Easy Integration :( (1 Viewer)

[lolokay]

I got the same answer (basically) and wolfram gives it exactly as you have it http://integrals.wolfram.com/index.jsp?expr=(25-x^2)^(1/2)&random=false

 

[shaon0]

shaon0's working is correct.

What would have been wrong with my answer from lyounamu's p.o.v?

 

[shaon0]

I got the same answer (basically) and wolfram gives it exactly as you have it http://integrals.wolfram.com/index.jsp?expr=(25-x^2)^(1/2)&random=false

Woah mad...do you have to pay for this?

 

[tommykins]

What would have been wrong with my answer from lyounamu's p.o.v?

I don't know.

How did you approach this question anyhow ?

 

[lyounamu]

What would have been wrong with my answer from lyounamu's p.o.v?

Sorry, I didn't know that you could approach the question in that way. But tell me how you approached it.

 

[lolokay]

a way you could approach it, apart from using trig substitution would be to observe that the function is a semicircle, then find an expression for the definite integral (area of that segment the circle) from 0 to some value, b. Then just add a constant.

 

[Just.Snaz]

Not entirely correct. It's also 25/2[sin2@/2 + @] , not 25/4.

You need to revert the @'s back into x's.

ah sorry yeah.. my mistake.. been doing english all last week.. literally..

 

[shaon0]

Sorry, I didn't know that you could approach the question in that way. But tell me how you approached it.

Ok i did.
S sqrt(25 - x²) dx

Let x = 5sin(θ)
dx = 5cos(θ) dθ

= S sqrt(25 - [5sin(θ)]²) 5cos(θ) dθ
= 5 S sqrt(25 - 25sin²(θ) cos(θ) dθ
= 5 S sqrt(25(1 - sin²(θ) cos(θ) dθ
= 5 S 5sqrt(1 - sin²(θ) cos(θ) dθ
= 25 S sqrt(1 - sin²(θ) cos(θ) dθ
= 25 S sqrt(cos²(θ) cos(θ) dθ
= 25 S cos(θ)cos(θ) dθ
= 25 S cos²(θ) dθ
= 25 S (1 + cos(2θ)/2 dθ
= 25/2 S (1 + cos(2θ) dθ
= 25/2 (θ + sin(2θ)/2) + C
= 25/2 (θ + cos(θ)sin(θ)) + C

You know how x = 5sin(θ)
Then, θ = arcsin(x/5)

= 25/2 [arcsin(x/5) + cos(arcsin(x/5))sin(arcsin(x/5))] +C
= 25/2 [arcsin(x/5) + (x/5)cos(arcsin(x/5))] +C
= 25/2 [arcsin(x/5) + (x/5)sqrt[1 - (x/5)²]] +C
= 25/2 [arcsin(x/5) + (x/5)sqrt(1 - x²/25)] + C

....Yea that took me 20 minutes to type up and do. lol (back to physics)
So that was my working
btw, if your file is to big too put on after you scan it and compress it. how would you put it on?
PLUS, i didn't know if my answer was correct because i taught calculus to myself, so sorry if it is kind of wrong at the top.

 

[tacogym27101990]

wow you taught yourself how to do that?
impressive

 

[shaon0]

wow you taught yourself how to do that?
impressive

mainly by looking at examples and reading up on concepts.

 

[12o9]

mainly by looking at examples and reading up on concepts.

Holy crap. That's pretty hectic. .

 

[shaon0]

Lol, he obviously does it for a response. Just ignore him, he's all talk.

shaon0 - great job, I thought you found out a new way, didn't know you used trig subs, awesome work nonetheless.

Thanks. Substitution is the easiest way but i kind of figured out another way.
Can you use integration by parts? since, (25-x^2)=(5-x)(5+x)

 

[shaon0]

Holy crap. That's pretty hectic. .

Thanks...i kind of skipped doing all my maths homework ie. plane geometry, to learn calculus. I've finished volumes of revolution (or solids of revolution) now.
Can't think of whats next.

 

[tommykins]

Thanks. Substitution is the easiest way but i kind of figured out another way.
Can you use integration by parts? since, (25-x^2)=(5-x)(5+x)

It'd be a huge hassle I think.

You need u = sqrt(25-x^2) u' = differentiate that.

then v' = 1, v = x and that just adds another x, making it more complex than required.

 

[tacogym27101990]

Thanks. Substitution is the easiest way but i kind of figured out another way.
Can you use integration by parts? since, (25-x^2)=(5-x)(5+x)

no i dont think i.b.p would work here
but even if it did it would be really unneccesarry

what were you planning to do
let u=rt(5-x) and v'=rt(5+x)dx???

 

[shaon0]

no i dont think i.b.p would work here
but even if it did it would be really unneccesarry

what were you planning to do
let u=rt(5-x) and v'=rt(5+x)dx???

yeah, something like that.

 

[tacogym27101990]

yeah its not very nice
just stick to the substitution
but im impressed that your teaching yourself four unit integration techiniques

 

[Aaron.Judd]

Damn I got lost.

 

[Pwnage101]

Damn I got lost.

lol @ 2U ppl not knowin wat u 4U guys are doin, next ur gonna invade the general maths forums, where u wont be welcomed lmao

 

[Aerath]

lol @ 2U ppl not knowin wat u 4U guys are doin, next ur gonna invade the general maths forums, where u wont be welcomed lmao

Haha, well that question was in our 2U past paper (we have trials tomorrow), and we were like, OMGWTF - haven't learnt how to integrate that yet.