Assignment 21

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Your trapezoidal approximation is good. However your Simpson's Rule approximation is not correct. You used

'1/12[same sequence as above]'

but the sequence isn't the same as the above. The values used are the same as they were in the trapezoidal rule, but they have different coefficients.

Your exact value of the integral is also incorrect. See below for the integration.

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

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. **

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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.....repeated for each of x values above

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. **

The Simpson's Rule approximation, if the correct coefficients are used, should come out very close to this.

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This is going to be an arc length integral.

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 your result will probably differ after a few significant figures from this result. **

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In some cases you didn't do the right integral, so be sure to see my notes.

You also appear to have used incorrect coefficients in your application of Simpson's Rule.

Assignment 21

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You need to be more specific about what you are taking the limit of.

** The integral as stated here diverges.

You need to take the limit as t -> infinity of INT(x^2 e^(-x^3), x from -t to t ).

Using the obvious substitution we see that the result is the same as the limiting value as t -> infinity of INT( 1/3 e^(-u), u from -t to t ). Using -1/3 e^(-u) as antiderivative we get -1/3 e^(-t)) - (-1/3 e^(-(-t))); the second term is 1/3 e^t, which approaches infinity as t -> infinity. The first term approaches zero, but that doesn't help. The integral approaches infinity.

Note that the integral from 0 to infinity converges: We take the limit as t -> infinity of INT(x^2 e^(-x^3), x from 0 to t ), which using the same steps as before gives us the limit as t -> infinity of -1/3 e^(-t) - (-1/3) e^0. The first term approaches zero, the second is just 1/3. So the limiting value is 1/3. **

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