Question

q

Answer 1

@katie00

Answer 2

For
\[\int\limits_{3}^{4} \frac{ 2 }{ \sqrt{3x-7} }dx\]

i would just put u = 3x - 7 , du = 3

then you wont have to change the interval of the definite integral

Answer 3

u = 3 x - 7
dy = 3 dx, or dx = du/3

Answer 4

\[\int\limits \frac{ \cos~x }{ \sin^4 x }dx\]

u misstyped a power when substituting...

u = sin x
du = cos x dx

\[\large \int\limits \frac{ 1 }{ u^4 }du = \frac{ -1 }{ (3*u^3) } = \frac{ -1 }{ 3*\sin^3(x) }\]

Answer 5

\[\int\limits (x+1)\sqrt{x-2}~~dx\]

i would just put
u = x -2 ---> x = u + 2
du = dx

\[\large \int\limits_{3}^{6}[u + 3]*u^{1/2}du = \large \int\limits_{3}^{6}[ u^{3/2} + 3u^{1/2} ]du\]


no change in integration interval

Answer 6

All the probs look good though,
doesnt matter how you get to the answer if it is right i guess, just was a couple things i noticed..

Answer 7

no, the method does matter. The prof grades the method the method is more imp than the answer.

Answer 8

so what would d be after i put u=3x-7 and du=3? @DanJS

Answer 9

That is how i remember to do it, just less writing maybe

Answer 10

did they show you probs replacing stuff under a root with a squared term like that t^2

Answer 11

no, not that I rember. But can you complete d through out if you think that's the answer than plz complete your problem so I can see what I did worng

Answer 12

\[\sqrt{t^2 }=\left| x \right|\] ,

yeah was just looking at wha tyou did again

Answer 13

\[\int\limits\limits_{3}^{4} \frac{ 2 }{ \sqrt{3x-7} }dx = \int\limits\limits_{3}^{4}\frac{ 2 }{ \sqrt{u} }*\frac{ du }{ 3 }\]

Answer 14

when u = 3x-7 and du = 3 dx

Answer 15

\[\frac{ 2 }{ 3 } *\int\limits_{3}^{4}u^{-1/2}*du\]

Answer 16

you get
\[(2/3)*2u^{1/2} \] from 3 to 4

Answer 17

I though there wont be a u in the answer

Answer 18

right , put back in the sub, u = 3x - 7

\[\frac{ 4 }{ 3 }\sqrt{3x - 7}\]

Answer 19

evaluated for the interval 3 to 4, looks the same final answer you have

Answer 20

hm okay for the next one that u mentioned. u said I missed a power. can you do that one too plz.

Answer 21

i never get the cos/sin ones anyway so

Answer 22

TO decide where a u-sub may be good for a prob, if you notice things in the function that are derivatives of other things , that is a good hint

remember d/dx (sin x) = cos (x)

Answer 23

so if you let the u= sin, the derivative will be cos, and take out the numerator term with the du,


u = sin(x)
du = cos(x) dx

\[\int\limits \frac{ 1 }{ u^4 }du = \int\limits u^{-4}du \]

Answer 24

= (-1/3) * u^-3

= -1/( 3 * sin^3(x))

Answer 25

okay so in this case what do i plug in at the end? or do i even d that?

Answer 26

is it a def or indef integral?

Answer 27

indef

Answer 28

yeah just simplify the function as much as you can , that is the answer ,

Answer 29

so -1/3(sin^3 x) is my answer

Answer 30

yeah,
\[\huge \frac{ -1 }{ 3\sin^3(x) }\]

you had the process right, you just put a u^3 in the bottom to integrate, shoul dbe a 4

Answer 31

okay last one that u mentioned (F) def 3-6

Answer 32

tysm for pointign out my erros :) I would have never realized it ahah but tysm for everything

Answer 33

lets finish the last one so I can work on my hw for other classes after we are doen with clac

Answer 34

ok, they arent huge errors, you have the right idea pretty much

Answer 35

yeah, you just did that let t^2 sub thing again for the quantity under the root,

Answer 36

let
u = x -2 --->x= u + 2
du = dx

Answer 37

\[\int\limits_{3}^{6}(u + 2 + 1)\sqrt{u}*du\]

Answer 38

multiplies out to the integral of

u ^ (3/2) + 3u^(1/2)

Answer 39

what answer did u got I thin k i did the whole prob wrong

Answer 40

no, yours actually looks good, probably less writing too in this case

Answer 41

so do you want me to keep it the way it is? I am changing my work to what you are showing me here

Answer 42

\[\large \frac{ 2 }{ 5 }u^{5/2}+2u^{3/2}\]

replace u=x-2

evaluated from 3 to 6

=132/5

Answer 43

either way looks fine

Answer 44

show them to your teacher, and see what they have to say

Answer 45

oh okay tysm yay done! alright thanks man. I will send you few next time. thnaks bye tc

Answer 46

you are changing the prob from the domain in x, to domain in u

you can change it to whatever you want, change from x to t^2 , just have to keep eveyuthing consistent and defined

Answer 47

alright sure I will try to do that!

Answer 48

subbing variables by how you change from x-space, to another space (u)

so you can have a simpler function to integrate in another domain

Answer 49

okay will try to keep that in mind :)

Answer 50

you just have to keep your Map from one to the othe rconsistent

Answer 51

k goodluck

Answer 52

okay I will but the next assignmnet is going to be hard I am worried about that one