How does change of variables during integration fit in with view of integral as area under the curve?

Question

I'm trying to understand/visualize, how change of variables during integration fits in with view of integral as area under the curve. Assuming we have a definite integral: $$\int _a^bx^3dx$$ and we decide to substitute $$u = x^3$$ $$x = u^{\frac{1}{3}}$$ $$\frac{d}{dx}u=3x^2$$ $$dx\:=\frac{du}{3x^2}=\frac{du}{3u^{\frac{2}{3}}}$$ We end up with: $$\int _a^bx^3dx = \int _{a^3}^{b^3}\frac{1}{3}u^{\frac{1}{3}}du\:\:$$

$x^3$ and $\frac{1}{3}u^{\frac{1}{3}}$ are two very different curves. Is there geometric relationship between the two areas, which could be used to visualize their equivalence?

Answer 1

You can think of the substitution as a distortion of your region.

You distort the curve with the $u(x)$ input to your function. At the same time you simultaneously accounting for the impact of that distortion with the multiplication of $u'(x)$.

A simple example $u(x) = 2x$

$f(u(x))$ dilates everything horizontally by a factor of 2 (when expressed in terms of u.

$2 dx = du$ says that since everything has been uniformly dilated, you must half-weight the area of this region.

Answer 2

When you change the variables, you do not just change the curve, you also change the interval over which you integrate.

To make this concrete, suppose you were going to integrate $x^3$ from $x = 2$ to $x = 3.$ That is, the integral is

$$\int_2^3 x^3 \, dx. $$

Now you make a change of variables, $u = x^3,$ and you end up with the integral $$\int_8^{27} \frac13 u^{1/3}\,du.$$

Interpreted as areas between curves and the $x$ axis, bounded in each case by two vertical lines, these integrals are modeled by regions that not only have very different shapes, they do not even touch each other. They happen to have the same area measurement, however.

For an analogy, take a tall, narrow glass of water and pour it in a bowl. The shape of the water changes dramatically, but it still has the same volume, doesn't it?