Trapezoildal Sum Question

[april19]

v(t) = 40 + 6* sq root(t)

As n (number of increments) goes to infinity, using our calculator, we found the limit of the sums goes to 468.

Now I need to find a number X for which the trapezoidal sum is within .01 unit of this limit when n > X.

Without using a calculator, is there a systematic way to find this number X?

Thanks.

 

[galactus]

You can use the formula for the error estimate for the trapezoidal rule.

\(\displaystyle |E_{T}|=\frac{(b-a)^{3}K_{2}}{12n^{2}}\)

Where \(\displaystyle K_{2}\) is the max value of the second derivative, \(\displaystyle |f''(x)|\), on the interval [a,b].

You did not state, but your interval is [0,9], I believe. Because \(\displaystyle \int_{0}^{9}[6\sqrt{t}+40]dt=468\)

\(\displaystyle f''(x)=\frac{-3}{2t^{\frac{3}{2}}}\)

\(\displaystyle |E_{T}|=\frac{(9-0)^{3}\cdot K_{2}}{12n^{2}}\leq .01\)

It is difficult to find a max value on [0,9], and inequalities are of little value in finding an upper bound on the magnitude of the

error. Because, \(\displaystyle f''(x)\to {-\infty}\) as \(\displaystyle t\to 0^{+}\)

So, in cases where it is rough to find values of \(\displaystyle K_{2}\), they may be replaced with any larger constant if the constant is easier to find.

For instance, if \(\displaystyle K_{2}<K\), then \(\displaystyle |E_{T}|\leq \frac{(b-a)^{3}K^{2}}{12n^{2}}<\frac{(b-1)^{3}K}{12n^{2}}\)

So, the right side is also an upper bound on the value of \(\displaystyle |E_{T}|\), though it is larger and not quite as good as using \(\displaystyle K_{2}\).

Try some values for K and solve for n.