Should the Units of a Constant Match the Integral Boundary Conditions?

[Another]

Homework Statement

Integrate an exponential function with a constant (1/cm) from 0.8 um to 1.8 um.

Relevant Equations

int A exp(-(B/(x+x^2+x^3)))

I want to integrate this function

$\int_{0.8um}^{1.8um} A e^{B/E(x)} \, dx$

But A has a unit as $1/cm$. Should I change $1/cm$ to $1/um$ by multiplying $1/10^{4}$

For this function, I decided to integrate using the online numerical integral, This side .

I am just curious that the unit of the constants must be the same as boundary condition for integration.

 

[PeroK]

Homework Statement: Integrate an exponential function with a constant (1/cm) from 0.8 um to 1.8 um.
Relevant Equations: int A exp(-(B/(x+x^2+x^3)))

I want to integrate this function

$\int_{0.8um}^{1.8um} A e^{B/E(x)} \, dx$

But A has a unit as $1/cm$. Should I change $1/cm$ to $1/um$ by multiplying $1/10^{4}$

For this function, I decided to integrate using the online numerical integral, This side .

I am just curious that the unit of the constants must be the same as boundary condition for integration.

The integral can only have one value. And that must coincide with the calculation where the units match.

 

[Another]

The integral can only have one value. And that must coincide with the calculation where the units match.

So, Do I have to change the unit of a constant to be the same?
Thankyou very much.

 

[vela]

Homework Statement: Integrate an exponential function with a constant (1/cm) from 0.8 um to 1.8 um.
Relevant Equations: int A exp(-(B/(x+x^2+x^3)))

I want to integrate this function

$\int_{0.8um}^{1.8um} A e^{B/E(x)} \, dx$

If the function $E(x)$ is $E(x) = x + x^2 + x^3$, you need to fix that first. You clearly can't add quantities with different dimensions and units, i.e., $\rm{m, m^2, and\ m^3}$.

But A has a unit as $1/cm$. Should I change $1/cm$ to $1/um$ by multiplying $1/10^{4}$

You don't technically have to change it, but at the very least you have to keep track of the units as you should always do.

On the one hand, you could have an answer that has units of $\rm \mu m/cm$ but most likely you're expected to give a unitless answer. It's kind of like answering the question "if you're traveling 5 m/s, what distance will you travel in 1 min?" by saying "5 m min/s" whereas the expected and useful answer is "300 m". On the other hand, I can imagine scenarios where the units $\rm \mu m/cm$ might make more sense semantically.

 

[Steve4Physics]

If the function $E(x)$ is $E(x) = x + x^2 + x^3$, you need to fix that first. You clearly can't add quantities with different dimensions and units, i.e., $\rm{m, m^2, and\ m^3}$.

In $E(x) = x + x^2 + x^3$ we are not told the unit for $x$; this is missing critical information.

For illustration, suppose that $E(x)$ is potential energy in joules as a function of distance $x$ is metres. Then:
$E(x) = x + x^2 + x^3$
is valid and is in effect shorthand for
$E(x) = Ax + Bx^2 + Cx^3$
where the 3 constants are $A = 1J/m, B = 1J/m^2$ and $C = 1J/m^3$.
The constants and their units are 'hidden' because the numerical value of each constant happens to be 1.

If we wish to change so that $x$ is in cm, the modified equation would simply be:
$E(x) = \frac x{100} + (\frac x{100})^2 + (\frac x{100})^3$

Ideally the units should be indicated in the function definition, e.g.
$E(x/m) = x + x^2 + x^3$ and
$E(x/cm) = \frac x{100} + (\frac x{100})^2 + (\frac x{100})^3$

 

[Another]

If the function $E(x)$ is $E(x) = x + x^2 + x^3$, you need to fix that first. You clearly can't add quantities with different dimensions and units, i.e., $\rm{m, m^2, and\ m^3}$.You don't technically have to change it, but at the very least you have to keep track of the units as you should always do.

On the one hand, you could have an answer that has units of $\rm \mu m/cm$ but most likely you're expected to give a unitless answer. It's kind of like answering the question "if you're traveling 5 m/s, what distance will you travel in 1 min?" by saying "5 m min/s" whereas the expected and useful answer is "300 m". On the other hand, I can imagine scenarios where the units $\rm \mu m/cm$ might make more sense semantically.

Thank you for answering, I know that there are coefficients on the front of the polynomials but I just typed the very simple form of the equation by ignoring these coefficients (make it very easy to read and I'd like to focus on A and the unit of it) . Finally, the completely correct form of this equation will be calculated in the program by myself, and the unit of E(x) will be canceled by the unit of B.

Thank you very much for the wonderful advice.