anonymous
  • anonymous
I don't know how to approach this one... Help would be very much appreciated!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Zale101
  • Zale101
18x-x^2 is a standard quadratic equation. If you converted (18x-x^2) into the vertex form, you can easily apply u-sub.
Zale101
  • Zale101
Have you tried that?

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More answers

anonymous
  • anonymous
Ohhh just a sec! Im gonna give that a try :)
anonymous
  • anonymous
So is it 81 - (x - 9)^2 ?
Zale101
  • Zale101
|dw:1438583101045:dw|
Zale101
  • Zale101
Correct
anonymous
  • anonymous
Ah yes! Great! :)
Zale101
  • Zale101
|dw:1438583282353:dw|
anonymous
  • anonymous
Ohh I see! So then it turns into \[\int\limits \sqrt{-u^2 +81} du = \int\limits (-u^2+81)^{1/2}du = \int\limits (-u+9)du = -\frac{ 1 }{ 2 }u^2 + 9u\] Is that right?
anonymous
  • anonymous
I think my 3rd step is wrong, isn't it?
anonymous
  • anonymous
yea theres something wrong with the power
anonymous
  • anonymous
I knew it xD lol How would I simplify it?
anonymous
  • anonymous
|dw:1438584976279:dw|
anonymous
  • anonymous
Right!
anonymous
  • anonymous
I mean is my step from (-u^2 + 81)^(1/2) to (-u+9) correct?
anonymous
  • anonymous
i remember you have to divide by 18+1
anonymous
  • anonymous
\[\int\limits_{}^{}(ax+b)dx=\frac{ 1 }{ a }\times \frac{ (ax+b)^{a+1} }{ a+1 }\]\[a \neq-1\]
anonymous
  • anonymous
\[\int\limits_{}^{}(ax+b)^ndx\]oh god the lag with the equation is horrible
anonymous
  • anonymous
Lol xD I'm kinda lost
anonymous
  • anonymous
the x^2 is the messy thing
anonymous
  • anonymous
I don't exactly know what to do with it xD but it totally looks wrong
anonymous
  • anonymous
where did ya power go?
anonymous
  • anonymous
you need to add 1 to 1/2 and divide it by 1+1/2
anonymous
  • anonymous
I considered it as (-u^2)^(1/2) and 81^(1/2) which is completely wrong, haha
anonymous
  • anonymous
@saseal but then what about inside the brackets?
anonymous
  • anonymous
I can? But isn't that just like breaking up a root like this... \[\sqrt{a+b} = \sqrt{a} + \sqrt {b} \] And I though that was incorrect.. Is it not?
anonymous
  • anonymous
thought*
Astrophysics
  • Astrophysics
\[\int\limits \sqrt{-u^2+81}du \implies \int\limits \sqrt{81-u^2}du \] do another substitution, requires trig.
Jhannybean
  • Jhannybean
Oh, maybe a double substition could work?
Jhannybean
  • Jhannybean
Dang it.
anonymous
  • anonymous
yea i feel i should have gone \[\int\limits_{}^{}\sqrt{18x-x^2}dx = \int\limits_{}^{}(18x-x^2)^{\frac{ 1 }{ 2 }}dx\]\[u=x^2\]\[du=2x\]
Astrophysics
  • Astrophysics
This is not a pleasant integral lol Your next substitution is \[u = 9\sin(t)\]
anonymous
  • anonymous
Haha! Right!! Something to do with these... ? \[\sin^2x+\cos^2x =1\] and \[\tan^2x +1 = \sec^2x \] Or am I on the wrong track again xD lol
anonymous
  • anonymous
@Astrophysics i tried that with a calculator and the answer looks a little crazy
anonymous
  • anonymous
@Astrophysics how'd you get 9sin(t) ? :O
anonymous
  • anonymous
@saseal which one did you try?
Jhannybean
  • Jhannybean
\[\int \sqrt{-u^2+81}du \rightarrow \int (-u^2+81)^{1/2}du \rightarrow \\ v=-u^2+81 ~,~ dv=-\frac{2}{3}u^3du\rightarrow -\frac{3dv}{2} \\ \sf \text{ it gets messy with algebraic subs :(...}\]
