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Determine whether each integral is convergent or divergent. Evaluate those that are convergent.$$\int_{-\infty}^{\infty} x e^{-x^{2}} d x$$

The integral converges to 0

Calculus 2 / BC

Chapter 6

TECHNIQUES OF INTEGRATION

Section 6

Improper Integrals

Integration Techniques

Campbell University

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Harvey Mudd College

University of Michigan - Ann Arbor

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in this problem were given the improper integral X times each the negative X squared DX from negative infinity to infinity and were asked to find whether this is convergent or divergent. So since we have two bounds that go to infinity, I'm going to rewrite this as two inch girls breaking up that bound. So one is going to have the bounds Negative infinity to zero of our function plus the bounds zero to infinity of our function. So because we have now to and girls ah, with the same terms I'm going to begin by just solving the indefinite integral X times eats the negative x squared d X that we can use it for both of our integral. So, in order to solve this, we're gonna use U substitution where you equals negative X squared and d u equals negative two x d x. So if we plug this in two are integral, we get a negative 1/2 on the side and then we have the integral e to the u D year, and this will just equal 1/2 each the you pussy. If we steps to back in our value for you, that gives us negative 1/2 e to the negative X squared plus c. And finally, uh, we're gonna put this into our two intervals. And instead of writing those infinities, I'm going to use the term T for our first in a girl. I'm gonna have t go to negative infinity in 1st 2nd in a girl. I'm gonna have a go to infinity, and this is going to give us negative 1/2 e to the negative x squared from t 20 plus negative 1/2 e to the negative X squared from zero to T where in our first term to goes negative infinity and in our second t goes to infinity. So from here, I'm gonna take the limit of both of these for a first term, the limit is going to be as t approaches negative infinity. And if we plug in our bounds, we get negative 1/2 plus 1/2 each the negative t squared, plus the limits as t approaches infinity of negative 1/2 e to the negative cheese squared plus 1/2. So if we plug in our limit, both of these terms are going to go to zero because we get e. It's the negative infinity for both. And if we look at a really simple sketch of each, the X will notice that as e approaches native infinity, it approaches zero. So this leaves us from both of our terms. Negative 1/2 minus 1/2. Which leaves us with negative 1/4. Oh, I'm sorry. I got my negatives off here. This leaves us with negative 1/2 plus 1/2 which gives us zero. And in both cases, this gives us a convergent integral.

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