| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I am having a problem integrating the following:
1 ____________
S \/ 1 - x**2 dx
-1
I can determine geometrically that the area is PI/2. Can anyone
integrate this.
(BTW, sorry for the notation I do not know how to make it look
like an normal integral sign. )
Mike
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 441.1 | What's wrong with geometry? | CLT::STAN | Stanley Rabinowitz | Tue Feb 11 1986 19:58 | 2 |
This is half the area of a unit circle, so the area is pi/2 as you
said.
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| 441.2 | Use trigonometry | CLT::STAN | Stanley Rabinowitz | Tue Feb 11 1986 20:06 | 5 |
Now to find the indefinite integral of sqrt(1-x^2), let x=sin t
so that dx=cos t dt.
Then S sqrt(1-x^2) dx = S cos^2 t dt = S (1 - cos 2t)/2 dt =
S 1/2 dt - S (cos 2t)/2 dt = t/2 - (sin 2t)/4.
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| 441.3 | is it me or the function | NACHO::MCMENEMY | Tue Feb 11 1986 20:09 | 7 | |
Nothings wrong with geometry, I wanted to check my work by
integrating the function. However, when I couldn't integrate it
I was concerned whether the function could be integrated or
whether my integratable abilities were lacking.
Mike.
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| 441.4 | Solution using MAPLE | METOO::YARBROUGH | Wed Feb 12 1986 14:27 | 7 | |
MAPLE yields:
x*sqrt(1-x^2) + arcsin(x)
-------------------------
2
Lynn Yarbrough
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| 441.5 | integrated bank accounts? | AUSSIE::GARSON | Mon Jan 25 1993 00:40 | 12 | |
Slightly related to this topic...
Walking past the bank the other day I glanced at a poster in the window. On
it was a handful of mathematical symbols. Past experience led me to expect
that it would be meaningless but I looked more closely anyway. The poster
contained
/1 10
132 | x(1-x) dx = 1
/0
which to my surprise is correct.
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| 441.6 | STAR::ABBASI | free like a bird | Mon Jan 25 1993 05:25 | 43 | |
> /1 10
>132 | x(1-x) dx = 1
> /0
i wanted to do this from first pricple, i took a remman partion,
and used the inf sum as the reimman sum (since we "know" this
is reimman integrable ;-), and i get this:
let P be the partition set {x_0,x_1,...,x_n} on [1,2]
that partitions [1,2] into n equal parts. each part has width
1/n.
S(f,P)= sum(i=1,n) Area_i
but Area_i = (1/n) x_i(1-x_i)^10 <--- taking the smaller sum
so S={ term_1 + term_2 + ..... +term_n}
where
term_1 = 1/n (1/n) (1-(1/n))^10 <-- x= 0+1/n
term_2 = 1/n (2/n) (1-(2/n))^10 <-- x= 0+ 1/n + 1/n
term_3 = 1/n (3/n) (1-(3/n))^10 <-- x= 0+ 1/n + 1/n + 1/n
...
term_n-1 = 1/n (n-1)/n (1- (n-1)/n)^10
term_n = 1/n (n/n) (1-(n/n))^10 = 0
so S= 1/n 1/n( (1-(1/n))^10 +2(1-(2/n))^10) +3(1-(3/n))^10 +..
(n-1)(1-(1-((n-1))/n))^10
)
after simplify more, i get S=
1
---- {(n-1)^10 + 2(n-2)^10 + 3(n-3)^10 + ...+(n-2)(n-(n-2))^10 + (n-1)(1)}
n^12
1/n^12 { sum(k=1,n-1) k(n-k)^10 }
unless i made a stupied mistake, the limit to the above S expression
must then go to 1/132 as n->oo .
i leave the proof to show that this limit goes to 1/132 as an excerise
for the reader ;-)
\nasser
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