# 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

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…

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