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I greatly appreciate it if someone could help me with this problem:

Use the Error Bound to find the least possible value of $N$ for which $Error(S_N)\le 1 \times 10^{-9}$
in approximating the integral $\int\limits_0^1 3e^{x^2}dx$.

Thanks so much!

EDIT:

The following is my work,

We already know that Error(Sn) <= (K(b-a)^5)/(180N^4)

Thus, the first step is to find the fourth derivative: f'(x) = 6xe^(x^2) f''(x) = 6(2x^2 + 1)e^(x^2) f'''(x) = 12e^(x^2)x(2x^2+3) f''''(x) = 12e^(x^2)(4x^4 + 12x^2 + 3)

K = f''''(1) = 12*19e = 228e

Then I just plugged K in and solved for N:

(228e(1-0)^5)/(180N^4) <= 10^-9

And I eventually came up with 242 <= N, so the least possible value is 242, but apparently that's not right.

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  • $\begingroup$ Did you try using the error bound? What did you get and why are you stuck? Perhaps we can help with that. $\endgroup$ – Patrick Da Silva Sep 17 '13 at 9:46
  • $\begingroup$ What method are you using? Rectangle method / Trapezoidal rule / Simpson's rule / other Newton-Cotes quadrature rule ? $\endgroup$ – Oleg567 Sep 17 '13 at 9:49
  • $\begingroup$ I'm using Simpson's Rule, and I edited the problem description my work and what I did. $\endgroup$ – user95297 Sep 17 '13 at 10:27
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You have everything correct, except the least n value would be the next even whole number which is 244.

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  • $\begingroup$ Ah, that was what was wrong. But do you mind actually explaining why it's the next even number? $\endgroup$ – user95297 Sep 18 '13 at 6:03
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Your calculations are correct...

Exact value of integral: $4.38795~52377~21544~82641~...$ .

Denote $N=242$, $~~h = 1/N=1/242$ ($N$ is even).

Denote $x_j = j/N$;   $f_j = e^{x_j^2}$   ($j=0,1,2,...,N$).

When you'll use Simpson's Rule, you'll calculate this value:

$ I = \dfrac{h}{3}\Bigl((f_0 + 4f_1+f_2) + (f_2+4f_3+f_4)+ ... + (f_{N-2}+4f_{N-1}+f_N)\Bigr) = \dfrac{h}{3}\Bigl(f_0 + 4f_1+2f_2+4f_3+2f_4+4f_5+2f_6+4f_7+ ... + 2f_{N-2}+4f_{N-1}+f_N\Bigr) = 4.38795~52379~8572... $

Error is $\varepsilon \approx 0.00000~00002~642 = 2.642 \times 10^{-10} < 1 \times 10^{-9}$.

All seems to be correct.

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