The portal has been deactivated. Please contact your portal admin.

Question Video: Determining Whether the Improper Integration of a Rational Function with Infinite Limits of Integration Is Convergent or Divergent Mathematics • Higher Education

Determine whether the integral ∫_(βˆ’βˆž) ^(0) 1/(3 βˆ’ 4π‘₯) dπ‘₯ is convergent or divergent.

05:53

Video Transcript

Determine whether the integral from negative ∞ to zero of one divided by three minus four π‘₯ with respect to π‘₯ is convergent or divergent.

We’re given a definite integral, and we’re 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 ∞. And if either of our limits of integration are positive or negative ∞, then we say that our integral is improper. So, we’re going to need to use our rules for evaluating improper integrals. We need to recall the following. The integral from negative ∞ to a constant π‘Ž of 𝑓 of π‘₯ with respect to π‘₯ will be equal to the limit as 𝑑 approaches negative ∞ of the integral from 𝑑 to π‘Ž of 𝑓 of π‘₯ with respect to π‘₯. 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 we’re 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 π‘Ž is zero. This gives us that the improper integral given to us in the question should be equal to the limit as 𝑑 approaches negative ∞ of the integral from 𝑑 to zero of one divided by three minus four π‘₯ with respect to π‘₯. And the question wants us to determine the convergence or divergence of this limit. Before we do this, there’s a few things worth pointing out.

First, 𝑑 is approaching negative ∞. In particular, this tells us that eventually 𝑑 will be less than zero. So, we can just assume 𝑑 is less than zero, our upper limit of integration. In particular, 𝑑 will be a finite number. So, we can now see our interval of integration is now the closed interval from 𝑑 to zero. So now, we’re just trying to evaluate a regular definite integral. So, we can use any of our tools to help us evaluate this integral. We’re going to want to do this by using substitution. We want to use the substitution 𝑒 is equal to three minus four π‘₯.

To use this substitution, we need to differentiate both sides with respect to π‘₯. We get d𝑒 by dπ‘₯ is equal to negative four. And although d𝑒 by dπ‘₯ is not a fraction, when we’re 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 d𝑒 is equal to negative four dπ‘₯. 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 they’re 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 π‘₯ would be three over four. So, our integrand is continuous everywhere except where π‘₯ is equal to three over four. And since 𝑑 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 𝑒. Let’s start with the new upper limit of integration. We substitute π‘₯ is equal to zero into our expression for 𝑒. We get 𝑒 is equal to three minus four times zero. And we can just calculate this expression; it’s equal to three. And we can do the same for the lower limit of integration. This time, we substitute π‘₯ is equal to 𝑑. We get 𝑒 is equal to three minus four 𝑑. We’re now ready to start using our substitution.

First, we’ll work the new upper and lower limit of integration. The new upper limit is three, and the new lower limit is three minus four 𝑑. Next, in our integrand, we’ll replace three minus four π‘₯ with 𝑒. This gives us a new integrand of one over 𝑒. Finally, we want to replace dπ‘₯. We can see in terms of differentials we have d𝑒 is equal to negative four dπ‘₯. So, we’ll divide both sides of this equation through by negative four. This gives us negative a quarter d𝑒 is equal to dπ‘₯. Now, we can replace dπ‘₯ with negative one-quarter d𝑒. This gives us the limit as 𝑑 approaches negative ∞ of the integral from three minus four 𝑑 to three of one over 𝑒 times negative one-quarter with respect to 𝑒.

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 𝑒 with respect to 𝑒, and we know how to do this. The integral of one over 𝑒 with respect to 𝑒 will be the natural logarithm of the absolute value of 𝑒 plus the constant of integration. And of course, we don’t need to include our constant of integration because this is a definite integral. This gives us negative one-quarter multiplied by the limit as 𝑑 approaches negative ∞ of the natural logarithm of the absolute value of 𝑒 evaluated at the limits of integration three minus four 𝑑 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 𝑑 approaches negative ∞ of the natural logarithm of the absolute value of three minus the natural logarithm of the absolute value of three minus four 𝑑. 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.

Let’s look at the first term. The first term is the natural logarithm of the absolute value of three. This is a constant; it doesn’t change as 𝑑 changes. Now, let’s take a look at our second term. We’re subtracting the natural logarithm of the absolute value of three minus four 𝑑, and our value of 𝑑 is approaching negative ∞. As 𝑑 approaches negative ∞, three minus four 𝑑 is growing without bound. And if it’s growing without bound, it’s approaching ∞.

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

Therefore, because this limit does not converge, we were able to show the integral from negative ∞ to zero of one divided by three minus four π‘₯ with respect to π‘₯ is divergent.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.