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Evaluate the following integrals using integration by parts:1" sin(u) drIn _tAl2r sil ()dr...

Question

Evaluate the following integrals using integration by parts:1" sin(u) drIn _tAl2r sil ()dr

Evaluate the following integrals using integration by parts: 1" sin(u) dr In _ tAl 2r sil ()dr



Answers

Evaluate the integrals in Exercises $1-24$ using integration by parts.
$$\int\left(r^{2}+r+1\right) e^{r} d r$$

Okay, So the inner girl that we have here to evaluate is the integral of the quantity R squared plus our plus one all multiplied by E T v r D r. And this is a standard in the integration by parts problem. So just to remind ourselves of the formula for integration by parts we have here, the integration by parts tells us that the integral of anything of the form you Devi on Girl of this form can be re written equivalently as you times v minus the integral of VD you and to just get a quick understanding quick glimpse of sort of the visual intuition behind why integration by parts. It works the way that it does. We have here a little graphic which can give you that If you examine it, you can pause a video, take a look. You can see how all the pieces fit together to give us this formula. But moving on to our integral we I want to find a part of this expression to be Are you in another part of this expression to be our Devi And naturally we want to pick a you that will differentiate down to something simpler so we wouldn't want to pick the heat of the R because that would just give us another exponential function When we don't differentiate it, that would be just the same thing. So in instead of that, we're going t take this polynomial here as being Are you? So we write, let you equal r squared plus our plus one then d you will be the derivative of this thing. Times d are so that's two are plus one d r Devi, in this case is going to equal you to the e r Do you are? And then RV will eagle by integration Integral of the r d r is just e to the r so And they worked that out Let's get on with our integral We have the integral of our square plus our plus one you to the e r g r equals you times v So you is our square plus our plus one and ah v is really are minus the integral of VD You ve again is either the r and D u is to our plus one d are So when we put these together are integral will look like two are plus one times e v e R D r. Now, in order to evaluate the second, Integral will have to perform another application of integration by parts to simplify it a little further for our u will want to pick two Arpels one and then we'll have either. The r D r r D v So working that out, we say, let you equal to our plus one. D'You in this case equals two times D. R and R d v equals e T R D r and R. V Again is the interval of eating the R D r equals each of the R already then. So writing this out we have you Tim's free is to our plus one times easier minus the integral of Edie you where v is if they are and you is to d r so putting that all together we have the role of to you we are d r. And then ah, putting this all together again. We can distribute this negative over the sun and get positive the integral of to eat of the r d r. Now this integral is just to eat of the are so we can write you out like that. And then finally, we can add a constant now to simplify our answer. We can collect like terms since this heat of the R appears and all of these factors in all of these terms. And so we just add all these polling no meals together in front of it and get our squared. Plus are minus two are gives us minus r, and then plus one minus one plus two will give us plus two. So that times either they are plus C and that is our solution.

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Here were to find the integral from one. The negative one of the quantity are plus one quantity squared. We are now. There's lots of ways you could do this problem. One is to just simply expand our post one quantity squared. U u r squared plus two are plus one g r, which is going to give you 1/3 are cubed plus R squared plus are evaluated from one to negative one, which is gonna get you a negative 1/3 plus a one minus one minus quantity. 1/3 plus one plus one look, well, thes cancel and you're left with essentially negative 1/3 minus 1/3 minus one minus one, which is going to get you negative 8/3 for your answer. Another way you could do this problem is if at the beginning, you let visited and color that you equal are plus one. So then therefore do you equals d armed this emptiness institute except the bounce. Okay, so this has become a U squared. Do you question however you need to plug in your bound so one plus one would be to and negative one plus one would be zero. This gets us 1/3 you cubed evaluated from 20 And if you plug in a zero first, you get zero minus 1/3 time's two cubed you negative 8/3 which matches our bottom answer there.

Okay, This question wants us to evaluate this integral. So to start, we'll have to do an anti derivative which you could solve this using you substitution. But since we're just squaring this, it's easy just to foil it out. So expanding out are plus one squared. We get r squared plus two are plus one. And now this is something that we can find the anti derivative of easily so r squared. The anti derivative is our cubed over three to our has an anti derivative of two r squared over two. So just r squared and then one has an anti derivative are then we're plugging in from one to negative one. So plugging in negative one we get negative one third plus one minus one and then subtracting are lower limit of one third plus one plus one. So simplifying each of these Now we get negative one third from our top limit. And, well, this is three thirds. If this is three thirds, so we get three plus three plus one, which seven thirds and then negative one third minus seven thirds would be negative. Eight thirds. And that's our answer.


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