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.

∫ x⁵ e³ˣ dx

Accepted Answer

Welcome back everyone. In this problem we want to compute the integral of x 7 multiplied by e x 4 with respect to x. Now before we solve here, it may be easier for us to first start by rewriting the integral. OK, we can simplify the integrand by recognizing that a part of the power of x x cubed is related to the derivative of the exponent. X 4th, OK. So in other words. We can take our integral of x 7th multiplied by e x 4th with respect to x and rewrite it as the integral of x 4 multiplied by e x 4 multiplied by x cubed with respect to x and now it will make it easier to solve by substitution because in substitution we know that at some point we'll have to differentiate some term somewhere or another. So how can we do that? Well, let's let you be equal to x 4. Then that means the derivative of u with respect to x would be equal to 4 x cubed, OK. Which means then that the derivative of you. OK, or DU would be equal to 4 x cubed dx, or we could say then that a quarter of duu, if we divide both sides by 4, then a quarter of duu, OK. Would be equal to x cubed dx. So now we have x 4 in terms of u and we have x cubed dx in terms of duu. So by substitution, OK. Then we could rewrite. Or integral, yeah. That is X4 multiplied by E X 4 multiplied by X cubed DX, OK. As the integral of u multiplied by e to the u multiplied by a quarter of duu or as a quarter of the integral of u e to the u duu. OK, now this is a bit easier for us to integrate because now we can integrate you by parts. What do we know in integrating by parts? Well, recall that in integrating biparts we basically say that the integral of WDV OK. The integral of WDV would be equal to wv minus the integral of VDW. So if we can set one of these terms to be w and the other to be dV, then we should be able to go ahead and substitute and solve using integration by parts, and then we can rewrite it in terms of x. So let's do that Let's let W be equal to you. And let's let dV be equal to e to the uDU. Well, in that case, if W equals u, then the derivative of w with respect to u would be 1, so dw would be equal to duu. On the other hand, if we integrate both sides of dV equals e to the uDU, then we end up with V being equal to e to the u. So By applying this to our integration by parts formula, this means that the integral of u e to the uDu would be equal to WV, that is u e to the u minus the integral of VDW, that's e to the uDu. I know when we simplify here, that's UE to U minus E to the U because that's the integral of E to the U and we can factor out Eu here to get U minus 1. So since this is our indefinite integral, if we use it back into what we got before, OK, then. This means that a quarter of the integral of you. EuDU would be equal to a quarter of eu multiplied by u minus 1. Remember that u equals x 4. So really then our integral is a quarter of e x e x 4 multiplied by x 4 minus 1 plus some constant c. This is our integral of this right there. This is our integral of x 7 multiplied by e x 4th with respect to x. 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.∫ x⁵ e³ˣ dx

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Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫ x⁵ e³ˣ dx