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Integration by parts. f x?(x 5 4)} dx...

Question

Integration by parts. f x?(x 5 4)} dx

integration by parts. f x?(x 5 4)} dx



Answers

Using integration by parts. $$\int x^{5} e^{x} d x$$

Okay, So this question we're taking the integral of X times five to the X power DX and the first steps to establish what our you and Dean V are going to be. So in this case, you is gonna be equal. Thio X and Devi is going to be five to the X power. Now let's find the other parts that we need in order to plug into the formula. So do you just gonna be the druid of both sides? So it's gonna be equal to D X. And now to find V, we take the integral of both sides, and I'm just gonna go over quickly on the side here, a property of exponential functions when you take the dreaded because I feel like this property isn't the most common. So if you have a function f of X is equal to a to the X power on. In this case, a cannot be zero or one, so f prime of X is going to be equal to Ellen of a time's A to the X power. So we're gonna use this property in order to find V missus, we're going backwards, and we're taking the integral he is going to be equal to five to the X power divided by ln of five. Some sort of most flying gonna have divided. Okay, so now that we have, all of our components were ready to plug in. So this is gonna be equal to your times. V So x times five to the ex over. Ellen of five, minus the Inter Girl of five to the ex over Ellen of five d X. Now we've already had taken this integral. Um, so we just did five to the X power. Just we're here. We have a constant of one over Ellen X in front or one over. Ellen, five in front. So, um, we take this into girl is that you re writes this first part where we have X times five to the X over Ellen of five minus five to the ex over ln 05 squared. And we have to remember to add a C since is an indefinite integral, um, and this is the final answer

So for this integral we're gonna use integration by parts and I'm gonna set you here equal to x squared minus five X. Since if I set u equal E to the X. Um it's not really gonna do much do, you would just be equal to E to the X. And would kind of be left with an even more complicated function. And so if we let you hear equals X squared minus five X. Then D. You is gonna be two X minus five. T. V. Is equal to E. To the X. So V is also equal to E. To the X. And we have this is equal to U. Times V. So each of the X times X squared minus five X. And then minus the integral of E times D. You So that would be two X minus five times E. To the X. Dx. And so what we're gonna do here is we're gonna do just another integration by parts. Um to figure out the value of this integral here and we're gonna let you equal to x minus five then D. U. Is equal to just two and tv is still equal to E. To the X. So then V. Is also equal to E. To the X. And go ahead and just move these over here. And so this is equal to um negative one times in parentheses, um U. Times V. So either the X times two X minus five and then minus the integral V times D. You just to eat the X. Dx. And so the integral of to eat the exes just itself to eat the X. So this is equal to E. To the X times X squared minus five. X minus E. To the X times two X minus five. And then we have minus but we have this negative sign here also that we need to actually distribute. So this is going to be plus um and it's just gonna be to E. To the X. And then lastly we have plus C.

We have the integral from 0 to 5 and three is the height from 0 to 3. And then why equals X is the height from three and beyond. Then we want to find the area under the two curves. So the area under this rectangle here would just be three times three or nine and the area under this blue. Why, it was X line there. It's going to be what we could do this with a trapezoid, Um, and so that would end up being a base of two. Actually, you know, let's just do a rectangle on the triangle instead. So that would be two times three. So that would be an area of six there, and then that would be a height of two with of 22 times 24 divided by two. So that would be in the area of two up top. So just so you can see both of those, that's to that six. And that's of course, our first nine. So the total area under the curve there is going to be 17 9 plus six plus two. Therefore, that's what the Senate girl evaluates dio

So we want to find the following integral. Using integration by parts. So I've got the formula over here to put right. Um So in our formula we need to choose a you and a D. V. That makes up this entire and a role here. Um So let's let you be equal to X. And let's let DV Be the rest. So we can rewrite this as two X -5 to the negative one half power. Mm hmm the X. So now we need to differentiate you. Um So d'you is just going to be equal to dx and we need to integrate both sides over here to get ve so V is going to be equal to the integral uh two X minus five to the negative one half power D. X. So to solve this integral we need to use some sort of U substitution. But we already have you going on here. So let's uh use a different variable. So let's let z be equal to two X -5. Then dizzy is equal to two D. X. So now we can rewrite are integral. We've got the integral of the sea to the negative 1/2 power. And then in order to put a DZ here we need to put it to so we can multiply that out front. Um So we've got to times the in the role of Z to the negative one half. So we add one to the exponents Z. To the one half. Um And we need to multiply another two out front there. So this is equal to four square root of Z. So now we have everything we need we can use our formula. So our original integral here is equal to U. V. So X. Times for square root dizzy. Uh Sorry we don't want to keep seeing here. So yes this is for scrubber dizzy but we uh need to substitute X. Um back in in terms of X. So this is four square root of two X -5. So let's write that here two X -5. Uh that U. V minus. And then we need to integrate V. D. You so V. This four can come out front and then we've got square root of two x -5. Mhm. Uh D. U. Which is D. X. So now we just need to integrate this guy, let's rewrite our first term four X. Square root two x -5 minus up to the side here let's do our integral Square root of two x -5 D. X. Um So we need to do another substitution. We can use anything. So let's Collins Odetta Be equal to two X -5. So again D Data is equal to two d. x. Um So are integral becomes the integral of square root of data. And to put a. D. Data here, we need to add this to out front. So this is going to be equal to two times in a role of data to the one half. So what do we have enough data to the one half plus one, which is three halves. Then we need to multiply by 2/3 out front. So this is going to be equal to 4/3. We can replace data with our original two X -5 Uh to the 3/2 power. So that's going to be our inner girl here. So we've got four Times This 4/3. So we really have uh 16 3rd times two X -5 to the 3/2 power. And then we have to add our constancy.


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