$f\left( x \right) = {x^3} + x$, then $\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $ 
If $f\left( x \right) = {x^3} + x$, then
$$\int\limits_1^2 {f\left( x \right)dx}  + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $$
is________.

My approach is as follows:
$$g = {f^{ - 1}} \Rightarrow g\left( x \right) = {f^{ - 1}}\left( x \right)$$
$$g\left( {2x} \right) = {f^{ - 1}}\left( {2x} \right)$$
$$2y = {8x^3} + 2x$$
$${f^{ - 1}}\left( {{x^3} + x} \right) = x$$
$$\int\limits_1^2 {f\left( x \right)dx}  + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $$
I am not able to proceed further.
 A: Draw the graph of $f$ in the $x$-interval $[1,2]$. The graph extends from $(1,2)$ to $(2,10)$. The first integral is easy to compute. The second integral is
$$\int_1^5f^{-1}(2x)\>dx={1\over2}\int_2^{10}f^{-1}(y)\>dy\ .$$
The essential point now is to recognize the last integral as a surface area that is already visible in your figure.
Note that it is impossible to obtain a simple expression for $y\mapsto f^{-1}(y)$. Furthermore: Do not make a mix between the different variables $x$ and $y$.
A: I agree with Christian Blatter's answer.  My response is intended to expand the OP's intuition.  This is a trick question, which can be solved without attempting to compute $f^{(-1)} x.$  Here, I am using $f^{(-1)} x$ to represent the inverse function of $f(x),$ rather than the reciprocal of $f(x)$.
The easiest way to see this is to consider a similar but much simpler problem. Suppose that $f(x) = x^2$ and you are asked to determine the area to the left of the curve $y = f(x)$ as $y$ ranges from $0 \to 4.$
The direct approach is to compute $x = g(y)$ where $g$ is the inverse function of $f$.  Then the area is computed as $I = \int_0^4 g(y) dy.$  This approach might be construed as integrating horizontally.
The alternative approach is:
(1) Consider the rectangle formed by $x=0, x=2, y=0,$ and $y=4.$ This rectangle has an area of 8 square units.
(2) Consider the vertical integration of $J = \int_0^2 f(x) dx.$
Clearly, $I + J = 8,$ so $I$ can be computed as $8 - J.$
Using the alternative approach (directly above) facilitates indirectly computing a horizontal integration when you are given $f(x)$ and wish to avoid having to compute $g(y) = f^{(-1)}x.$
A: $$f(x)=x^3+x$$
Let $$K=\int_{1}^{2} f(x) dx+2\int_{1}^{5} f^{-1}(2x) dx=I+J$$
$$ I=\int_{1}^{2} (x^3+x) dx=\frac{21}{4}$$
$$J=2\int_{1}^{5} f^{-1}(2x)dx=\int_{2}^{10} f^{-1}(z) dz$$
Let $f^{-1}(z)=t \implies z=f(t) \implies dz=f'(t) dt$, then
$$J=\int_{1}^{2} t f'(t) dt=\int_{1}^{2}t(3t^2+1)dt=\frac{51}{4}.$$
Finally $$K=21/4+51/4=18.$$
A: After the substitution $u=2x$ in the second integral, your expresion becomes
$$\int\limits_1^2 {f\left( x \right)dx}  + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx}= \int\limits_1^2 {f\left( x \right)dx} + \int\limits_{f(1)=2}^{f(2)=10} {{f^{ - 1}}\left( {u} \right)du}$$
Now, you can use an integration rule for the inverse function and obtain
$$\int\limits_1^2 {f\left( x \right)dx} + \int\limits_{f(1)=2}^{f(2)=10} {{f^{ - 1}}\left( {u} \right)du} = 2\cdot f(2) - 1\cdot f(1) =18$$
