}(x))}$$ thus

$$\frac1{f'(y)}dy=dx$$

Answer 2

Hint: If $f\colon [a,b]\to [c,d]$ is an increasing bijection, then $$ \int_a^b f(x)\, dx + \int_c^d f^{-1}(y)\, dy = bd - ac. $$ (See here.)

Answer 3

find area enclosed between f(x) and y=x ( DEFINITE INTERGRAL) also find area between y=x and and x axis. (trapezium) now use the fact that f (x) and inverse if f(x) are symmetric about y=x.