The integrals in Exercises 1–34 converge. Evaluate the integra...

Question

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.

∫₋∞^∞ 2x e^(−x²) dx

Accepted Answer

Welcome back everyone. In this problem we want to evaluate the improper integral of xe to the negative x2 with respect to x between the bones of negative infinity and infinity. A says it's 1, B-1, C0, and D says the integral diverges. Now, to figure this out, first, let's examine the examine the integral, OK? So let's let F of X be equal to XE to the negative X2. If we replace x with negative x, then f g x would be equal to x multiplied by e to the g x 2, which would be equal to x e x2. And now if we take a look here, you notice that this is actually negative f of x. So here fx equals -fx, so that means the function is odd. And for any other function. OK, for any odd function. Then when the integral converges over negative infinity to infinity. Then the integral of fx with respect to x would be equal to 0. So if we're going to evaluate this integral, we must split it into two separate integrals at x equals 0 and then use limits. In other words, we have to rewrite our definite integral. OK. As or integral from negative infinity to 0, OK, of x e to the negative x 2 with respect to x plus our integral from 0 to infinity of xe to the negative x 2 with respect to x and then if we were to rewrite this integral with a limit then. That means it's going to be the limit as A approaches negative infinity. Of the integral from a to 0 of x e to the negative x2 with respect to x plus the limit k as b approaches infinity of the integral from 0 to b of x e to the negative x 2 with respect to x. So notice here we have a common term for each of these expressions. Let's see if we can integrate its indefinite integral and use that to help us solve. So solving for the integral, OK, of Xe to the negative x2 with respect to x using substitution, let's let you be equal to x2. That means DUDX will be equal to -2X, so -5 of DU, well, let me write that out clearly here. So we can find a derivative of u with respect to x and thus rewrite duu in terms of x, that is, we can say negative half of duu would be equal to x dx. So in that case if we substitute, then our integral of x e to the negative x 2 with respect to x would be equal to 12 of the integral of e to the u with respect to u. And when we integrate here, that's -52 of E to U + C, which is going to be 52, OK, of E to the X2 + C. Now we can evaluate both definite integrals since we have our result, because no, OK, this means that our integral starting with a from a to 0 of xe to the negative x 2 d x. Will be equal to 12 of e to the g x 2 between the bones of a and 0. And that would be equal to -52 of e to the 02 minus e to the A2, which when we simplify that would be equal to 52. OK, of 1 minus E to the negative A2. On the other hand, OK. Or integral from 0 to B of XE to the negative X 2 DX would be negative 12. Of E to the negative X2 again from 0 to B, and that would be equal to the 12 of E to the negative B2. Minus e to the 02, which will give us 5 of e to the b2 minus 1. Now that we have both of these terms, we can take our limits. So starting with the first term, OK. Then as A approaches negative infinity, OK, the limit, OK, as it approaches negative infinity of our term. -12 of 1 minus E to the negative A2, OK. It is going to be equal to -5 multiplied by 1 or a limit as a approaches negative infinity of 1 will be 1, but on the other hand, as a approaches negative infinity, e to the negative a squad will approach 0. So that's really 1 minus 0, which would give us -5. No, the limit, OK, as b approaches infinity of -522 multiplied by e to the neb 2 minus 1 then would be equal to -52 multiplied by 0. OK, because as B approaches infinity, E to the negative B2 would approach 0 as well, minus 1, OK, and 1/2 multiplied by a -1 equals 5. So now, when we combine both limits. OK. Then the limit. As a approaches negative infinity, well, the limit as a approaches negative infinity between 0 and a of xe to the negative x2 with respect to x plus the limit as b approaches infinity of the integral from 0 to b of xe to the negative x2 d x. Equals 1/2 plus 1/22, which equals 0. Therefore, our improper integral is 0. If we look back at our answer choices, that means that C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.∫₋∞^∞ 2x e^(−x²) dx

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The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₋∞^∞ 2x e^(−x²) dx