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I'm having trouble finding the error bound of this function. My professor says it's "trivial" and thus, he refused to offer me any help beyond a simple hint I kind of knew anyway.

I am given this integral and I need to find it's error when we calculate it using the trapezoidal rule.

$$\int_{0.75}^{1.3}((sinx)^2 - 2xsinx + 1)dx$$

So I know

$$\epsilon = |\frac{h^3}{12}f''(\xi)|$$ where $\xi \in (0.75,1.3)$

Now, I find the second derivative:

$$f'(x) = 2sinxcosx - 2xcosx - 2sinx$$ $$f''(x) = -8sinxcosx + 6sinx + 2xcosx$$

Now, the only hint my professor did give me was that since we can't get the maximum of $f''(\xi)$ analytically we can take advantage of the property that both $sinx$ and $cosx$ are bounded by 1 from above. This means the worst that any of these trig functions could possibly be is 1. The issue is that $sinx$ and $cosx$ are 1 at different values.

Can anyone give me any hints on what I need to do to make the leap to understanding what is going on? I feel really lost.

EDIT: What I truly need help on is understanding how to maximize $f''(\xi)$

What I have done is (given $h=0.55)$:

$$\epsilon = |\frac{0.55}{12} * -8sin(\xi)cos\xi + 6sin\xi + 2(\xi)cos\xi|$$

and knowing that on $(0.75,1.3)$ $f''(\xi)$ is maximized at 1.3 just plug that in

$$\epsilon = |\frac{0.55}{12} * -1.3sin(1.3)cos1.3 + 6sin1.3 + 2(1.3)cos1.3| = 0.0612099$$

but the book says

$$\epsilon = 0.02444080544$$

and I have no idea how they got this.

Thanks!

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  • $\begingroup$ You want the modulus and |a+b|<=|a|+|b| $\endgroup$ – Paul Nov 22 '14 at 9:05
  • $\begingroup$ @Paul if you're talking about using the triangle inequality to split $\frac{h^3}{12}f''(\xi)$ I know that. $\endgroup$ – John Nov 22 '14 at 17:49
  • $\begingroup$ @Paul I have posted my work $\endgroup$ – John Nov 22 '14 at 18:03
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\begin{align*} \epsilon &\le \frac{h^3}{12} \max_{\xi\in[a,b]}\left| f''(\xi) \right| \\ &= \frac{0.55^3}{12} \max_{\xi\in[0.75,1.3]} \left| 2\xi\sin(\xi) -4 \cos(\xi)+2\cos(2\xi) \right|\\ &=\frac{0.55^3}{12}\left|2 \cdot 0.75 \sin(0.75)-4\cos(0.75)+2\cos(2\cdot 0.75)\right|\\ &\approx 0.02444080544 \end{align*}

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  • $\begingroup$ Thank you for your answer, however the book says the error bound should be 0.02444080544. Plugging in $h=0.55$ you get 0.39416 using your solution, which is rather large. How did you get the numbers you did? $\endgroup$ – John Nov 22 '14 at 17:53
  • $\begingroup$ I have posted my work $\endgroup$ – John Nov 22 '14 at 18:04
  • $\begingroup$ I just created an upper without putting any effort into minimizing it. I will take a look at it later. $\endgroup$ – slo Nov 22 '14 at 19:50
  • $\begingroup$ thank you, I'm looking forward to it. I'm still pretty lost on it. $\endgroup$ – John Nov 22 '14 at 22:19
  • $\begingroup$ I have updated my answer to match the book. $\endgroup$ – slo Nov 23 '14 at 12:13

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