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When we integrate a function with respect to dx, we are breaking the area under the function into many pieces of width dx and finding the area of each piece and then take the limit of the sum as dx becomes 0. In this case dx is constant so this is pretty intuitive.

When we use u-substitution, we replace dx with du. But clearly du is not constant as it changes with x. To me, this doesn't make sense because how do we integrate with respect to something that is not constant? Would appreciate if someone can give a visualisation of what is going on (graphically) when we integrate by substitution. For instance if we graph out x, g(x) and f(x) parametrically, how can we show that ∫f(u)du and ∫f(g(x)*g'(x)dx are the same thing?

I also found this related thread: Is there a way to graphically visually integration by substitution?, but I don't really understand the answer.

Luo Zeyuan
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  • In the integration notation, you can interpret "dx" to mean as "with respect to x" and not as a constant. "dx" may be thought of as a constant when you use the Sum notation only as you have said. The concept of area under the curve is only valid in bound integral with limits, otherwise, the integration that is indefinite, is related to anti-derivative. A more precise history and terminology can be found in the Fundamental Theorem of Calculus. Substitution works because when you choose u(x) = sin(x), you integrate u(x) with respect to "u" , so we use "du". – NoChance Nov 25 '18 at 12:46
  • Yes. In this case, I think it would be more convenient to treat dx as a constant, since it is easier to visualise on a graph. – Luo Zeyuan Nov 25 '18 at 12:46
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    Not everything can be easily visualized and "dx" IS NOT a constant in integration. A constant is like 5,4...etc. – NoChance Nov 25 '18 at 12:48
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    I don't recommend thinking of $dx$ as being necessarily constant. To integrate a function over an interval, you chop the interval up into tiny pieces, compute the contribution of each piece, and sum up these individual contributions. (The integral is a limit of sums like this.) The tiny pieces do not all have to have the same size. In fact, it is unnatural to require the tiny pieces to all have the same size. – littleO Nov 25 '18 at 14:02

1 Answers1

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Of course the indefinite integrals $\int f(x)\>dx$ and $\int f\bigl(g(x)\bigr)\>g'(x)\>dx$ are not the same. As an example, let $f(x):=x$ and $g(x):=x^2$. Then $\int f(x)\>dx={1\over2}x^2+C$, and $\int f\bigl(g(x)\bigr)\>g'(x)\>dx={1\over2}x^4+C$.

The correct way to write this substitution formula for indefinite integrals is $$\int f\bigl(g(t)\bigr)\>g'(t)\>dt=\int f(x)\>dx\biggr|_{x:=g(t)}\ .\tag{1}$$ Formula $(1)$ is an immediate consequence of the chain rule. I don't think there is a graphical visualization.

It is another thing with the substitution formula for definite integrals: $$\int_a^b f\bigl(g(t)\bigr)\>g'(t)\>dt=\int_{g(a)}^{g(b)}f(x)\>dx\ .$$ Here we can invoke Riemann sums and draw narrow rectangles. In this way we arrive at the following chain of approximations: $$\eqalign{\int_{g(a)}^{g(b)}f(x)\>dx&\approx\sum_{k=1}^N f(\xi_k)(x_k-x_{k-1})\cr &\approx\sum_{k=1}^N f\bigl(g(\tau_k)\bigr)\>g'(\tau_k)\>(t_k-t_{k-1})\cr &\approx\int_a^b f\bigl(g(t)\bigr)\>g'(t)\>dt\ .\cr}$$ Here we have assumed that the partition points are $x_k=g(t_k)$, and the sampling points are $\xi_k=g(\tau_k)$. Furthermore we have used the MVT formula $x_k-x_{k-1}\approx g'(\tau_k)(t_k-t_{k-1})$ in order to handle the width of corresponding rectangles on the $x$-, resp., the $t$-axis.

Christian Blatter
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