I don't know how to approach this one... Help would be very much appreciated!!

- anonymous

I don't know how to approach this one... Help would be very much appreciated!!

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

##### 1 Attachment

- 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

Have you tried that?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

Ohhh just a sec! Im gonna give that a try :)

- anonymous

So is it 81 - (x - 9)^2 ?

- Zale101

|dw:1438583101045:dw|

- Zale101

Correct

- anonymous

Ah yes! Great! :)

- Zale101

|dw:1438583282353:dw|

- 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

I think my 3rd step is wrong, isn't it?

- anonymous

yea theres something wrong with the power

- anonymous

I knew it xD lol How would I simplify it?

- anonymous

|dw:1438584976279:dw|

- anonymous

Right!

- anonymous

I mean is my step from (-u^2 + 81)^(1/2) to (-u+9) correct?

- anonymous

i remember you have to divide by 18+1

- anonymous

\[\int\limits_{}^{}(ax+b)dx=\frac{ 1 }{ a }\times \frac{ (ax+b)^{a+1} }{ a+1 }\]\[a \neq-1\]

- anonymous

\[\int\limits_{}^{}(ax+b)^ndx\]oh god the lag with the equation is horrible

- anonymous

Lol xD I'm kinda lost

- anonymous

the x^2 is the messy thing

- anonymous

I don't exactly know what to do with it xD but it totally looks wrong

- anonymous

where did ya power go?

- anonymous

you need to add 1 to 1/2 and divide it by 1+1/2

- anonymous

I considered it as (-u^2)^(1/2) and 81^(1/2) which is completely wrong, haha

- anonymous

@saseal but then what about inside the brackets?

- 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

thought*

- Astrophysics

\[\int\limits \sqrt{-u^2+81}du \implies \int\limits \sqrt{81-u^2}du \] do another substitution, requires trig.

- Jhannybean

Oh, maybe a double substition could work?

- Jhannybean

Dang it.

- 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

This is not a pleasant integral lol
Your next substitution is \[u = 9\sin(t)\]

- 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

@Astrophysics i tried that with a calculator and the answer looks a little crazy

- anonymous

@Astrophysics how'd you get 9sin(t) ? :O

- anonymous

@saseal which one did you try?

- 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

astro's trig integral

- anonymous

Hmmm... @Jhannybean I'm trying to figure out what you did there lol

- Jhannybean

Its called a double substitution.

- Jhannybean

But don't use it, it's not going to be pretty.

- Jhannybean

i got it.

- Jhannybean

\[\sqrt{81-u^2}du \longrightarrow \sqrt{a-x^2}dx ~\therefore~ x=a\sin(t)\]

- anonymous

@Astrophysics I tried putting in wolfram's answer before, but it said it was incorrect >.<

- Jhannybean

It's a trig substitution rule that you learn in calculus.

- anonymous

you wan symbolab's answer?

- anonymous

https://www.symbolab.com/solver/step-by-step/%5Cint%5Csqrt%7B18x-x%5E%7B2%7D%7Ddx/?origin=suggestion

- Astrophysics

Oooh nice

- anonymous

always good to have more options

- Astrophysics

Yeah try that answer, it uses the exact approach I was talking about

- anonymous

Alrighty... thats a lot to type in xD haha hold up xD

- Astrophysics

I mean maybe there is a simpler way, I don't know.

- anonymous

i don't know how i forgot much of my integration stuff right after the test

- anonymous

##### 1 Attachment

- anonymous

I don't know am I typing it in incorrectly or what? XD

- Astrophysics

You put it in wrong, it's all over 2

- anonymous

Ohhh shoooot! >.<

- anonymous

Still wrong...

##### 1 Attachment

- Astrophysics

Your doing the wrong integral

- anonymous

it looks like you have enter your answer on a different question

- Astrophysics

bahaha!

- anonymous

XDDDDD

- anonymous

Oh my goodness ahahaha

- anonymous

dont blame me its almost 1 am

- anonymous

pretty cool you can enter the answer over and over again

- anonymous

hehehehe
My bad :P

##### 1 Attachment

- 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

yay!

- anonymous

much wow, such correct

- Astrophysics

xD

- anonymous

I know! Lol but i still don't understand how they get that >.<

- Astrophysics

Go over the symbolab result I guess

- Astrophysics

it shows all the steps

- anonymous

thats why i chose em over wolfram

- Astrophysics

here is the trick though...http://www.sosmath.com/calculus/integration/trigsub/trigsub.html

- anonymous

Yeah, i guess i should! It sucks that wolfram doesn't show all the steps unless you go Pro XD

- Astrophysics

Yeah :(

- Astrophysics

Nice work everyone! Collective work is always good, especially for bad integrals!

- anonymous

Yeah! Thanks all!! @Jhannybean @saseal @Astrophysics :D

- Astrophysics

and @Zale101 :P

- anonymous

I was going all nuts over this one lol

- anonymous

Oh yes!! And thanks to @Zale101 as well! :D
Thanks for reminding me! Haha

Looking for something else?

Not the answer you are looking for? Search for more explanations.