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I am confused as to what $dx$ truly is. I am doing some u-substitution problems and this is what I came across:

$$\int 2x(x-1)^{1/2}\,dx$$

$u=x-1$ and therefore $du=1$

when we substitute we get: $$2 \int (u^{3/2}+u^{1/2}) \, du$$ (here the du simply replaces the dx because our variable changed)

In another example: $$\int 4x^5(x^2+1)^{1/3} \, dx$$ $$u=x^3+1$$ $$du=3x^2$$ therefore it becomes: $$\int 4(u-1)(du/3)(u)^{1/3}\,$$ -here my teacher didn't put $d$x at the end, she just left it off So my question is this: why is it that sometimes $dx$ and $du$ are treated as values that can be multiplied to other terms in the integrand and sometimes they are simply treated as a command (do "blank" with respect to $x$, or $u$ or whatever is used)?

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    $\begingroup$ If $u=x^3+1$, then you should have $du=3x^2\,dx$, not just $3x^2$. Similarly if $u=x-1$ then $du=dx$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Feb 15 '15 at 4:05
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    $\begingroup$ Here's an answer to a similar question: math.stackexchange.com/questions/200393/… ${}\qquad{}$ $\endgroup$ – Michael Hardy Feb 15 '15 at 4:09
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    $\begingroup$ your teacher might have just forgotten to put $dx$ after the $3x^2.$ in the first substitution, you may be better off doing $x - 1 = u^2$ instead of the $x-1 = u$ you tried. $\endgroup$ – abel Feb 15 '15 at 4:12
  • $\begingroup$ What I'm asking though is in the second example a du appeared when converting so the dx was not put at the end. Why can we just leave off a dx when it really is like it being multiplied to the integrand? In the first example, the dx was actually du so we replaced it accordingly. Why the inconsistency? $\endgroup$ – King Squirrel Feb 15 '15 at 4:19
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    $\begingroup$ The truth is that the $\mathrm{d}x$ is entirely notation, telling you which symbol is your variable of integration (or a "designator", as you say). However, like many things in intro calculus, the truth is suppressed in favour of methods. The way you manipulate differentials (quantities with the $\mathrm{d}$ in front) leads you to believe it is an algebraic quantity, but it's only notation. However, treating it as an algebraic object makes remembering methods much easier, so that's why you can do it. $\endgroup$ – Ducky Feb 15 '15 at 4:25
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It really depends. It can be seen both as a measure and a differential form, depending on whether you are integrating over measurable subsets of a set, or over chains, respectively. Moreover, the two concepts are not mutually exclusive.

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Without advanced techniques, the best explanation I can think of (which is not the most useful for calculations) is the following. If $g$ is an integrable function on $[a,b]$ and $f$ is a one-to-one correspondence (or bijection) between $[a,b]$ and $[c,d]$ such that $f'$ is continuous and never zero, then

$$\int_a^b g(x) dx = \int_c^d g(f^{-1}(u)) \left | (f^{-1})'(u) \right | du.$$

This is the change of variables formula. It is slightly different from how we usually do things in calculus, because both sides are being integrated from left to right, whereas usually if $f$ is decreasing, the right side is integrated from right to left.

But in this case when $f$ is decreasing, $\left | (f^{-1})'(u) \right | = -(f^{-1})'(u)$. This extra minus sign can be canceled by integrating from right to left. So regardless of whether $f$ is increasing or decreasing, we get the more familiar formula

$$\int_a^b g(x) dx = \int_{f(a)}^{f(b)} g(f^{-1}(u)) (f^{-1})'(u) du.$$

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Here is how I think about differentials and infinitesimals: Warning: this might confuse you.

I see $d$ as an operator that makes a normal sized expression (like -5, 2, 2^10) into an infinitesimal number. I even found rules to deal with higher-order infinitesimals like $d^3 x$ and $(dx)^3$, but what you need to know are: "normal sized number * infinitesimal = infinitesimal" and "infinity * infinitesimal = normal sized number".

$\int$ repeatingly adds an expression infinite number of times, and if you say something like $\int (x+1)^2$, then the answer doesn't converge info a normal-sized number because normal sized number * infinity = infinity.

However, the expression $\int (x+1)^2 dx$ makes sense because infinity * infinitesimal = normal sized number.

What about $\int (x+1)^2 ds$? This integral doesn't reduce to anything if the variable $s$ isn't defined, but what if we explicitly define $s$ is defined in term of $x$? For example, $s = x+1$, then $\int (x+1)^2 ds = s^3/3 + C = (x+1)^3/3 + C$.

I also treat $d$ as an operator that binds to an expression, not variable. We can say $ds = d[x+1]$. So the integral becomes $\int (x+1)^2 d(x+1) = (x+1)^3/3+C$.

Now let's see how to do your integral without introducing a variable u:

$\int 4x^5 (x^2 + 1)^{1/3} dx = \int (4x^3/3) (3 x^2dx) (x^2 + 1)^{1/3}$

Since $d[x^3 + 1]/dx = 3x^2$, then $d[x^3 + 1] = 3x^2 dx$:

$\int (4x^3/3) (3 x^2dx) (x^2 + 1)^{1/3} = \int (4x^3/3) d[x^3 + 1] (x^2 + 1)^{1/3}$

Since $(4/3)(([x^3+1] - 1)[x^3+1]^{1/3} = (4/3)x^3[x^3+1]^{1/3}$, we can rewrite:

$\int (4x^3/3) d[x^3 + 1] (x^2 + 1)^{1/3} = \int (4/3)([x^3+1] - 1)[x^3+1]^{1/3} d[x^3 + 1] $.

If we replaces all instances of $[x^3+1]$ into $u$ (remember, $d[x^3+1]$ becomes $du$), then we get integration by u-subsitution.

See also: http://www.math.vanderbilt.edu/~schectex/commerrs/#Differentials

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Consider $$\begin{array}{lll} \displaystyle\int 4x^5(x^2+1)^\frac{1}{3}dx&=&\displaystyle\int 4x^5(x^2+1)^\frac{1}{3}dx\\ &=&\displaystyle2\int (x^2)^2(2x)(x^2+1)^\frac{1}{3}dx\\ &=&\displaystyle2\int ((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}dx\\ \end{array}$$ Next, let's make the substitution $u=x^2+1$, implying that $du/dx=2x$. We do this by multiplying by $1$.

$$\begin{array}{lll} &=&\displaystyle2\int ((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}dx\cdot\frac{\frac{du}{dx}}{\frac{du}{dx}}\\ &=&\displaystyle2\int \frac{((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}\frac{dx\cdot du}{dx}}{\frac{du}{dx}}\\ &=&\displaystyle2\int \frac{((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}du}{2x}\\ \end{array}$$ Notice how I interpreted $du/dx$ as a ratio of differential in the numerator, but as a derivative in the numerator. This "inconsistency" isn't a bad thing, for I think that everyone can remember in middle/high school how useful it was to interpret a division as a multiplication (e.g. $5\div2=5\times\frac{1}{2}$) while manipulating fractions. Extending this technique of "multiple interpretations" to calculus, can only make your mathematical skills stronger.

Continuing $$\begin{array}{lll} &=&\displaystyle2\int \frac{((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}du}{2x}\\ &=&\displaystyle2\int ((x^2+1)-1)^2(x^2+1)^\frac{1}{3}du\\ &=&\displaystyle2\int (u-1)^2u^\frac{1}{3}du\\ &=&\dots\\ \end{array}$$

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