Estimate area between x-axis and curve y= (x^2+16)^.5?

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DW2000

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You wish to estimate the area between the x-axis and the curve y= (x^2 +16)^.5 (the square root) on the interval [0,3].

How many rectangles should be used so that the difference between the left-hand and right-hand sums is no more than 0.0001?

I have tried to figure this out and I am really lost. I tried using x values of 0, 1, 2, and 3, but the differences between the sums were 1.0, which is obviously more than 0.0001. Any help is appreciated!
 
There are error estimates for the midpoint, trapezoid, and simpson's rule.

This is probably what you are expected to use.

Let's take the midpoint rule:

The error estimate is \(\displaystyle |E_{M}|\leq \frac{(b-a)^{3}K_{2}}{24n^{2}}\)

\(\displaystyle K_{2}\) is the max value of the second derivative on the given interval [0,3]. So, in this case, \(\displaystyle f''(x)=\frac{16}{(x^{2}+16)^{\frac{3}{2}}}\)

It's max value is 1/4 at x=0.

So, subbing into the formula with \(\displaystyle E_{M}=.0001, \;\ K_{2}=1/4, a=0, \;\ b=3\), we solve for n:

\(\displaystyle .0001=\frac{3^{3}(1/4)}{24n^{2}}\).

This is how many subdivisions we need in order to attain the desired accuracy.
 
Thanks for the response, however, this formula was never taught in class. Therefore, I'm expected to perform this problem in a different way. The class was taught left and right-hand sums for taking estimates. The few formulas taught were

?t = (b-a)/n where [a,b], n= the number of subdivisions, and ?t= the amount between each subdivision

we also learned that the difference in estimates is the absolute value of (f(b)-f(a))?t
 
Well, use the formula then. Show off a little. :D

Other than that, it is guess and check.

You're going to need more than 2, 3, or 4 subdivisions.

The actual solution is 13.0451774445

So, you need to be within .0001 of that.
 
DW2000 said:
You wish to estimate the area between the x-axis and the curve y= (x^2 +16)^.5 (the square root) on the interval [0,3].

How many rectangles should be used so that the difference between the left-hand and right-hand sums is no more than 0.0001?

I have tried to figure this out and I am really lost. I tried using x values of 0, 1, 2, and 3, but the differences between the sums were 1.0, which is obviously more than 0.0001. Any help is appreciated!
Click to expand...

You tried n= 3

Now try n = 30

Generalize the error expression for any 'n'.
 
Could I say that ?t = 3-0/n, so ?t=3/n ?

If the difference between the sums can be no more than 0.0001, is it safe to say that ?t can be no more than 0.0001? If that's the case, ?t, or 3/n ? 0.0001.

Solving for n, you get 30,000. So n, or the number of rectangles, needs to be more than or equal to 30,000.

I'm trying to put together everything that has been posted into processes that make sense to me. Thanks again!
 
Really?. I must have messed up then. I get around 53.

Sorry, I will have to see what I done wrong. :oops:

I used the error estimate for midpoint.

\(\displaystyle .0001=\frac{27(1/4)}{24n^{2}}\Rightarrow n\approx 53\)

I even checked via the midpoint rule and the difference was withing the parameter of .0001.

Oh well, back to the drawing board. :D
 
Oh,the difference between the left and right hand sums is no more than .0001.

I reckon I had pay better attention. OOps :oops: :roll:
 
Subhotosh Khan said:
Per my calculation 29,950 < n < 30,000
Click to expand...

from yahoo (but I concur)

Suppose that we split the interval [0, 3] into n sub-intervals of equal width. The width of each sub-interval is then (3 - 0)/n = 3/n.

The left-hand sum is:
LHS = sum(i=0 to n-1) f(a + i?x)?x
= sum(i=0 to n-1) ?[(0 + 3i/n)^2 + 16](3/n)
= (3/n)sum(i=0 to n-1) ?(9i^2/n^2 + 16)
= (3/n)[-5 + sum(i=0 to n) ?(9i^2/n^2 + 16)]
= -15/n + (3/n)sum(i=0 to n) ?(9i^2/n^2 + 16),

while the right-hand sum is:
RHS = sum(i=1 to n) f(a + i?x)?x
= sum(i=1 to n) ?[(0 + 3i/n)^2 + 16](3/n)
= (3/n)sum(i=1 to n) ?(9i^2/n^2 + 16)
= (3/n)[-4 + sum(i=0 to n) ?(9i^2/n^2 + 16)]
= -12/n + (3/n)sum(i=0 to n) ?(9i^2/n^2 + 16).

Note that both the left and right-hand sum expressions both have (3/n)sum(i=0 to n) ?(9i^2/n^2 + 16); this is crucial to find the difference because this part cancels out when you take the difference.

At this point, we see that the difference between these two is -12/n - (-15/n) = 3/n. In order for the difference to be no more than 0.0001:
3/n ? 0.0001 ==> n ? 30,000.

Therefore, 30,000 or more rectangles are required in order for the left and right-hand sums to be within 0.0001.

More simplistic calculation (like DW did) will give the result n= 90,000
 
I understand now. I used the difference in estimate equation to find that ?t=0.0001 and solved for n using b-a/n. n is 30,000

Thanks to everyone for all the help
 
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