Evaluate the improper integrals in Exercises 53–62.
∫ fr...

Question

Evaluate the improper integrals in Exercises 53–62.

∫ from −∞ to ∞ of (1 / (4x² + 9)) dx

Accepted Answer

Welcome back everyone. In this problem we want to evaluate the improper integral of 1 divided by 9 x2 + 16 with respect to x between the bones of negative infinity and positive infinity. A says it's 1/12 of pi, B 1/8 of pi, C 1/6 of pi, and D a quarter of pi. Now, to solve this improper integral, we, we can treat the infinite boundaries by expressing the integral as a limit, but notice here that our integral goes from negative infinity to infinity. So what can help us to make this a bit simpler for us to apply our limit? Well, Let's let FX. Be equal to 1 divided by 9 x2 + 16 over integral, right? Notice that OK. Here, FFX is an even function. And since it's an even function, what that basically means is that fx is equal to f of x. So with that in mind, we can simplify our calculation to say that our definite integral of fx with respect to x between negative infinity and infinity would be the same thing as 2 multiplied by the integral between 0 and infinity of fx with respect to x. And now if we apply our limit here, then really we can rewrite this as 2 multiplied by the limit as b approaches infinity of our integral between 0 and b of fx with respect to x. So now it's a bit easier for us to solve in this form. So let's go ahead and try to find the antiderivative and then apply our limits. Now notice, OK. Notice that If we take our integrant 1 divided by 9 X2 + 16, we could rewrite it, OK, as 1 divided by 3 x 2 plus 42. And that's important because recall, OK, that in integration, the integral. Of 1 divided by U2 plus a 2, OK, with respect to you. It will be equal to 1 divided by a multiplied by the inverse tangent of u divided by a. So if you compare both of these forms, you notice that it's similar. It's 1 divided by the sum of 2 squares. So in this case, if we let u be equal to 3 x. Then that means DUDX, the derivative of u with respect to x, would be equal to 3, OK. And if we wanted to solve here for DX, that means DX is going to be equal to 1/3 of DU. So now we can rewrite our definite or sorry or indefinite integral, OK, uh, that is 1 divided by 3 x 2 plus 42, OK, with respect to x as 1/3 of the integral of 1 divided by u2 plus 42 with respect to u. I know when we evaluate that. It turns out to be a 3rd, OK, of. A quarter, OK, because here notice that A is 4, so that's a third of a quarter of the inverse tangent of u divided by 4. And now if we simplify and rewrite it with x in our expression, that's really 1/12 of the inverse tangent of 3/4 of x plus c. Since this is our indefinite integral, we can use it with our definite integral. No, we can evaluate the definite integral with limits. So applying the limits for the symmetric integral, that means it becomes 2 multiplied by the limit as b approaches infinity of 1/12 of the inverse tangent of 3/4 of x, OK. Between the limits of 0 and b If we go ahead here, so that will be 2 multiplied by 1/12 multiplied by the limit as b approaches infinity of the inverse tangent of 3/4 of b, OK. Minus the inverse tangent of 0. No, 2 multiplied by 1/12 is 1/6, OK. And we know that the limit as b approaches infinity of the inverse tangent of 3/4 of b would be equal to 1/2 of pi. So this is 1/6 multiplied by 1/2 of pi minus 0 because the inverse tangent of 0 is 0. I know when we simplify here we get this to be 11/12 of pi. Therefore, the integral of 1 divided by 9 x2 + 16 with respect to x between negative infinity and infinity is 1/12 of pi. If we look back at our answer choices, that means a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Evaluate the improper integrals in Exercises 53–62.∫ from −∞ to ∞ of (1 / (4x² + 9)) dx

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Evaluate the improper integrals in Exercises 53–62.
∫ from −∞ to ∞ of (1 / (4x² + 9)) dx