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
Best Answer
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:
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.