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Obtain an upper bound on the absolute error when we compute $\displaystyle\int_0^6 \sin x^2 \,\mathrm dx$ by means of the composite trapezoid rule using 101 equally spaced points.

The formula I'm trying to use is:

$$ I = \frac{h}{2} \sum_{i=1}^n \Big[f(x_{i-1}) + f(x_i)\Big] - \frac{h^3}{12} \sum_{i=1}^n f^{''}(\xi_i) $$

But I'm lost on how to calculate the error and find a value for $\xi$. What's the general way of finding the error like this? Thanks for any help :)

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To find Upper Bound of Error using Trapezoidal Rule

No. of sub intervals = $n$

Given integral is $$\int_0^\pi \sin(2x)\,\mathrm {d}x$$

$$\implies f(x)=\sin(2x), a=0,b=\pi$$

$$f'(x) = 2\cos(2x)$$

$$f''(x)=-4\sin(2x)$$

The maximum value of $|f''(x)|$ will be 4

$M=4$

The upper bound of error,

$$|e_T|\le \frac{M(b-a)^3}{12n^2}$$

$$|e_T|\le \frac{\pi^3}{3n^2}$$

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  • $\begingroup$ This isn't the same integral. Was he just giving a example? Or are you supposed to change the integral from $x^2$ to $2x$ in the problem for some reason... $\endgroup$ – Shammy Nov 3 '15 at 2:55
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You are not supposed to compute the exact values of the $\xi_i$. Instead, what can you say about $f''$? Can you find a simple expression $g(x)$ with $|f''(\xi)|\le g(\xi)$? And then find a simple estimate for the sum, given that each $\xi_i$ is in a known interval?

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