Question

Find the area bounded by the inverse of bijective function $f\left(x\right)=4x^3+6x$, the $x$-axis and the ordinates $x=0$ and $x=44$

Answer 1

Step 1. Find the inverse of the function

Given bijective function is: $f\left(x\right)=4x^3+6x$

$$\begin{gathered} f\left(0\right)=4·{\left(0\right)}^3+6·\left(0\right) \\ \Rightarrow f\left(0\right)=0 \\ \Rightarrow 0=f^{-1}\left(0\right) \\ f\left(2\right)=4·{\left(2\right)}^3+6·\left(2\right) \\ \Rightarrow f\left(2\right)=44 \\ \Rightarrow 2=f^{-1}\left(44\right) \end{gathered}$$

Step 2. Find the area bounded

So the required area of the inverse bijective function is

$A={\int }_0^{44}{f}^{-1}\left(x\right)dx$

Let $x=f\left(t\right)$

Therefore $dx=f\text{'}\left(t\right)dt$

$A={\int }_{f^{-1}\left(0\right)}^{f^{-1}\left(44\right)}{f}^{-1}\left(f\left(t\right)\right)f\text{'}\left(t\right)dx={\int }_0^{2}tf\text{'}\left(t\right)dt$

So, the area bounded by the inverse of bijective function $f\left(x\right)$ is :

$$\begin{gathered} A={\int }_0^{2}t\left(12t^2+6\right)dt\left[\because f\left(t\right)=4t^3+6t\right] \\ ={\int }_0^2\left(12t^3+6t\right)dt \\ ={\left[3t^4+3t^2\right]}_0^2 \\ ={\left[3{\left(2\right)}^4+3{\left(2\right)}^2\right]}_{}^{}-0-0 \\ =\left[48+12\right]-0 \\ =60 \end{gathered}$$

Hence, the area bounded by the inverse bijective function $f\left(x\right)=4x^3+6x$, the $x$-axis and the ordinates $x=0$ and is $60sq.\mathrm{units}$