The first integral well, we can actually use the comparison test, because cosine of x is between negative 1 and positive 1 point. So therefore, cosine squared x must be less than or equal to 1 now, since x, squared plus 6 is a positive number. Now, if we divide both sides by x, squared plus 6, the inequality does not change. So if we integrate this from 1 to infinity- and that should be less than the integral of this 1- and in other words we are comparing 2 d now we have to show that this integral the d is actually converging the anti derivative of 1 over x. Squared plus 6 is 1 over 6 times out 10 x over root 6. Now, if we evaluate from 1 to infinity now, if we replace x by infinity, we have artan infinity so arc. Tan infinity is pi over 2, because tan pi over 2 is infinity. So therefore, we get 1 over root 6 times pi over 2 for the first part, and when we replace the x variable by 1 here. Well, it is pretty simple: we get 1 over root, 6 acton 1 over root 6 right now. This is finite. This is finite, so the subtraction is finite, so we do have a finite number, so this integral is less than a finite number. So therefore, this is converging and we made that conclusion by comparing it to dwell. The second 1 is well. It is exactly the same thing right, it is the d. Now we showed that d is conversing in the earliest step, so therefore this integral should be converging and well. I suppose we can use a d here as well and for the third 1 using this fact. We can say that e to the negative x, divided by x, squared is less than or equal 1 of x squared. So therefore, this integral is less than that integral we are actually using j okay. Now we have to show that this is convergent. So how do you show that this 1 is converging? The j is converging well te anti derivative of 1 over x squared is avener x, and if we evaluate that we will get 1, so this integral equals to 1 point. So that means this integral this improper integral is less than or equal to 1 point. So therefore, it is also converging- and we use j, to show that all right for the fourth 1 now x, over root x to the 6 plus 6, must be less than or equal to x, divided by root x. To the 6 point right, when we reduce the denominator, then the ratio is going to get…

# SOLVED: For each of the improper integrals below, if the comparison test applies, enter either A or B followed by one letter from C to K that best applies. If the comparison test does not apply, enter

Comparison Test for Improper Integrals

Comparison Test for Improper Integrals