Check out this dialogue between two calculus students (based on a true story):

Devyn

Riley! We should be able to figure some integrals geometrically using transformations of functions.

Riley

That sounds like a cool idea. Maybe, since we know the graph of \(f(x) = \sqrt {1-x^2}\) is a semicircle, we get an ellipse defined on \([-2,2]\) just by stretching the graph of \(f\) by a factor of \(2\) horizontally. The equation of this ellipse would be

\[ g(x) =\sqrt {1-\left (\frac {x}{2}\right )^2} \]

Devyn

Exactly! So since we know that

\[ \int _{-1}^1 \sqrt {1-x^2} dx = \frac {\pi }{2} \]

geometrically…

Riley

And we know that the area under \(g\) from \([-2,2]\) is twice the under \(f\)…

Devyn and Riley

We must have

\[ \int _{-2}^2 \sqrt {1-(x/2)^2} dx =\pi ! \]

Devyn and Riley

Jinx!

Devyn

It is kind of like we just stretched out our whole coordinate system, and that helped us solve an integral.

Riley

In this case, everything got stretched out by a constant factor of \(2\) in the horizontal direction. I wonder if we could ever say anything useful about cases where we stretch the \(x\)-axis by a different amount at each point?

Devyn

Whao, that is a wild thought. That seems really hard. Since derivatives measure how much a function stretches a little piece of the domain, maybe the derivative will come into play here?

Riley

Hmmmm, but I do not see exactly how. Maybe we should ask our TA about this?