If $f(x)= x^3 + 3x + 4$ then the value of $\int_{-1}^1f(x)dx + \int_{0}^4 f^{-1}(x)dx$ equals 
If $f(x)= x^3 + 3x + 4$ then the value of $$\int_{-1}^1f(x)dx + \int_{0}^4 f^{-1}(x)dx$$ equals

So in this question I got stuck with the part of inverse of the function so I referred the solution and the solution was given as

$$\int_{-1}^0f(x)dx + \int_{0}^1f(x)dx + \int_{0}^4 f^{-1}(x)dx = \int_{0}^1f(x)dx $$

What does this step mean , I'm new to Calculus and don't have much experience and so couldn't understand it. Please help :)
 A: The solution is not clear to me. But I have a suggeation for dealing with the inverse function. Note that $f(x)$ is monotone in $[0,4]$, so the inverse makes sense. Let $y=f^{-1}(x)$ so $y$ goes from $-1$ to $0$ and $x=f(y)$ and thus 
$$\int_0^4f^{-1}(x)\,dx=\int_{-1}^0yf'(y)\,dy$$
and now everything is clear.
A: Some observations:


*

*$f'(x)=3x^2+3 > 0$ so the function is strictly injective.

*$f(-1)=0$

*$f(0)=4$
That should allow you to say something about the two integrals on the left side of the latter equation.
A: Not related to your question directly, but if the limits of integration are changed then this result holds:
$\int_{a}^{b}f^{-1}(x)dx +\int_{f^{-1}(a)}^{f^{-1}(b)}f(x)dx=bf^{-1}(b)-af^{-1}(a)$
Try proving this yourself(Hint: consider $\int_{a}^{b}f^{-1}(x)dx$ take $f^{-1}(x)=t \implies x=f(t)$ change the limits of Integration accordingly applying integration by parts and rearrange terms and you should get the result)
Note: this method can be used in your question as well, but only that  $\int_{f^{-1}(a)}^{f^{-1}(b)}f(x)dx$ term doesn't appear, so you need to evaluate that explicitly, and also the other integral of f(x)
A: There are some interesting properties graphically. The area under $f(x)$ from $-1$ to $1$ is identical to the area under the curve between the inverse of $f$ and the y axis on $[0,4]$ which has a negative value. This leaves only the integral of $f(x)$ from 0 to 1. This is expressed analytically by the answers above.
