DanJS
  • DanJS
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Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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DanJS
  • DanJS
@katie00
DanJS
  • DanJS
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
DanJS
  • DanJS
u = 3 x - 7 dy = 3 dx, or dx = du/3

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DanJS
  • DanJS
\[\int\limits \frac{ \cos~x }{ \sin^4 x }dx\] u misstyped a power when substituting... u = sin x du = cos x dx \[\large \int\limits \frac{ 1 }{ u^4 }du = \frac{ -1 }{ (3*u^3) } = \frac{ -1 }{ 3*\sin^3(x) }\]
DanJS
  • DanJS
\[\int\limits (x+1)\sqrt{x-2}~~dx\] 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
DanJS
  • DanJS
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..
anonymous
  • anonymous
no, the method does matter. The prof grades the method the method is more imp than the answer.
anonymous
  • anonymous
so what would d be after i put u=3x-7 and du=3? @DanJS
DanJS
  • DanJS
That is how i remember to do it, just less writing maybe
DanJS
  • DanJS
did they show you probs replacing stuff under a root with a squared term like that t^2
anonymous
  • anonymous
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
DanJS
  • DanJS
\[\sqrt{t^2 }=\left| x \right|\] , yeah was just looking at wha tyou did again
DanJS
  • DanJS
\[\int\limits\limits_{3}^{4} \frac{ 2 }{ \sqrt{3x-7} }dx = \int\limits\limits_{3}^{4}\frac{ 2 }{ \sqrt{u} }*\frac{ du }{ 3 }\]
DanJS
  • DanJS
when u = 3x-7 and du = 3 dx
DanJS
  • DanJS
\[\frac{ 2 }{ 3 } *\int\limits_{3}^{4}u^{-1/2}*du\]
DanJS
  • DanJS
you get \[(2/3)*2u^{1/2} \] from 3 to 4
anonymous
  • anonymous
I though there wont be a u in the answer
DanJS
  • DanJS
right , put back in the sub, u = 3x - 7 \[\frac{ 4 }{ 3 }\sqrt{3x - 7}\]
DanJS
  • DanJS
evaluated for the interval 3 to 4, looks the same final answer you have
anonymous
  • anonymous
hm okay for the next one that u mentioned. u said I missed a power. can you do that one too plz.
anonymous
  • anonymous
i never get the cos/sin ones anyway so
DanJS
  • DanJS
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)
DanJS
  • DanJS
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 \[\int\limits \frac{ 1 }{ u^4 }du = \int\limits u^{-4}du \]
DanJS
  • DanJS
= (-1/3) * u^-3 = -1/( 3 * sin^3(x))
anonymous
  • anonymous
okay so in this case what do i plug in at the end? or do i even d that?
DanJS
  • DanJS
is it a def or indef integral?
anonymous
  • anonymous
indef
DanJS
  • DanJS
yeah just simplify the function as much as you can , that is the answer ,
anonymous
  • anonymous
so -1/3(sin^3 x) is my answer
DanJS
  • DanJS
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
anonymous
  • anonymous
okay last one that u mentioned (F) def 3-6
anonymous
  • anonymous
tysm for pointign out my erros :) I would have never realized it ahah but tysm for everything
anonymous
  • anonymous
lets finish the last one so I can work on my hw for other classes after we are doen with clac
DanJS
  • DanJS
ok, they arent huge errors, you have the right idea pretty much
DanJS
  • DanJS
yeah, you just did that let t^2 sub thing again for the quantity under the root,
DanJS
  • DanJS
let u = x -2 --->x= u + 2 du = dx
DanJS
  • DanJS
\[\int\limits_{3}^{6}(u + 2 + 1)\sqrt{u}*du\]
DanJS
  • DanJS
multiplies out to the integral of u ^ (3/2) + 3u^(1/2)
anonymous
  • anonymous
what answer did u got I thin k i did the whole prob wrong
DanJS
  • DanJS
no, yours actually looks good, probably less writing too in this case
anonymous
  • anonymous
so do you want me to keep it the way it is? I am changing my work to what you are showing me here
DanJS
  • DanJS
\[\large \frac{ 2 }{ 5 }u^{5/2}+2u^{3/2}\] replace u=x-2 evaluated from 3 to 6 =132/5
DanJS
  • DanJS
either way looks fine
DanJS
  • DanJS
show them to your teacher, and see what they have to say
anonymous
  • anonymous
oh okay tysm yay done! alright thanks man. I will send you few next time. thnaks bye tc
DanJS
  • DanJS
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
anonymous
  • anonymous
alright sure I will try to do that!
DanJS
  • DanJS
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
anonymous
  • anonymous
okay will try to keep that in mind :)
DanJS
  • DanJS
you just have to keep your Map from one to the othe rconsistent
DanJS
  • DanJS
k goodluck
anonymous
  • anonymous
okay I will but the next assignmnet is going to be hard I am worried about that one

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