### Video Transcript

Determine whether the integral from negative โ to zero of one divided by three minus four ๐ฅ with respect to ๐ฅ is convergent or divergent.

Weโre given a definite integral, and weโre asked to determine whether this definite integral is convergent or divergent. The first thing we need to notice is the lower limit of integration in our definite integral is negative โ. And if either of our limits of integration are positive or negative โ, then we say that our integral is improper. So, weโre going to need to use our rules for evaluating improper integrals. We need to recall the following. The integral from negative โ to a constant ๐ of ๐ of ๐ฅ with respect to ๐ฅ will be equal to the limit as ๐ก approaches negative โ of the integral from ๐ก to ๐ of ๐ of ๐ฅ with respect to ๐ฅ. And this is, of course, provided that our limit converges. This is what the question means when it asked us if this limit is convergent or divergent. We just need to check the convergence or divergence of this limit.

So, the first thing weโre going to want to do is construct this limit for the improper integral given to us in the question. We can see in our case, the upper limit of integration is zero. So, our value of the constant ๐ is zero. This gives us that the improper integral given to us in the question should be equal to the limit as ๐ก approaches negative โ of the integral from ๐ก to zero of one divided by three minus four ๐ฅ with respect to ๐ฅ. And the question wants us to determine the convergence or divergence of this limit. Before we do this, thereโs a few things worth pointing out.

First, ๐ก is approaching negative โ. In particular, this tells us that eventually ๐ก will be less than zero. So, we can just assume ๐ก is less than zero, our upper limit of integration. In particular, ๐ก will be a finite number. So, we can now see our interval of integration is now the closed interval from ๐ก to zero. So now, weโre just trying to evaluate a regular definite integral. So, we can use any of our tools to help us evaluate this integral. Weโre going to want to do this by using substitution. We want to use the substitution ๐ข is equal to three minus four ๐ฅ.

To use this substitution, we need to differentiate both sides with respect to ๐ฅ. We get d๐ข by d๐ฅ is equal to negative four. And although d๐ข by d๐ฅ is not a fraction, when weโre using integration by substitution, it can help to treat it a little bit like a fraction. We get the equivalent statement in terms of differentials d๐ข is equal to negative four d๐ฅ. And there is one thing worth checking here for our substitution. We want our integrand to be continuous on the interval of integration.

And to do this, we need to notice that our integrand is a rational function. Rational functions are continuous everywhere where theyโre defined. And a rational function will be defined everywhere except where its denominator is equal to zero. And we can solve the denominator being equal to zero. We would get that ๐ฅ would be three over four. So, our integrand is continuous everywhere except where ๐ฅ is equal to three over four. And since ๐ก is negative, this is not in our interval of integration. So, we can just carry on using our substitution.

Now, we need to find the new limits of integration. To find our new limits of integration, we need to substitute our upper limit and our lower limit into our equation for ๐ข. Letโs start with the new upper limit of integration. We substitute ๐ฅ is equal to zero into our expression for ๐ข. We get ๐ข is equal to three minus four times zero. And we can just calculate this expression; itโs equal to three. And we can do the same for the lower limit of integration. This time, we substitute ๐ฅ is equal to ๐ก. We get ๐ข is equal to three minus four ๐ก. Weโre now ready to start using our substitution.

First, weโll work the new upper and lower limit of integration. The new upper limit is three, and the new lower limit is three minus four ๐ก. Next, in our integrand, weโll replace three minus four ๐ฅ with ๐ข. This gives us a new integrand of one over ๐ข. Finally, we want to replace d๐ฅ. We can see in terms of differentials we have d๐ข is equal to negative four d๐ฅ. So, weโll divide both sides of this equation through by negative four. This gives us negative a quarter d๐ข is equal to d๐ฅ. Now, we can replace d๐ฅ with negative one-quarter d๐ข. This gives us the limit as ๐ก approaches negative โ of the integral from three minus four ๐ก to three of one over ๐ข times negative one-quarter with respect to ๐ข.

And now, we want to simplify this expression. First, we see we have a constant factor of negative one-quarter. Since this is a constant, we could just take this outside of our integral. But then, this is a constant factor, so we could also take it outside of our limit. And this gives us the following expression. We see we now need to evaluate the integral of one over ๐ข with respect to ๐ข, and we know how to do this. The integral of one over ๐ข with respect to ๐ข will be the natural logarithm of the absolute value of ๐ข plus the constant of integration. And of course, we donโt need to include our constant of integration because this is a definite integral. This gives us negative one-quarter multiplied by the limit as ๐ก approaches negative โ of the natural logarithm of the absolute value of ๐ข evaluated at the limits of integration three minus four ๐ก and three.

All we need to do now is evaluate this at the limits of integration. Doing this, we get negative one-quarter multiplied by the limit at ๐ก approaches negative โ of the natural logarithm of the absolute value of three minus the natural logarithm of the absolute value of three minus four ๐ก. Remember, the question wants us to determine whether this integral is convergent or divergent. And the convergence or divergence of this integral is exactly the same as the convergence or divergence of this limit. So, all we need to do is determine whether this limit converges or diverges.

Letโs look at the first term. The first term is the natural logarithm of the absolute value of three. This is a constant; it doesnโt change as ๐ก changes. Now, letโs take a look at our second term. Weโre subtracting the natural logarithm of the absolute value of three minus four ๐ก, and our value of ๐ก is approaching negative โ. As ๐ก approaches negative โ, three minus four ๐ก is growing without bound. And if itโs growing without bound, itโs approaching โ.

We then need to recall that the limit as ๐ฅ approaches โ of the natural logarithm of ๐ฅ is also equal to โ. In other words, if weโre taking the natural logarithm of something which grows without bound, then this will also grow without bound. Therefore, weโre taking the limit of a constant in some integral that grows without bound. We can conclude that this must not converge. And itโs important to reiterate here, when a limit evaluates to give us positive or negative โ, it still does not converge; itโs a divergent limit. However, saying itโs equal to positive or negative โ can give us useful information.

Therefore, because this limit does not converge, we were able to show the integral from negative โ to zero of one divided by three minus four ๐ฅ with respect to ๐ฅ is divergent.