
Welll done to Andrei from Tudor Vianu for his solution for the first question.
To prove the formula, I represented a sketch of the graph of a function f(x):

Now, I identify the parts of the formula on this graph:
1. $\int_a^bf(x)dx$ is the area between the x-axis and the graph, on the domain of x between a and b. This is the green area below.
2. $\int_{f(a)}^{f(b)} f^{-1}(t) dt$ is the area between the y
axis and the part of the graph situated between f(a) and f(b). This
is yellow below.
This becomes more evident by by considering the graph of
$f^{-1}(t)$, which is the above, but reflected in the line
$y=x$
3. $bf(b)$ and $af(a)$ are the areas of the rectangles in my diagram, of sides $b, f(b)$, and $a, f(a)$ respectively.
The difference of the two areas is exactly the sum of the yellow and green areas.

Derek Wan continues to finish the problem:
2) Find the value of $\int _1^4 \sqrt t dt$, firstly by evaluating the integral directly, and secondly by using the formula above with $f(x)=x^2$.


3) Use the formula to evaluate $\int_0^1\sin^{-1}t dt$

The last few ideas remain unanswered, can you add to them?
What other functions can you integrate more easily using this formula than by other means? Why must $f(x)$ be increasing in the interval $a\leq x\leq b$? How could you evaluate a similar integral if $f(x)$ is decreasing?Published September 2004,July 2009,August 2009.