Question

Integration by parts

Answer 1

\[\int\limits 2x \sqrt{2x-81}\]

\[solve for dv = \sqrt{2x-81} dx\]

\[solve for u = 2x-81\]

Answer 2

ok so....

u = 2x
du = xdx

\[dv = \sqrt{2x-81} or {2x-81}^{1/2}\]

v = \[\frac{3(2x-81)^\frac{3}{2}}{2}\]

Answer 3

why on gods green earth would you integrate this by parts? oh well, if that's what they want...

Answer 4

oh wait 2nd part was solve by substitution u = 2x-81 :P

Answer 5

we can use a u-sub?
cuz that would be way easier

Answer 6

first part wants by parts, 2nd wants u-sub :).

Answer 7

am I correct so far for the parts part?

Answer 8

in all it's ugliness.....

Answer 9

yeah I think so, pain in the butt
let me work it on paper

Answer 10

no you got your du wrong already

Answer 11

A challenge for THE Turing... This is going to be rough... :P

Answer 12

oops 2dx :P

Answer 13

I keep moving too fast :!!!! :(

Answer 14

\[\frac{3(2x-81)^\frac{3}{2}}{2} * 2x - \frac{6}{2} \int\limits (2x-81)^\frac{3}{2} *dx\]

Answer 15

\[\frac{3(2x-81)^\frac{3}{2}}{2} * 2x - 3\int\limits (2x-81)^\frac{3}{2} dx\]

Answer 16

ok first off\[dv=(2x-81)^{1/2}dx\implies v=\frac13(2x-81)^{3/2}\]

Answer 17

ah crap, I switched it around by accident, I made sure to not do that too! :(

Answer 18

isn't it 2/3?

Answer 19

damn that ruins my plan!!!

Answer 20

yeah it's actually not so bad
just be careful, that integral requires a u-sub, and you have your fraction upside-down

Answer 21

yeah.. So it's 2/3 then?

Answer 22

1/3

Answer 23

why 1/3?

Answer 24

I thought you take the n+1 and divide by it? Like u^2 would be u^3/3 and in this case it was u^(1/2) which changed to u^(3/2) so wouldn't it be 2/3 * u^(3/2)?

Answer 25

there is a mini u-sub happening here that you are supposed to do mentally\[dv=(2x-81)^{1/2}dx\]\[\theta=2x-81\implies d\theta=2dx\implies dx=\frac{d\theta}2\]so we get\[\frac12\int u^{1/2}du\]

Answer 26

oh crap the chain rule huh? So it ends up as 2/3 * 1/2 which is 2/6 = 1/3?

Answer 27

\[\frac12\int u^{1/2}du=\frac12\cdot\frac23u^{3/2}=\frac13u^{3/2}\]

Answer 28

GRAWR!

Answer 29

yes, the chain rule is the inverse of the u-sub in a sense

Answer 30

I always forget that GRR!!! :)

Answer 31

\[\frac{3(2x-81)^\frac{3}{2}}{2} * 2x - \frac{2}{3} \int\limits (2x-81)^\frac{3}{2} *dx then?\]

Answer 32

this is where you have to really buckle down in calc so you can do this in your head
the last integral can be done with the exact same sub and then we are done

Answer 33

yeah I am moving too fast trying to get this done and forgetting basic stuff... :(

Answer 34

actually... I didn't think there wa sa chain rule when doing integration... from dv to v...?

Answer 35

so you're saying even though we went from dv to v we still used u sub huh?

Answer 36

yeah, but they call those "simple u-subs" as opposed to the tricky u-sub that is required to solve the problem without integration by parts.
you should get to the point where you can do these "simple" u-subs in your head by the time you are towards the end of calcII

Answer 37

Weird........

Answer 38

I guess that makes sense since we are integrating that part but doing a sub like that hmmm....

Answer 39

I guess it's still needed technically... I will remember that :D.

Answer 40

\[\frac{3(2x-81)^\frac{3}{2}}{2} * 2x - \frac{2}{3} \int\limits (2x-81)^\frac{3}{2} *dx\]

u = (2x-81)
du = 2dx

1/2 du = dx

\[\frac{3(2x-81)^\frac{3}{2}}{2} * 2x - \frac{2}{6} \int\limits (u)^\frac{3}{2} *du\]

\[\frac{3(2x-81)^\frac{3}{2}}{2} * 2x - \frac{1}{3} \int\limits (u)^\frac{3}{2} *du\]

\[\frac{3(2x-81)^\frac{3}{2}}{2} * 2x - \frac{2}{15} \int\limits (u)^\frac{5}{2} +c?\]

Answer 41

\[\frac{3(2x-81)^\frac{3}{2}}{2} * 2x - \frac{2}{15} (u)^\frac{5}{2} +c?\]

Answer 42

\[\frac{3(2x-81)^\frac{3}{2}}{2} * 2x - \frac{2}{15} (2x)^\frac{5}{2} +c?\]

Answer 43

stupid integration symbol getting in my way! :P

Answer 44

besides the spurious integral sign, (and always remember your du)
it looks okay

Answer 45

I had my du's, no? :P

Answer 46

hmm not right... what did i DOOOOOOOOOOO :(

Answer 47

oh derp... I did 2x not 2x-81...

