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In integration by substitution, we change the variable. For example: $$\int_{x^{2}=0}^{x^{2}=9} x^{2} d(x^{2})=\int_{x=0}^{x=3} x^{2} \dfrac{d(x^{2})}{dx}dx$$ Here since we have two different variables I find it impossible to graphically visualize how integration by substitution works.

Is there any other way to visualize this. Any help will be appreciated.

stack exchange
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  • Draw both graphs and mark the limits. Then you get two equal areas (At least in the given case) – Peter Nov 26 '16 at 14:56
  • That's right. But that doesn't give a graphical intuition on why we are getting equal areas. – stack exchange Nov 26 '16 at 15:07
  • A graphical intuition would only be possible by changing the scale of the axis, but in thic case, the change of the scales would not be intuitive. I do not think that such an intuition you want is possible. – Peter Nov 26 '16 at 16:21

1 Answers1

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The plot below shows, in a Cartesian $(x, u, y)$-space, the change-of-variables identity $$ \int_{x = a}^{x = b} f(g(x)) g'(x)\, dx = \int_{u = g(a)}^{u = g(b)} f(u)\, du \tag{1} $$ for the region $R$ defined by $0 \leq y \leq f(u)$ for $g(a) \leq u \leq g(b)$ in the $(u, y)$-plane.

The integral on the left side of (1) represents the parametrization of $R$ by the region $R_{0}$ defined by $0 \leq y \leq f(x)$ for $a \leq x \leq b$ in the $(x, y)$-plane. To get the area of $R$ as an integral over $R_{0}$, send the point $(x, y)$ to $(u, y) = (g(x), y)$. As the shaded areas indicate, the vertical strip of $R_{0}$ at location $x$ is scaled in width (hence in area) by $g'(x)$.

The geometry of change of variables

Andrew D. Hwang
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  • What is meant by parametrization of $R$ by $R_{0}$? How is it that the integral on the left side of (1) represents the parametrization of $R$ by $R_{0}$? – stack exchange Nov 27 '16 at 09:59
  • The area of $R_{0}$ is $\int_{a}^{b} f(g(x)), dx$. Each point $(x, y)$ of $R_{0}$ corresponds to a unique point $(u, y) = (g(x), y)$ of $R$, but (in the language of infinitesimals) this mapping sends the thin vertical strip of height $f(g(x))$, between $x$ and $x + dx$, to the thin vertical strip of height $f(g(x))$ between $g(x)= u$ and $g(x + dx) = u + du = u + g'(x), dx$. The area of $R$ is therefore $\int_{a}^{b} f(g(x)) g'(x), dx$, the integral of the areas of the images of the thin strips. – Andrew D. Hwang Nov 27 '16 at 12:25
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    I get it. What seemed impossible turned out to be very simple. Thanks a million. – stack exchange Nov 28 '16 at 14:15
  • +1. This is great! What tool do you use for the figure? –  Jan 17 '22 at 13:35
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    @ripples I'm happy the answer was useful! The figure was made with ePiX. – Andrew D. Hwang Jan 17 '22 at 14:25