# Video: Determining Whether the Improper Integration of a Function with Infinite Limits of Integration Is Convergent or Divergent

Determine whether the integral β«_(0)^(β) 1/β(1 + π₯) dπ₯ is convergent or divergent.

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### Video Transcript

Determine whether the integral from zero to β of one divided by the fourth root of one plus π₯ with respect to π₯ is convergent or divergent.

In this question, weβre given a definite integral and weβre asked to determine if this definite integral is convergent or divergent. And we can immediately notice something about our integral. The upper limit of integration is β. And when one of our limits of integration are positive or negative β, we know that our integral is improper.

So to answer this question, we first need to recall how we evaluate integrals where the upper limit of integration is β. We recall the integral from π to β of π of π₯ with respect to π₯ is equal to the limit as π‘ approaches β of the integral from π to π‘ of π of π₯ with respect to π₯ provided that this limit exists. And this is what the question means when it asks us about the convergence or divergence of our integral. It means the convergence or divergence of this limit.

And there is one thing worth pointing out. Since our values of π‘ are approaching β, we can just assume our values of π‘ are bigger than π. So the first thing weβre going to need to do is construct this limit for the integral given to us in the question. We set our function π of π₯ to be the integrand given to us in the question. And we set our lower limits of integration π equal to zero. This gives us the integral from zero to β of one divided by the fourth root of one plus π₯ with respect to π₯ is equal to the limit as π‘ approaches β of the integral from zero to π‘ of one divided by the fourth root of one plus π₯ with respect to π₯ provided that this limit exists.

So we need to determine the convergence or divergence of this limit. And to do this, weβre going to need to evaluate our definite integral. To do this, we first need to notice since weβre taking the limit as π‘ approaches β, eventually our values of π‘ will be bigger than zero. So our interval of integration will be the closed interval from zero to π‘. And in fact, we can see our integrand is continuous on this interval. Our integrand is the composition of continuous functions. Itβs a linear function composed with a power function. And the composition of continuous functions is continuous on its entire domain.

In particular, we can see the top integrand is defined for all values of π₯ greater than or equal to zero. So we can evaluate this integral by using any of our rules of integration. Weβre going to do this by using substitution. Weβre going to use the substitution π’ is equal to π₯ plus one. To integrate by substitution, weβre first going to need to find an expression for dπ’ by dπ₯. We need to differentiate both sides of our expression for π’ with respect to π₯. We get dπ’ by dπ₯ is equal to one. Then although dπ’ by dπ₯ is not a fraction, when weβre using integration by substitution, we can treat it a little bit like a fraction. We get the equivalent statement in terms of differentials dπ’ is equal to dπ₯.

Weβre not done yet though. Remember, this is a definite integral. So we need to find the new limits of integration. To do this, we need to substitute these values of π₯ into our expression for π’. Letβs start with the upper limit of integration. We substitute π₯ is equal to π‘ into our expression for π’. We get π’ is equal to π‘ plus one. And we can do the same for the lower limit of integration. This is when π₯ is equal to zero. We get π’ is equal to zero plus one, which is of course just equal to one.

Weβre now ready to use our substitution π’ is equal to one plus π₯. First, we already showed the new upper limit of integration is π‘ plus one and the new lower limit of integration is one. Next, in our integrand, we replace one plus π₯ with π’. This gives us a new integrand of one divided by the fourth root of π’. Finally, in terms of differentials, we showed that dπ’ is equal to dπ₯. So we can replace dπ₯ with dπ’. This gives us the limit as π‘ approaches β of the integral from one to π‘ plus one of one over the fourth root of π’ with respect to π’.

And now weβre almost ready to evaluate our integral. To make this easier, weβll use our laws of exponents to rewrite our integrand as π’ to the power of negative one-quarter. Now, we can just evaluate this integral by using the power rule for integration. We want to add one to our exponent of π’, giving us a new exponent of three over four, and then divide by this new exponent of three over four. And then of course we still need to evaluate this at the limits of integration.

And we can simplify this expression. Instead of dividing by three over four, we can multiply by the reciprocal of three over four. And we know the reciprocal of three over four is four over three. Doing this, we get the limit as π‘ approaches β of four π’ to the power of three over four divided by three evaluated at the limits of integration π’ is equal to one and π’ is equal to π‘ plus one. All we need to do now is evaluate our antiderivative at the limits of integration.

Doing this, we get the limit as π‘ approaches β of four times π‘ plus one all raised to the power of three over four all over three minus four times one to the power of three over four divided by three. And at this point, we can just evaluate this limit. Our limit is as π‘ is approaching β. In our first term, we know that π‘ plus one is approaching β. Then because three over four is a positive number, this means that π‘ plus one all raised to the power of three over four is also approaching β. Then, multiplying this by three over four is not going to change its value. Itβs still going to grow without bound. Therefore, this approaches β.

We do still need to check our second term, however, but we can see that this is just a constant. It doesnβt change as the value of π‘ changes. Therefore, weβve shown this limit grows without bound. Itβs equal to β. But remember, saying the limit is equal to β still means that itβs divergent. Therefore, we were able to show the integral from zero to β of one divided by the fourth root of one plus π₯ with respect to π₯ is divergent.