# The test says that if r a g x dx converges then so

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The test says that if R a g ( x ) dx converges, then so does R a f ( x ) dx , and if R a f ( x ) dx diverges then so does R a g ( x ) dx . This should make sense: if g ( x ) has finite area, then so must f ( x ) since f ( x ) is smaller, and if f ( x ) has infinite area, then so does g ( x ) since g ( x ) is bigger. In truth, we can apply this test to all sorts of improper integrals provided that there is only one improper bound and it happens at an endpoint. However, the functions must be greater than or equal to zero . 12.2.1 What to compare to: the p-test Notice that, from the statement of the theorem, if we want to prove an integral converges, we need to find a bigger function which we know converges. Similarly, if we want to prove an integral diverges, then we must find a smaller function which we know diverges. This means that when you first encounter an integral, you have to make an educated guess as to whether you think the integral converges or diverges, because this determines which way you need to bound the function. To make these guesses, you want to keep the following in mind: (1) Integrals of the form Z a 1 x p dx, where a > 0 (must be strictly positive), converge when p > 1 and diverge when p 1. (2) Integrals of the form Z a 0 1 x p dx, where a > 0 is not infinity, converge when p < 1 and diverge when p 1. Notice nothing good ever happens with p = 1, and that everything here is the (3) Now consider Z a e kx dx, where a is not -∞ . This converges if k < 0 and diverges if k 0. Exponential decay is as good as it gets. (4) Similarly, if we consider Z a -∞ e kx dx, now with a not , this converges if k > 0 and diverges if k 0. With these last two, notice that all you need is for the graph of the exponential to be tending to 0 as you approach the infinite bound. That should make it easy to remember. 12.2.2 Examples Example 12.11. Determine whether R 1 1 x 4 +1 dx converges or diverges. 29
Math 31B Notes Sudesh Kalyanswamy Solution. This is not an integral we know how to compute, so we use the comparison test. First, we need to make a guess, as this will determine which way we want inequalities to go. Notice that as x → ∞ , the bottom just behaves like x 4 , since the extra 1 is insignificant. So the fraction behaves like 1 x 4 , and R 1 1 x 4 dx converges by case (1) of the integrals above. So we can make the guess that the integral converges. One thing you should try and do is “remove” the terms you don’t care about, assuming the inequality goes in the proper direction when you do. In this case, we’d like to remove the 1 from the denominator, so is Z 1 1 x 4 + 1 dx Z 1 1 x 4 dx ? Well yes, since x 4 + 1 x 4 , meaning 1 x 4 +1 1 x 4 , which is what we wanted. The other way to think about it is that the denominator of 1 x 4 +1 is bigger than that of 1 x 4 , so the fraction is smaller. The integral R 1 1 x 4 dx converges by the p -test, so by comparison, the original integral converges.