# Why no change in limits of integration here?

Gold Member

## Homework Statement

Please see attached image for the full scope of the problem, and to see the work drawn out by the text.
My question lies with line 3 as it is clear that u-substitution was used on a definite integral, but the limits of integration were not changed.

## The Attempt at a Solution

I think that line 3 should read:

##=100+\left[100e^{0.02t}\right]|_0^{0.2}##

My reasoning is that when t=0, u=0 and when t=10, u=0.2

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No u-substitution was necessary. It's a easily confirmable fact that
$$\int_a^b A e^{Bt} dt = [\frac{A}{B} e^{Bt}]^{b}_a$$

However, even if a u-substitution were used, if you change back to your original integration variable, you should change the limits back.

\begin{align*} \int_0^{\pi/2} \sin(\theta) \cos(\theta) d\theta &= \int_0^1 u\, du\\ &= [ \frac{1}{2} u^2]^1_0\\ &= [\frac{1}{2} \sin^2(\theta)] ^{\pi/2}_0\\ &= \frac{1}{2} \end{align*}

opus
Gold Member
Thank you for the reply. So if we change back, what's the point of changing them in the first place?
Maybe my confusion lies in the fact that I'm not really sure why we change them to begin with.

Thank you for the reply. So if we change back, what's the point of changing them in the first place?
Maybe my confusion lies in the fact that I'm not really sure why we change them to begin with.
We generally make a u-substitution to make integration easier, and more straightforward. For example, in the integral I gave above, it's not immediately obvious how to integrate it, until you make the substitution ##u = \sin(\theta)##. Then it becomes trivial. Generally, for a definite integral, there's no reason to switch back, unless calculating the new limits is for some reason very difficult, which is pretty rare. For an indefinite integral, you pretty much always switch back, because you want the answer in terms of your original variable -- but in that case, there are no limits of integration, so it's a bit of a moot point.

opus
Gold Member
So then if I use u-substitution I will always have to change the limits of integration, and once I do so, I don't have to change them back. But if I don't use u-substitution, then I obviously don't have to change the limits of integration and I can just proceed with ##F(b)-F(a)##?

Yes, that is correct.

opus
Mark44
Mentor
Thank you for the reply. So if we change back, what's the point of changing them in the first place?
Maybe my confusion lies in the fact that I'm not really sure why we change them to begin with.
The reason for a substitution (which isn't the question you asked) is to get an integral that's easier to evaluate. If the integral is a definite integral, you can change the limits of integration to new limits, or you can leave them unchanged, provided that you remember that the integral is in terms of one variable, and the limits are of a different variable.

Here's Dewgale's example, where the limits of integration are changed:
##\int_0^{\pi/2} \sin(\theta) \cos(\theta) d\theta = \int_0^1 u\, du\\
= [ \frac{1}{2} u^2]^1_0\\
= [\frac{1}{2} \sin^2(\theta)] ^{\pi/2}_0\\
= \frac{1}{2}##

Same example, with the limits not changed:
##\int_0^{\pi/2} \sin(\theta) \cos(\theta) d\theta = \int_{\theta = 0}^{\pi/2} u\, du\\
= [ \frac{1}{2} u^2]_{\theta = 0}^{\pi/2}\\
= [\frac{1}{2} \sin^2(\theta)] ^{\pi/2}_0\\
= \frac{1}{2}##

In the next to last line above, we "undid" the substitution, so the integral is again in terms of ##\theta##, so I no longer need to remind myself of this fact.

opus
Gold Member
Interesting so for the first example, line 2, you have an integral in terms of ##u##, and limits in terms of ##u##. Then in the next line, everything is back in terms of θ.
For the second example, line 2, you have an integral in terms of ##u## and limits in terms of θ. But once you change the integral back in terms of θ they both now match. This makes sense.
I was under the impression that once you used u-substitution, you had a new function and thus would need new limits of integration. But I can see, as you said, that either way they end up with the same solution.