hw asst 20

020.

Applied Calculus II

07-23-2008

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17:00:41

Query problem 6.5.12 (was 6.5.10) trapezoidal and Simpson's rules, n=4, integral 0 to 2 of x `sqrt(x2+1)

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RESPONSE -->

Using Trap

b - a / n = 2-0 / 4 = 1/2

x0 = 0, x1 = .5, x2 = 1, x3 = 1.5, x4 = 2

b-a /2n =1/4 so,

1/4 ( 0 'sqrt(0^2 +1) + 2( .5 'sqrt(.5^2 +1) + 2(1 'sqrt(1^2 +1) + 2(1.5 'sqrt 1.5^2 +1) + (2 'sqrt(2^2 +1)

1/4 ( 0+1.118+2.828+5.408+4.472)

=3.457

Using Simpson's

width again = 1/2

x0 = 0, x1 = .5, x2 = 1, x3 = 1.5, x4 = 2

b-a / 3n =1/6

1/6( 0 'sqrt(0^2 +1) + 4( .5 'sqrt(.5^2 +1) + 2(1 'sqrt(1^2 +1) + 4(1.5 'sqrt 1.5^2 +1) + (2 'sqrt(2^2 +1)

= 1/6 (0+2.449+2.828+10.817+4.472

= 3.427

both answers = 3.4

confidence assessment: 3

The differences you are looking at here occur at a much higher level of precision.

You would use x0=0, x1=1/2, x2=1 x3=3/2 x4=2.

The corresponding values of x^2 sqrt(x^2+1) are 0, 0.5590169943, 1.414213562, 2.704163456, 4.472135954.

Using these values:

Trap gives you 3.4567.

Simpson's rule gives you 3.3922.

The exact result, to five significant figures, is int(x sqrt(x^2+1),x,0,2) = 3.3934. This is based on the antiderivative 1/3 * (x^2+1)^(3/2)

According to these results Simpson's approximation is within .0012 while trap is within about .064. So the Simpson's approximation is .064 / .0012 = 50 times better, approx..

Theory says that it should be about n^2 = 4^2 = 16 times better. **

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To integrate x sqrt( x^2 + 1 ) let u = x^2 so that du = 2 x dx. The integrand becomes 1/2 sqrt(u) du with antiderivative 1/2 ( 2/3 u^(3/2) ) = 1/3 u^(3/2), which with the given limits, calculated to five significant figure, gives you the result 3.3934.

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This income stream would be $6000 at the beginning and $6400 at the end of the 4-year period. The income during this period would therefore be between $24,000 and $25,600. The present value of this income will be somewhat less.

** The present value of [6000 + 200 `sqrt(t)] `dt, for a short time interval t, is [ 6000 + 200 `sqrt(t) ] * e^(-.07 t ) `dt.

Evaluating the integral INT( ( 6000 + 200 `sqrt(t) ) * e^(-.07 t ) with respect to t from 0 to 4) we obtain $21,836.98. **

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I don't think your derivative is correct. See the given solution below.

The x = 0 point of the curve is (0, 2/3) and the x = 1 point is (1, 0). The straight-line distance between these points is thus sqrt(1^2 + (2/3)^2) = sqrt(13/9) = 1.2 or so. The straight line being the shortest distance between two points, we expect the arc distance to be more than 1.2. However unless the curve gets pretty steep the distance won't be drastically greater than this.

** Integrate to find the arc length of the curve.

If a line with slope m lies above an interval `dx on the x axis then the 'run' of the segment above `dx is just `dx, while the rise is m * `dx. The hypotenuse is therefore `sqrt( `dx^2 + ( m `d)^2 ) = `sqrt( 1 + m^2 ) * `sqrt(`dx^2) = `sqrt(1 + m^2) `dx.

If a curve y = f(x) lies above the interval `dx then the average slope of the curve is a value of y ' for some x in the interval. After a little fancy reasoning we can prove that the arc length is in fact equal to `sqrt( 1 + y'^2) `dx, but proof or not it's easy enough to understand if you understand the thing about m.

This leads to the theorem that for a differentiable function y = f(x) the arc length between x = a and x = b is INT ( `sqrt( 1 + y' ^ 2) dx, x from a to b).

The derivative of the given function is 1/2 x^.5 - 1/2 x^-.5; the square of this expression is 1/4 ( x - 2 + 1/x) so you integrate `sqrt( 1 + 1/4 ( x - 2 + 1/x) ) from 0 to 1.

The integral gives you 1.33327, accurate to 6 significant figures. However you'll probably have to use an approximation technique so you probably won't get this exact result. **

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&#Good work. See my notes and let me know if you have questions. &#