Is it necessary to change the bounds when substituting in a definite integral?

[helixkirby]

Is it bad math to do this:

The definite integral from 1 to 2 of 1/(3x+1)

using the substitution u=3x+1, du=3dx
Then changing the integrand but keeping the bounds, then integrating, to (1/3)ln|u| from 1 to 2, then substitute back in u, so then I integrate with the bounds to get (1/3)ln|3x+1| from 1 to 2, getting (1/3)(ln(7)-ln(4)), is it bad math to not change your bounds if you're going to re-substitute anyway?

 

[Simon Bridge]

I lost you here:

...then integrating, to (1/3)ln|u| from 1 to 2, then substitute back in u, ...

... do you mean the result of the indefinite integral is ln|u|/3 then you... sub u=3x+1... then apply the limits to x?
That would be the normal approach ... you are implicitly changing the bounds twice without going to the trouble of writing it down.

 

[helixkirby]

Yes, pretty much, that's OK right?

 

[Simon Bridge]

Formally it goes like this:
Task: evaluate the following definite integral: $$\int_1^2\frac{\mathrm dx}{3x+1}$$

1. Substitute: $x+1 \to u : \mathrm dx = \frac{1}{3}\mathrm du$ into the indefinite integral and evaluate: $$\frac{1}{3}\int \frac{\mathrm du}{u} = \frac{1}{3}\ln\big|u\big|+c = \frac{1}{3}\ln\big|3x+1\big|+c$$ ...

2. Apply what you have learned (step 1) to the original problem: $$\int_1^2\frac{\mathrm dx}{3x+1} = \frac{1}{3}\bigg[ \ln\big|3x+1\big| \bigg]_1^2 = \frac{1}{3}\ln(7/4)$$

... it is a standard technique in mathematics to solve a problem by first working on a similar but easier problem in the hope that this will shed light on how to solve the original problem. It's like learning to swim in the sea by, first, swimming lengths in a pool.

 

[Mark44]

Is it bad math to do this:

The definite integral from 1 to 2 of 1/(3x+1)

using the substitution u=3x+1, du=3dx
Then changing the integrand but keeping the bounds, then integrating, to (1/3)ln|u| from 1 to 2,
then substitute back in u, so then I integrate with the bounds to get (1/3)ln|3x+1| from 1 to 2

This is one way to do it. It's good practice though to note that you are leaving the bounds as x values until you actually undo your substitution, like so:
$\int_1^2 \frac {dx}{3x + 1} = \frac 1 3 \int_{x = 1}^2 \frac{du}{u} = \frac 1 3 ln|u| |_{x = 1}^2$
Now you can undo the substitution and evaluate the antiderivative at 2 and 1.[/quote]

, getting (1/3)(ln(7)-ln(4)), is it bad math to not change your bounds if you're going to re-substitute anyway?

 

[certainly]

In regards to your original question:-

is it bad math to not change your bounds if you're going to re-substitute anyway?

Yes, because many times you can solve definite integrals with just these bounds. (i.e. without performing actual integration)
So if you don't change your bounds when you know your going to re-substitute, you might get into that habit.

 

[Mark44]

In regards to your original question:-

Yes, because many times you can solve definite integrals with just these bounds. (i.e. without performing actual integration)
So if you don't change your bounds when you know your going to re-substitute, you might get into that habit.

I wouldn't call this "bad math." If you're careful about keeping track of which variable the bounds are, as I showed in my work, you can get the correct value.

Evaluating a definite integral without actually doing the integration doesn't really come up that much, other than when you're first learning the techniques of integration.

 

[HallsofIvy]

It is not "bad math" in the sense of being mathematically invalid but it is, in my opinion, harder and more prone to error.

You have the integral $\int_1^2\frac{dx}{3x+ 1}$ and make the substation u= 3x+ 1 so that du= 3dx.
When x= 1, u= 4 and when x= 2, u= 7 so the integral become $\frac{1}{3}\int_4^7\frac{du}{u}= \left[\frac{1}{3}ln|u|\right]_4^7= \frac{1}{3}\left(ln(7)- ln(4)\right)= \frac{1}{3}ln\left(\frac{7}{4}\right)$.

But it is mathematically valid to do the indefinite integral $\frac{1}{3}\int \frac{du}{u}= \frac{1}{3}ln|u|+ C$, change back to x: $\frac{1}{3}\int \frac{du}{u}= \frac{1}{3}ln|3x+1|+ C$ and then evaluate at 1 and 2.

It would be bad notation to change the variable to u and leave the old limits of integration:
$\frac{1}{3}\int_1^2\frac{du}{u}$ is completely wrong!