prove that $\int_{a}^b f \ + \int_{f(a)}^{f(b)} \ f^{-1}=bf(b)-af(a)$ Let $0 <a <b\ $ let $f >0 $  be continuous and strictly increasing  function on $ [a,b]$. Prove that $$\int_{a}^b f \  + \int_{f(a)}^{f(b)} \ f^{-1}=bf(b)-af(a)$$
How to approach this problem . Any Hint? I am suppose to do it without using antiderivatives.
 A: Add any left Riemann sum with partition $(a,x_1,\ldots,x_{n-1},b)$ for the first integral to a right Riemann sum with partition $(f(a),f(x_1), \ldots,f(x_{n-1}),f(b))$ for the second integral to get
$$\sum_{j=1}^n f(x_{j-1})(x_j - x_{j-1}) + \sum_{j=1}^nf^{-1}(f(x_j))(f(x_j) - f(x_{j-1})) \\ = \sum_{j=1}^n f(x_{j-1})(x_j - x_{j-1}) + \sum_{j=1}^n x_j(f(x_j) - f(x_{j-1}))\\ = \sum_{j=1}^n (\, x_jf(x_j) - x_{j-1}f(x_{j-1})\, ) \\ = bf(b) - a f(a)$$
The LHS converges to $\int_a^b f + \int_{f(a)}^{f(b)} f^{-1}$ as the partition norm goes to $0$ and the RHS stays the same.
A: You can draw a picture. Draw a picture with $a$, $b$ marked on the $x$-axis. Draw some random increasing function in the first quadrant. Mark $f(a)$ and $f(b)$ on the $y$-axis. 

Now $\int_a^bf(x)\,dx$ is the area under the curve between $x=a$, $x=b$, and the $x$-axis.
And $\int_{f(a)}^{f(b)}f^{-1}(y)\,dy$ is the area left of the curve between $y=f(a)$, $y=f(b)$, and the $y$-axis. 
Adding these areas together gives an inverted L-shape. That is, a thing that is one larger rectangle with a smaller rectangle removed. The large rectangle is $b\times f(b)$, and the smaller is $a\times f(a)$.
A: Geometrical idea
Consider the graph of the strictly increasing function $y=f(x)$. This graph passes through the points $(a,f(a))$ and $(b,f(b))$. 
The area under the curve $y=f(x)$ from $x=a$ to $x=b$ is given by $\displaystyle \int_a^b f(x)\, dx$. Looking at the same graph "from" the $y-$axis, it will be the graph of the inverse function $f^{-1}$. The area under the inverse curve from $y=f(a)$ to $y=f(b)$ is given by $\displaystyle \int_{f(a)}^{f(b)} f^{-1}(y)\, dy$. The sum of these two areas is 
$$\int_a^b f(x)\, dx+\int_{f(a)}^{f(b)} f^{-1}(y)\, dy.$$
Now this same area can be found in the following manner as well:
Consider the rectangle formed by two corners at $(0,0)$ and $(b,f(b))$. The area of this is given by $bf(b)$. Now if we remove the area $af(a)$ of the smaller rectangle with corners at $(0,0)$ and $(a,f(a))$ we get 
$$bf(b)-af(a).$$
This is the same area as we have calculated above using the integrals. Hence they are equal.
