<s>[INST] Find Area bounded by Inverse of $f(x)=x^3+3x+2$ between $x=-2$ and $x=6$ Find Area bounded by Inverse of $f(x)=x^3+3x+2$ between $x=-2$ and $x=6$
My Try:
we need $$I=\int_{-2}^{6} f^{-1}(x) dx$$
Use substitution $x=f(t)$, then limits will change to $-1$ and $1$
So
$$I=\int_{-1}^{1}f^{-1}(f(t)) f'(t)dt=\int_{-1}^{1}tf'(t)dt=\int_{-1}^{1}t(3t^2+3)dt$$
Now since we need Unsigned area, Required area is
$$I=2\int_{0}^{1}t(3t^2+3)dt=\frac{9}{2}$$
But Answer is $\frac{5}{4}$...
 [/INST] Firstly note that as $f'(x) = 3x^2 + 3 > 0$ for all $x$, $f$ is an increasing function and so its inverse exists for all $x$. Denote this inverse by $f^{-1} (x)$.
Now, since $f(0) = 2$, then $f^{-1} (2) = 0$. As $f^{-1}(x)$ is an increasing function for all $x$ on its domain, we see that $f^{-1}(x) < 0$ for $x < 2$ and $f^{-1} (x) > 0$ for $x > 2$. The required area $A$ between the curve $f^{-1} (x)$, the $x$-axis, and the lines $x = -2$ and $x= 6$ will therefore be given by
$$A = \left |\int_{-2}^2 f^{-1} (x) \, dx \right | + \int_2^6 f^{-1} (x) \, dx.$$
To find the integral containing the inverse function the following result of
$$\int_a^b f(x) \, dx + \int_{f(a)}^{f(b)} f^{-1} (x) \, dx = b f(b) - a f(a),$$
will be used.
Firstly, by inspection we note that
\begin{align*}
f(-1) = -2 &\Rightarrow f^{-1} (-2) = -1\\
f(0) = 2 &\Rightarrow f^{-1} (2) = 0\\
f(1) = 6 &\Rightarrow f^{-1} (6) = 1
\end{align*}
So
\begin{align*}
\int_{-2}^2 f^{-1} (x) \, dx &= 0 \cdot f(0) - (-1) \cdot f(-1) - \int_{-1}^0 (x^3 + 3x + 2) \, dx\\
&= -2 - \left [\frac{x^4}{4} + \frac{3x^2}{2} + 2x \right ]_{-1}^0\\
&= -2 - \frac{1}{4} = -\frac{9}{4}
\end{align*}
And
\begin{align*}
\int_{2}^6 f^{-1} (x) \, dx &= 1 \cdot f(1) - 0 \cdot f(0) - \int_0^1 (x^3 + 3x + 2) \, dx\\
&= 6 - \left [\frac{x^4}{4} + \frac{3x^2}{2} + 2x \right ]_0^1\\
&= 6 - \frac{15}{4} = \frac{9}{4}.
\end{align*}
So the required area will be
$$A = \left |-\frac{9}{4} \right | + \frac{9}{4} = \frac{9}{2} \, \text{units}^2.$$
</s>