Let $f(x) =x^3+3x+1$. Find area bounded by $y=f^{-1}(x)$, the ordinates $x=-3$,$x=5$ and X-axis.

My Attempt: The required area will be equal to area enclosed by $y=f(x)$,the y-axis between the abcissae at $y=-3$ and $y=5$.

Required area$$=\int_{0}^{1}\left(5-f(x)\right)dx+\int_{-1}^{0}\left(f(x)-(-3)\right)dx=\frac{5}{2}$$

This was the given solution.But I am doubtful. Is there any other way.

  • $\begingroup$ What have you tried? First of all, what is $f^{-1}(x)$? You cannot get very far without knowing $f^{-1}$, as far as I can perceive. $\endgroup$
    – Crosby
    Jun 24, 2018 at 15:49
  • $\begingroup$ @Crosby. Ihave edited my post. $\endgroup$
    – Maverick
    Jun 24, 2018 at 15:54
  • $\begingroup$ How did you find the values $y=-2$ and $y=6$? $\endgroup$
    – Bernard
    Jun 24, 2018 at 16:09
  • $\begingroup$ @Bernard. This is what the solution was given. I too am little confused at this point $\endgroup$
    – Maverick
    Jun 24, 2018 at 16:11

1 Answer 1


First note that function is one-one and onto in the range $(-3,5)$. So the inverse does exist.

So if you want to find the area enclosed by $f^{-1}$ which is the reflection of $f(x)$ about the line $y=x$. So by symmetry area enclosed between $f^{-1}(x)$ and x axis is same as the area between $f(x)$ and $y$-axis. So you need to reflect $x=5$ about $y=x$ that gives you $y=5$ and similarly for $x=-3$.

Now when $y=5$ you will find that $f(x)-5=0$ gives the solution $x=1$ and $f(x)+3=0$ give solution $x=-1$. Now the first integral is subtracting the area under $y=5$ and $f(x)$ in positive region. And second integral is subtracting area above $y=-3$ and above $f(x)$. This gives you the area bound by between $y$-axis and the $y\in(-3,5)$ .


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