Evaluate the improper integrals in Exercises 53–62.
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Question

Evaluate the improper integrals in Exercises 53–62.

∫ from 0 to ∞ of (x² * e^(−x)) dx

Accepted Answer

Welcome back everyone. In this problem we want to compute the improper integral of x e to the negative x with respect to x between the bones of zero and infinity. A says it's 2, B1, C a half, and D says it's 0. No, to evaluate the improper integral, we must treat the infinite upper bound using a limit and solve the remaining definite integral using integration by parts. In other words, what we're saying is that we would take our integral or improper integral between zero and infinity of x e to the negative x with respect to x and rewrite it as the limit as b approaches infinity. Between the bones of the integral between the bones of zero and b of x e to the negative x with respect to x. So if we can integrate, then we should be able to figure out the definite integral and evaluate the limit as our variable B approaches infinity. So let's start by integrating by parts. Recall that to integrate by parts basically we say that the integral of UDV equals uV minus the integral of VDU. So if we can figure out what u and the DV are, we can use that to figure out du U and V and thus be able to integrate. So let's let you be equal to x. And dV equal to e to the negative xDX. Now if U equals x, that means the derivative of you with respect to x is 1, so DU would be equal to dx. And if we integrate both sides for dV equals e to the negative xDX, then we should get V to be equal to e to the negative x. Know that we have values for d u and V, we can substitute it into our formula for integration by parts because no, this means it will be equal to the integral. OK, or sorry, our integral will be equal to x multiplied by e to the neg nee to the negative x, that's UV. Minus the integral of VDU, so that's the integral of e to the negative XDX. If we simplify here, this will be xe to the negative x plus the integral of e to the g xDX. And when we integrate our term here. OK, then it will give us e to the negative X. And now if we factor out E to the negative X from our expression, it becomes X + 1 multiplied by E to the negative X. Now that we have that, we can apply the limits of integration. So for our limits then between 0 and B for our integrand. It's going to be equal to x + 1 multiplied by e to the g x between the bones of 0 and b. Using the fundamental theorem of calculus, we are going to substitute the upper and lower bones to give us b + 1 multiplied by e to the b minus the negative value of 0 + 1 multiplied by e to the 0. No, here our first term where we can rewrite it as the negative value of B + 1 divided by e to the B, and here 0 + 1 will be -1, E to the 0 is equal to 1, so we're subtracting -1 or rewriting it as +1. Now we're going to take the limit of this expression as B approaches infinity. That means we're finding the limit as B approaches infinity of negative. B + 1 I'm sorry, the negative value of B + 1, let me write that. A Little better here. Divided by E to B. All plus one. Now, we can evaluate each of these limits independently, right? To evaluate the limit as B approaches infinity of negative B + 1 divided by e to the b, we notice it's an indeterminate form of the type infinity divided by infinity. OK, so in that case we can apply Lapita's rule to differentiate the top and the bottom for our first term. To become the limit as b approaches infinity of 1, the derivative of b plus 1 divided by e to the b, the derivative of e to the b, and we know that that would be equal to 0. On the other hand, the limit as b approaches infinity of 1 is 1. So this will be 0 + 1, which equals 1. Thus this is the derivative of x e to the negative x with respect to x between the bounds of zero and infinity. If we look back at our answer choices, b is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Evaluate the improper integrals in Exercises 53–62.∫ from 0 to ∞ of (x² * e^(−x)) dx

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Evaluate the improper integrals in Exercises 53–62.
∫ from 0 to ∞ of (x² * e^(−x)) dx