Evaluate the integrals in Exercises 31–56. Some integrals do n...

Question

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.

∫ √x e√x dx

Accepted Answer

Welcome back everyone. In this problem, we want to evaluate the integral of root X multiplied by E to the 2 root X with respect to X. Now, before we evaluate this integral, we can substitute to make it a bit easier to read, OK, so by substitution. OK, a bit easier for us to solve that is so by substitution. Let's let you be equal to root x. Then that means x would be equal to u 2 and thus the derivative of x with respect to u would be equal to 2 u. Therefore, DU or DX, sorry, would be equal to 2 UDU. So by substituting here that means we could rewrite the integral of root x e to the two root x with respect to x. As the integral of u to the 2 u multiplied by 2 uu. Or even simpler, 2 multiplied by the integral of u squared multiplied by e to the 2 u with respect to u. Now how could we solve this problem? How can we solve for this integral, this indefinite integral here? Well, we can integrate by parts. Now recall that using integration by parts, the integral of WDV. Is equal to WV minus the integral of VDW. So if in this in this integral, OK, we let W equal u 2, and we let dV equal e to the 2 U DU, then we could use it to help us figure out what Dw and V would be, and that can help us to solve using integration by parts. Now if W equals u 2, that means the derivative of w with respect to u equals 2 u, so Dw would be equal to 2 uDU. On the other hand, if we integrate both sides of dV equals e to the 2u with DU, sorry, then that means we should get V to be equal to a half of e to the power of 2u. So using this information, this means then that the integral of u2 e to the 2 uDu would be equal to wv based on our equation or sorry or formula for integrating by parts. So that would be a half of u2 e to the 2 u minus the integral of vdw. So that's the integral of a half of e to the 2 u multiplied by 2 uDu. I know when we simplify here, we get this to be a half of U 2 e to the 2 u minus OK, the integral of u e to the 2U with respect to you. Now, you'll notice we're we're in the same problem that we had before. No, we'll need to integrate by parts again for us to integrate this product. So let's go ahead and do that, OK. So again, integrating by parts. OK. Integrating by parts again. OK, where. We Let P. Be equal to you. So let's call Pu and we let Q, sorry, DQ be equal to e to the 2 uDu, OK, then the derivative of p with respect to you would be 1. So DP would be equal to duu. While if we integrate both sides of d q equals e to the 2 u d u, then we'd get q to be equal to a half of e to the 2 u. And again we're just using p and q as arbitrary variables here to make it easier to read. OK, so again by this formula it means that the integral of u e to the 2 u d u would be equal to here pq. So that's a half of UE to the 2 U minus the integral of a half of E to the 2UDU, OK, so QDP. Now, in this case, if we simplify. The integral of, uh, let me rewrite here. The integral of a half of e to the 2 uuu would be equal to a half here of e to the 2 u. So this is a 12 multiplied by a half of e to the 2 u, which simplifies to a half of u e to the 2 u minus a quarter of e to the 2 u. Now, if we were to combine and back substitute all of this into our previous information. OK, then this means our integral of u2 e to the 2U with respect to you would be equal to a half of u2 e to the 2U, OK, minus. A half of you. E to the 2 u minus a quarter of e to the 2 u because now remember we're substituting the result of the second integration by parts back into the first and when we simplify that, that will be a half of u squared e to the 2 u minus a half of u e to the 2 u plus a quarter of e to the 2 u. Now remember we were multiplying by a factor of 2 from the original substitution. So when we do that here, this means about 2 multiplied by the integral of u2 e to the 2 u. DU is going to be equal to 2 multiplied by this entire expression, so I can pretty much copy and paste it here, so it's 2 multiplied by all of this. OK Multiplied by all of that. OK Plus or constant. Or integral constant C. OK, and now when we go ahead and multiply through by 2 here, we get U 2 E 2U minus UE 2U plus a half of E2U plus C. Finally, now, we can substitute u equals root x back into the expression. So really this will be equal to root x2 multiplied by e to the 2 root x. Minus root x multiplied by e to the 2 root x, OK, plus a half of e to the 2 root x plus C. Now when we simplify that, we'll get xe to the 2 root x minus root xe to the 2 root x plus a half of e to the 2 root x plus C and if we factor e to the 2 root x. Then we end up in bracket with x minus root x plus 12, OK, plus our integral constant c. Therefore, the integral of root x. OK, E2 root X with respect to X is this result. Thanks a lot for watching everyone. I hope this video helped.

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.∫ √x e√x dx

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Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫ √x e√x dx