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find the integral of the equation Y = 100 cos x / x^2 from 90 degrees to 270 degrees. please explain. am not very bright :(

2007-08-07 03:16:18 · 5 answers · asked by Sarah M 1 in Science & Mathematics Mathematics

find the integral of the equation Y = 100 cos x / x^2 from 90 degrees to 270 degrees. please explain. am not very bright :(



sorry what is parathesis also sahsjing would you please be able to explain it thoroughly?!

2007-08-07 03:28:33 · update #1

5 answers

It's not that you are not bright. Youve got a tiger by the
tail here because this function has no
elementary antiderivative! No wonder you are having
trouble!! Anyway, let's try to solve it.
Ignoring the 100 for now, let's work
I = ∫cos x/ x² dx by parts.
Let u = cos x dv = 1/x²
du = -sin x v = -1/x
So I = -cos x/x - ∫ sin x/x dx.
Now let's work on the definite integral:
First cos x is 0 at both π/2 and 3π/2, so we
have to compute ∫ sin x/x dx = Si(x)
between π/2 and 3π/2.
Here there is no elementary antiderivative, but
there are several ways to proceed. I'll let you
work out the details:
1). Expand sin x/x in a Taylor series and plug in
the values.
2). Evaluate the integral by Simpson's rule
or the trapezoidal rule.
3). Find the values of Si(x) on a graphing calculator.
In any case ∫(x = π/2..3π/2) sin x/x dx works out to about
0.0548 so the value of your original integral
is about -5.48.

2007-08-07 04:54:10 · answer #1 · answered by steiner1745 7 · 0 0

The integral is negative, as noticed earlier. There is no closed form for the antiderivative, also noticed earlier. It is not clear to me why one should use parts before using a calculator, although if a series solution is used, parts first simplifies the series.
I get -23.761, rounded for both the integral of cos(x)/x^2 and and -sin(x)/x from pi/2 to 3pi/2. The only way I could get the other two answers given was to put my calculator in degree (rather than radian) mode.

2007-08-14 04:11:53 · answer #2 · answered by hemidemisemiquaver 2 · 0 0

I = ∫100 cos x / x^2 dx, x from pi/2 to 3pi/2 = 42.39

2007-08-07 10:22:01 · answer #3 · answered by sahsjing 7 · 0 0

Steinert´s answer is very good.
The other answer is wrong because the integrand in the interval is negative so the integral can´t be positive

2007-08-07 15:44:00 · answer #4 · answered by santmann2002 7 · 0 0

Could you clarify with parenthesis please?

2007-08-07 10:20:58 · answer #5 · answered by Mitchell 5 · 0 0

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