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Show that $ \displaystyle \int_0^\infty x^2 e^{-x^2}\ dx = \frac{1}{2} \displaystyle \int_0^\infty e^{-x^2}\ dx $.

The limit is 0

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 8

Improper Integrals

Integration Techniques

Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

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The problem is show that integral from zero to infinity, Esquire comes into negative X squared DX is equal to one half times integral from zero to infinity e two negative X squared DX So now integral from zero to infinity X square YouTube negative X squared dx A definition this is equal to limit a goes to infinity integral from zero to a of the function X square Eat negative square the X We compute this definite integral first Yeah, we can write this function it interior from zero to a x square So here we can write X square as X and negative two x e two negative x choir the X and yeah, we need two times negative one half So we just write to ex choir as negative one half x times Negative two X Now we use my theory with the integration by parts This is Echo two negative one half integral from zero to a X The yeah e to the negative X square. Use my third of the integration by parts. This is negative. One half Sam's ex comes each negative X square from zero to a minus integral from zero to a you too Negative At square the X This is equal to make you one half. Eight times into negative is power has one half integral from zero to a e two negative x choir d x Now what a goes to infinity. This part is the same as this one. So this is one half integral from zero to infinity into negative X squared dx. So we need to prove the first part is equal to zero. One day goes to infinity. So for this one just to use Nope. It has rules that limit a over e two a square It goes to infinity Use the rope industrial So this is equal to the limit A goes to infinity one over you to a square and to a this is zero So the first part is zero second month interested the right hand side. So these two integral I e co

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