What is the total mass of the rope of length $L$ when its linear density varies as $H(x) = e^{x/2}$?

Question

The question appears as a simple integration question but I want to ask the real concept of this integration. Whether I would do $$M = \int_0^L xe^{x/2}\ dx$$ or do $$M=\int_0^L e^{x/2}\ dx$$ directly and how is it analogous to areas under bounded regions where we used to do $\int y\ dx$ directly. If the latter is a correct choice then why this extra $x$ term is introduced and as we know $dx$ also has units of length and $x$ also so its the units become $[\textrm{Length}]^2$? Really confused about how the quantities varying as mathematical functions

Answer 1

Dylan has already given the short and simple answer in a comment. To clarify the analogy with areas under curves, define a function $m(x)$ which gives the mass of the rope up to the point $x$. What is the derivative of that function? It is the limit of the ratio $\frac{m(x+\Delta x)-m(x)}{\Delta x}$ as $\Delta x \rightarrow 0$. This is exactly the quantity we call linear density - the mass contained between $x$ and $x+\Delta x$, divided by the length of that segment, which is $\Delta x$.

But if $H(x) = m'(x)$, then $m(x)$ is an integral of $H(x)$. This is exactly the same as if you have a curve $y= f(x)$; then the area under the curve (from some fixed left bound to a variable right bound) is given by an integral $\int_{x_0}^x y ds = F(x)$, where the function $F(x)$ is such that its derivative is $F'(x) = f(x)$ and $F(x_0) = 0$. For the rope, the role of $f(x)$ - the height of the curve - is taken by the linear density $H(x)$, while the area function $F(x)$ becomes the mass function $m(x)$. The most important thing for the analogy is that $m'(x) = H(x)$ looks and works just like $F'(x) = f(x)$.

If we want the mass of the whole rope, we find $M = m(L) = \int_0^LH(x)dx$. For completeness, the first integral with the extra $x$ is equal to $Mx_{CM}$, where $x_{CM}$ is the center of mass of the rope (provided the rope is straight).

Answer 2

The treatment of this sort of problem in an elementary physics textbook generally goes like this:

Consider the mass $dm$ of an infinitesimal length $dx$ of rope. This is given by:

$dm = \lambda dx$, where $\lambda$ is the (varying) linear density.

Hence the total mass $M$ of a rope of length $L$ is $\displaystyle M = \int_0^M dm = \int_0^L \lambda dx = \int_0^L e^{\frac x2}dx$

I find the consideration of infinitesimal elements before performing the integration a highly intuitive way to understand the problem.