Evaluate the integrals in Exercises 1–24 using integration by ...

Question

Evaluate the integrals in Exercises 1–24 using integration by parts.

∫ p⁴ e^(-p) dp

Accepted Answer

Welcome back everyone. Evaluate the integral using integration by parts integral of x 2 e to the power of xDX. For this problem, let's remember that the integral of UDV is equal to UV minus integral of VDU. According to the formula for integration by parts, so what we can do in this problem is simply set u equals x2, and then dV is going to be e to the power of x dx. Now let's identify DU. The derivative of U with respect to x is the derivative of x 2, which is 2 x. So that the EU is going to be equal to 2x multiplied by DX. We can multiply both sides by DX, right? And L4V, it is basically the integral of DV, which is the integral of e to the power of x dx, and that's e to the power of x. And there is no necessity to add a constant of integration for now. So we can now show that the integral of x 2 e to the power of xDX is going to be u multiplied by V. So that's x2 multiplied by e to the power of x minus the integral of VDU. Specifically for DU, we can take out 2 and simply write XDX. And multiplying by V, we basically get -2 integral x. E to the power of x dx, right? So we have another integral that will be evaluated using integration by parts. For now, let's label our main integral with i and let's label the final integral that we want to evaluate with i1. That's basically integral of x eats the power of x dx. Again, we're going to use integration by parts using the same principle. Let's suppose that u equals x. And dV equals e to the power of x dx. Now d u is going to be dx because u equals x and we already know what v is, right? That's e to the power of x so we can basically solve and show that this is a u multiplied by v, so x multiplied by Eats the power of x. Minus the integral of VDU, so minus the integral of. E to the power of x. D X Which is x e to the power of x minus the integral of e to the power of x is e to the power of x itself, and we're going to add a constant of integration c1. Well done, and now we can substitute this result into the main integral. So that our main integral i is going to be equal to. X 2 e to the power of x minus 2 multiplied by Xe to the power of x minus 2 e to the power of x we're distributing, right. Notice that we're multiplying by -2. So the first term is going to be -2 x e to the power of x, and the second term is going to be e to the power of x since we're multiplying by -2, right? And we have a negative sign in front of e to the power of x. And we can basically add a combined constant of integration C. And from here we can factor out e to the power of x to get x 2 minus 2 x. Plus 2 multiplied by e to the power of x plus c. This is our final answer, and thank you for watching.

Evaluate the integrals in Exercises 1–24 using integration by parts.∫ p⁴ e^(-p) dp

Watch next

Evaluate the integrals in Exercises 1–24 using integration by parts.
∫ p⁴ e^(-p) dp