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I've had a very hard time wrapping my mind around u-substitution.

I understand how the chain rule applies with the following intuition: Say I have some car whose position function is defined as: $x=3t$ Now say we have another car whose position changes with respect to the first cars position with the following equation: $y=(x)^2$. I understand that when taking the derivative of the second car with respect to time, we would take the derivative of $y$ with respect to $x$ and get $\frac{dy}{dx}= 2x$. To take the derivative of $x$ with respect to $t$ and get $\frac{dx}{dt}= 3$. Now to find the derivative of $y$ with respect to $t$ we would multiply both quantities (I know it's not 100% formal, but it is intuitive) to get $\frac{dy}{dt} = 2x*3 = 6x$. Then we replace this x with the position defined by $3t$ (because we are talking about $t$ here, and we don't need an $x$: $\frac{dy}{dt} = 18t$.

However, I'm having a hard time extending this argument to an integral. I know how to do u-substitution, but I can't intuitively understand it. Especially this: why can't dx=du if they are both approaching 0? Can someone please walk me through an intuitive explanation?

Thanks

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Believe it or not, u-substantiation is nothing more than doing the chain rule backwards. In other words, when you're doing u-substitution, you're going in the opposite direction. Take a look at the following example where we're about to take the derivative of the function $u(x)$:

$$\int f(u)\frac{du}{dx}\,dx=h(x)\Longleftrightarrow h'(x)=f(u)\frac{du}{dx}\\$$

I hope you agree that what we have on the left is equivalent to what we have on the right. Now, let's go further back in the opposite direction. We have differentiated $h(x)$ to get $f(u)$ and $\frac{du}{dx}$ is just about to appear, but let's freeze things for a second here:

$$h'(x)=f(u)\implies\int f(u)\,du=h(x)$$

Of course, $u$ is a function of $x$ there. What we've got on the left side is a new integral that should be integrated with respect to the variable $u$. The idea here is that the new integral we got will hopefully be easier to integrate than the one we had at the beginning. So, in short, when you're doing u-substitution, you're transforming the original problem into an incomplete chain rule problem.


Let's do a simple but concrete example to illustrate this process:

$$ \int(x+1)^2\,dx=h(x)\Longleftrightarrow h'(x)=(x+1)^2\\ \int(x+1)^2\cdot1\,dx=h(x)\Longleftrightarrow h'(x)=(x+1)^2\cdot1\\ \int(x+1)^2\frac{d}{dx}(x+1)\,dx=h(x)\Longleftrightarrow h'(x)=(x^2+1)^2\frac{d}{dx}(x+1) $$

Let's replace $x+1$ with $u$ to make things easier to read:

$$\int u^2\frac{du}{dx}\,dx=h(x)\Longleftrightarrow h'(x)=u^2\frac{du}{dx}$$

Let's take one more step back. We have differentiated $h(x)$ to get $u^2$ and it's just about time to take the derivative of $u(x)$, but we freeze things instead:

$$\int u^2\,du=h(x)\Longleftrightarrow h'(x)=u^2$$

And now, the integral that we've got is easy to differentiate because it's one of those elementary table integrals that we all know how to do:

$$ \int u^2\,du=\frac{u^3}{3}+C=\frac{(x+1)^3}{3}+C. $$

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  • $\begingroup$ Thanks for the answer Mike. Any chance that you can walk me through an argument with concrete physical objects (like car position, etc.)? I am able to grasp this stuff better if someone walks me through with physical intuition. $\endgroup$ – Dude156 Feb 17 at 20:45
  • $\begingroup$ Well, I can't really think of any concrete physical example. Again, all I can say is that u-substitution is just the chain rule done backwards. I always think of the chain rule as a purely mathematical operation that stems from the problem of trying to find the derivative of a function that's a composition of two or more functions. $\endgroup$ – Michael Rybkin Mar 1 at 1:40
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Two ways to look at it:

First, the indefinite integral: $\int F'(x)\,dx = F(x)+C$ is the antiderivative. In this viewpoint, the substitution rule is just the chain rule written backwards: $\int F'(g(x))\cdot g'(x)\,dx = F(g(x))+C$.

Second, the definite integral as the area problem; $\int_a^b f(x)\,dx$ is the area under the graph of $f$ between $a$ and $b$. Here, a substitution will transform the interval we integrate over, and we'll need to stretch the function vertically in order to keep the area the same:

Graph of sqrt(1-x^2) and transformation

Our transformation stretches a small horizontal segment $dx$ to $du$, multiplying by $\frac{du}{dx}$. In order to keep the same area, we have to multiply the function values by the reciprocal $\frac{dx}{du}$. The example here $(f(x)=\sqrt{1-x^2},u=\sin x)$ has $\frac{du}{dx} > 1$, so the transformed graph is shorter and wider in this case.

If we slice up the whole interval this way, each slice of the area under the transformed graph has the same area as the corresponding slice of the area under the original graph. Sum them up, and the areas are the same.

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