Let $y=g(x)$ be the inverse of a bijective mapping $f:R\to R f(x)=3x^3+2x.$Then area bounded by the graph of $g(x),x-$ axis and the ordinate at $x=5$

Let $y=g(x)$ be the inverse of a bijective mapping $f:R\to R f(x)=3x^3+2x.$Then what is the area bounded by the graph of $g(x),x-$ axis and the ordinate at $x=5$

I could not find the $f^{-1}$ from the equation for $f(x)$ and cannot solve the question.How should i attempt this question?

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The graph of $g = f^{-1}$ is just the graph of $f$ reflected through the line $y = x$. That is, if you picked up the plane and flipped it on this line, so that the $x$-axis becomes the $y$-axis and vice versa, then the graph of $g$ becomes the graph of $f$. The region whose area you are to find becomes the region between the $y$-axis on the left, the horizontal line $y = 5$ on top, and the graph of $y = f(x)$ on the right and bottom.

Note that this is the area of a rectangle less the area below the graph of $y = f(x)$.

Required area is $$\int_{0}^{5}f^{-1}(x)dx$$ Put $x=f(t)$ we get

$$I=\int_{f^{-1}(0)}^{f^{-1}(5)}f^{-1}(f(t))f'(t)dt=\int_{f^{-1}(0)}^{f^{-1}(5)} tf'(t)dt$$

But since $f$ is One-One $f(0)=0$ $\implies$ $f^{-1}(0)=0$ and $f^{-1}(5)=1$

So $$I=\int_{0}^{1}t(9t^2+2)dt=\frac{13}{4}$$