Answer 48

you didn't sub in the right u

Answer 49

still not right HMMM :'(

Answer 50

That's what I get for not looking LOL :(.

Answer 51

yes it is

Answer 52

hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

Answer 53

if you put 2x-81 where you put 2x it is right

Answer 54

still says it's wrong.. hmm IDk lemme check...

Answer 55

well are you typing this into a computer to validate your answer?
if so you may need to simplify it or something

Answer 56

yeah I am...

Answer 57

oh no you still have some stuff upside-down

Answer 58

Checking the practice version... hmph.

Answer 59

oh that satupid 3/2...

Answer 60

it syas it's 2/3x though?

Answer 61

it should be

Answer 62

whjere does that x come from...?

Answer 63

oh 2x...

Answer 64

I forgot to change that shizz before :P.

Answer 65

Got all caught up on the other side of things haha :'(

Answer 66

hopefully you don't have to simplify, because the u-sub way of doing this integral leaves it looking very different
so you would have to do a lot of simplification to show that they are the same solution

Answer 67

wel,,, not \(very\) different, but different enough to make you think you messed up

Answer 68

=
2/3x(2x − 285)3/2 − 2/15(2x − 285)5/2 + C
= 2/15(2x − 285)3/2(3x + 285) + C

= 2/5(2x − 285)3/2(x + 95) +c

Using 285 instead of 81...

Answer 69

konfusing how theyr get to there...

Answer 70

?
so you wanna see the ugly simplification? ok here we go, be patient...

Answer 71

Idk, that's what they show lol :(.

Answer 72

yeah I guess they want it simplified boo!

Answer 73

\[\frac{2x}3(2x-81)^{3/2}-\frac2{15}(2x-81)^{5/2}+C\]agree so far?

Answer 74

mhm

Answer 75

factor out \((2x-81)^{3/2}\) (watch the exponents carefully)\[\frac{2x}3(2x-81)^{3/2}-\frac2{15}(2x-81)^{5/2}+C\]\[(2x-81)^{3/2}\left(\frac{2x}3(2x-81)^0-\frac2{15}(2x-81)^1\right)+C\]\[(2x-81)^{3/2}\left(\frac{2x}3-\frac{4x+162}{15}\right)+C\]\[(2x-81)^{3/2}\left(\frac{10x-4x-162}{15}\right)+C\]\[(2x-81)^{3/2}\left(\frac{6x-162}{15}\right)+C\]\[\frac25(2x-81)^{3/2}\left(x-27\right)+C\]whew!

Answer 76

in the practice version for some reason the (x-27) would be (x+27?)

Answer 77

oh and wow that's stupid ugly :'(

Answer 78

oh yeah, it's a +

Answer 79

lets try the next one and see if it's the same!

Answer 80

u = 2x-81

Answer 81

du = 2dx
1/2du = dx

Answer 82

factor out \((2x-81)^{3/2}\) (watch the exponents carefully)\[\frac{2x}3(2x-81)^{3/2}-\frac2{15}(2x-81)^{5/2}+C\]\[(2x-81)^{3/2}\left(\frac{2x}3(2x-81)^0-\frac2{15}(2x-81)^1\right)+C\]\[(2x-81)^{3/2}\left(\frac{2x}3-\frac{4x-162}{15}\right)+C\]\[(2x-81)^{3/2}\left(\frac{10x-4x+162}{15}\right)+C\]\[(2x-81)^{3/2}\left(\frac{6x+162}{15}\right)+C\]\[\frac25(2x-81)^{3/2}\left(x+27\right)+C\]

Answer 83

I have already done it the other way, we will have to simplify again to see that we get the same answer

Answer 84

\[\frac{1}{2} \int\limits 2x \sqrt{u} du\]

Answer 85

but to get the 2x we need to substitute tin for u

Answer 86

yo never want x's and u's in the integrand at the same time

Answer 87

your guess for u is wrong

Answer 88

so since u = 2x-81

2x = u+81
x = (u+81)/2?

so the 2x would be just u+81?

Answer 89

that will just take you in circles
try another guess for u

Answer 90

problem says u = 2x-81 :'(

Answer 91

actually that will work, you're right
mine was different but works too

Answer 92

! :)

Answer 93

I did u=sqrt(2x-81) and that does work mind you!

Answer 94

\[\frac{1}{2} \int\limits (u+81)\sqrt{u} du\] ???

w/e you say :D

Answer 95

but yours is easier :P

Answer 96

yes we would :D

Answer 97

can we throw that 81 onto the other side?

Answer 98

and why would we do that?