81. Possible and impossible integrals
Let Iₙ = ∫ xⁿ e⁻ˣ²...

Question

81. Possible and impossible integrals

Let Iₙ = ∫ xⁿ e⁻ˣ² dx, where n is a nonnegative integer.

d. Show that, in general, if n is odd, then Iₙ = -½ e⁻ˣ² pₙ₋₁(x), where pₙ₋₁ is a polynomial of degree n - 1.

Accepted Answer

Given JN equals the integral of X N E 3 X 2 D X, where N is a non-negative integer, determine a formula for JN in terms of PM X when N is odd. Here, PM of X denotes a polynomial of degree N. And make use of the results of the integral of U M E UU equals E U PM of U plus C. Let's go ahead and rewrite our integral to make this a little easier to solve. We can first factor out an X, because we can say N equals 2 N + 1. This is because we want odd ins. So, we have X to the N equals X 2 M plus 1. We can factor out that X to get X multiplied by X to the 2 N. Meaning we rewrite our integral. J N equals the integral of X multiplied by X to the 2 M. Multiplied by E to the negative 3 X 2 DX. So let's try a U substitution here. We will let U equals negative 3 X squared. This means DU is -6XDX. And XDX equals -16 DU. Now we can rewrite our integral. This will be written as negative 16 integral of X 2 M E U D U. We also know X 2 M is equivalent to X2 race the M. And we can make another substitution. Where we note that X equals negative view divided by 3. So, we have a new integral. Of n of 16. And the girl Nay, the view divided by 3 rated to the m. E to the U to you. Let's go ahead and simplify it even further. We will end up getting a new value. Of -1 Race to the M + 1. Divided by 6 multiplied by 3 to the M. All multiplied by the integral of U M, Eu to you. So this is an integration by parts. However, we do have the integral of U M E U already. We know that this is given by E U P M of U plus C. So, let's go ahead and make another substitution. We know J N equals the integral, X N E 3 X squared. The X Based on what we have previously, we can rewrite this as 1/6 E to the 3 X2. Q in minus 1 of X plus C. So, we can write out our final integral. J N equals -16. Race to the 3 x squared. PN minus 1 of X plus C. This is the final form of our integral. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

81. Possible and impossible integrals
Let Iₙ = ∫ xⁿ e⁻ˣ² dx, where n is a nonnegative integer.
d. Show that, in general, if n is odd, then Iₙ = -½ e⁻ˣ² pₙ₋₁(x), where pₙ₋₁ is a polynomial of degree n - 1.

Key Concepts

Integration by Parts

Properties of Exponential Functions

Polynomial Representation of Integrals

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