Explain this proof without words of integration by parts to me Here is a proof of integration by parts: http://www.math.ufl.edu/~mathguy/year/S10/int_by_parts.pdf
But I don't understand how it works. Specifically, I don't understand why $\int_r^s u \, dv$ equals one of the areas (and likewise for $\int_p^q v \, du$). 
To me it looks like this only works if $g$ is the inverse of $f$...
Thanks!
 A: If we play around with the graph in our head, we can look at it from two perspectives: $(U,V(U))$ and $(V,U(V))$. That is, if we look at it from how it is presented, we see that the area of the second box in the equation is $\int_{p}^{q}V(U)dU.$ In their notation, that's exactly $\int_{p}^{q}vdu$. This follows from the definition of a definite integral.
Now, if we flip the graph (turn your head sideways), reflect it to the left, and look at it from the perspective of $(V,U(V))$, we see that the area underneath the flipped curve is $\int_{r}^{s}U(V)dV$. This is, in their notation, $\int_{r}^{s}udv$.
This is why we have that the total sum of the areas is $\int_{r}^{s}udv+\int_{p}^{q}vdu.$

There is not a loss of rigor what so ever in this diagram. It is quite unsettling at first, but what we're really just doing is this:
Let $v: U \to V$ be a map where, for a $u \in U$, we have $u \mapsto v(u)$. The domain and range thus form the set of ordered pairs $\{(u,v(u)): u\in U\}$. Our other map, $v^{-1}:V \to U$, is the flip-then-reflect-left map with the mapping $v(u) \mapsto u$. It is the set of ordered pairs $\{(v(u),u): u\in U\}$. This is the inverse map of $v$ by definition.
Having that $u=f(x_0)$ for some $x_0$ and $v(u)=g(x_1)$ for some $x_1$, however, does not necessarily imply $g$ is the inverse of $f$.
A: The black curve should be viewed as parametric curve $(f(t),g(t))$, $a\le t\le b$.
A: The important part is that the axes themselves are the original $x,y$ axes after applying the functions $u$ and $v$ to them!
Perhaps this diagram from here is clearer. Because $u=f(v)$ is the height of the curve from the $v$ axis, the area of the bottom part is $\displaystyle \int_{v_1}^{v_2}u \space d v$. But note that $u_1=f(v_1),u_2=f(v_2)$.
Next, let the $v$ values change. Now $v$ is the height of the curve from the $u$ axis, the area of the top part is $\displaystyle \int_{u_1}^{u_2}v \space d u$.
The sum of these areas is a nice difference of areas of boxes. I.e. $u_2v_2-u_1v_1$. This is however the  $uv|_{(v_1,u_1)}^{(v_2,u_2)}$. So, symbolically
$\displaystyle \int_{v_1}^{v_2}u \space d v +\int_{u_1}^{u_2}v \space d u=uv|_{(v_1,u_1)}^{(v_2,u_2)}$.

A: The area between the curves and the axes adds together to give the area of the rectangle with area $qs$, with the square of area $pr$. So subtract the two areas and set them equal to the sum of the integrals. Thank you for this proof, I really like it!
A: He used the formula that the area of a region under the curve $y=g(x),x=v(x)$ is equal to $\int_{C} xdy=\int_{C}vdu$. This is from Green's formula. See this wikipedia article. Since the derivation of Green's formula does not involve integration by parts, this is a "proof without words". 
