Geometry and substitution

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Two students consider substitution geometrically.

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?

Say we know that $\int _1^4 f(x) dx = 5.$ Then, using this transformation idea, we can evaluate $\int _a^b f(3x+1) dx$ if $a= \answer {0}$ and $b=\answer {1}$ . The value of the integral on this interval is $\answer {5/3}$ .

Since we are “squishing” the integrand, the integral has a value of $5/3$ . As food for thought, how do we change the integrand so that the two integrals above are equal?