Which of the improper integrals in Exercises 63–68 converge an...

Question

Which of the improper integrals in Exercises 63–68 converge and which diverge?

∫ from −∞ to ∞ of (2 / (e^x + e^(−x))) dx

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Determine whether the improper integral, the integral evaluated from negative infinity to infinity of 6 divided by e to the power of x plus e to the power of x dx converges or diverges using the direct comparison test. OK, so it appears for this particular prompt we're asked to take the improper integral that is provided to us and we're asked to use the direct comparison test to determine whether or not this specific improper integral will either a converges or b diverges. So once again, we're ultimately going to use the direct comparison test to see whether or not this improper integral will converge or diverge. So with that in mind, our first step that we need to take is we need to recall and use the direct comparison test. So based on our prior knowledge, we should be able to recall that the direct comparison test states that if 0 is less than or equal to fx and if fx is less than or equal to g x or x is greater than or equal to a. And the integral evaluated from A to infinity of G of X DX converges, then that will mean the integral evaluated from A to infinity of F X DX will also converge. So keeping this in mind, we need to note that based on what our problem tells us, we can say that the integrand 6 divided by. E to the power of X plus E to the power of X, this integrand will be even, so I put in E in quotation marks. Because our integrand is even, we can therefore write that the integral evaluated from negative infinity to infinity of 6 divided by e to the power of x plus e to the power of x dx will be equal to 2 multiplied by 6 multiplied by the integral evaluated from 0 to infinity of. The x divided by e to the power of x plus e to the power of x, which will be equal to 12 multiplied by the integral evaluated from 0 to infinity of d x divided by e to the power of x plus e to the power of negative x. And we now need to apply our direct comparison test on the interval square bracket of 0, infinity parenthesis. So we need to note that 4 x is greater than or equal to 0. That will mean that as a result, we will have that e to the power of x plus e to the power of x will be greater than or equal to e to the power of x. Thus we can say, so thus we can say that 0 will be less than or equal to 1 divided by e to the power of x plus e to the power of x which will be less than or equal to 1 divided by e to the power of x. So now at this point, we need to compute the integral of the upper bound. So we need to take the integral as that's going to be evaluated from 0 to infinity of 1 divided by e to the power of xDX, which will be equal to the limit as B approaches infinity of the integral evaluate from 0 to B. Of e to the power of x dx, which will be equal to the limit as b approaches infinity of the integral evaluated from 0 to b of e to the power of x multiplied by d of x. Which will be equal to the limit as b approaches infinity of square brackets of e to the power of x evaluated from 0 to b of which will be equal to negative the limit as b approaches infinity of square brackets of 1 divided by e to the power of x evaluated from 0 to b. Which will be equal to negative dilemma as b approaches infinity of square brackets of 1 divided by e to the power of b minus 1 divided by e to the power of 0, which will be equal to negative square brackets of 1 divided by e to the power of infinity minus 1, which will be simply equal to 1. And we need to note that since the integral evaluated from 0 to infinity of 1 divided by e to the power of x. D x we need to note because this integral converges, so I'll put a C in quotation marks, because this integral converges, we can say that 12 multiplied by the integral evaluate from 0 to infinity of 1 divided by e to the power of x plus e to the power of negative x dx will also converge. So that means our final answer without a doubt will be converges. Hooray, we did it. So looking at our multiple choice answers, our correct and final answer without a doubt will be multiple choice answer letter A, which once again is converges, and that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Which of the improper integrals in Exercises 63–68 converge and which diverge?∫ from −∞ to ∞ of (2 / (e^x + e^(−x))) dx

Watch next

Which of the improper integrals in Exercises 63–68 converge and which diverge?
∫ from −∞ to ∞ of (2 / (e^x + e^(−x))) dx