anonymous
  • anonymous
astro's trig integral
anonymous
  • anonymous
Hmmm... @Jhannybean I'm trying to figure out what you did there lol
Jhannybean
  • Jhannybean
Its called a double substitution.
Jhannybean
  • Jhannybean
But don't use it, it's not going to be pretty.
Jhannybean
  • Jhannybean
i got it.
Jhannybean
  • Jhannybean
\[\sqrt{81-u^2}du \longrightarrow \sqrt{a-x^2}dx ~\therefore~ x=a\sin(t)\]
anonymous
  • anonymous
@Astrophysics I tried putting in wolfram's answer before, but it said it was incorrect >.<
Jhannybean
  • Jhannybean
It's a trig substitution rule that you learn in calculus.
anonymous
  • anonymous
you wan symbolab's answer?
anonymous
  • anonymous
https://www.symbolab.com/solver/step-by-step/%5Cint%5Csqrt%7B18x-x%5E%7B2%7D%7Ddx/?origin=suggestion
Astrophysics
  • Astrophysics
Oooh nice
anonymous
  • anonymous
always good to have more options
Astrophysics
  • Astrophysics
Yeah try that answer, it uses the exact approach I was talking about
anonymous
  • anonymous
Alrighty... thats a lot to type in xD haha hold up xD
Astrophysics
  • Astrophysics
I mean maybe there is a simpler way, I don't know.
anonymous
  • anonymous
i don't know how i forgot much of my integration stuff right after the test
anonymous
  • anonymous
anonymous
  • anonymous
I don't know am I typing it in incorrectly or what? XD
Astrophysics
  • Astrophysics
You put it in wrong, it's all over 2
anonymous
  • anonymous
Ohhh shoooot! >.<
anonymous
  • anonymous
Still wrong...
Astrophysics
  • Astrophysics
Your doing the wrong integral
anonymous
  • anonymous
it looks like you have enter your answer on a different question
Astrophysics
  • Astrophysics
bahaha!
anonymous
  • anonymous
XDDDDD
anonymous
  • anonymous
Oh my goodness ahahaha
anonymous
  • anonymous
dont blame me its almost 1 am
anonymous
  • anonymous
pretty cool you can enter the answer over and over again
anonymous
  • anonymous
hehehehe My bad :P
Astrophysics
  • Astrophysics
Lol I always hated answering questions on the computer...that's why I like assignments where you can hand it in, and yay!
anonymous
  • anonymous
yay!
anonymous
  • anonymous
much wow, such correct
Astrophysics
  • Astrophysics
xD
anonymous
  • anonymous
I know! Lol but i still don't understand how they get that >.<
Astrophysics
  • Astrophysics
Go over the symbolab result I guess
Astrophysics
  • Astrophysics
it shows all the steps
anonymous
  • anonymous
thats why i chose em over wolfram
Astrophysics
  • Astrophysics
here is the trick though...http://www.sosmath.com/calculus/integration/trigsub/trigsub.html
anonymous
  • anonymous
Yeah, i guess i should! It sucks that wolfram doesn't show all the steps unless you go Pro XD
Astrophysics
  • Astrophysics
Yeah :(
Astrophysics
  • Astrophysics
Nice work everyone! Collective work is always good, especially for bad integrals!
anonymous
  • anonymous
Yeah! Thanks all!! @Jhannybean @saseal @Astrophysics :D
Astrophysics
  • Astrophysics
and @Zale101 :P
anonymous
  • anonymous
I was going all nuts over this one lol
anonymous
  • anonymous
Oh yes!! And thanks to @Zale101 as well! :D Thanks for reminding me! Haha

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