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Suppose the function $f(x)$ has a Taylor series expansion. Then $$\int_a^bf(x)dx=\int_a^b(f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2+\cdots)dx=\\ \frac{f(a)}{1!}(b-a)+\frac{f'(a)}{2!}(b-a)^2+\frac{f''(a)}{3!}(b-a)^3+\cdots$$

and

$$\int_a^bf(x)dx=\int_a^b(f(b)+f'(b)(x-b)+\frac{1}{2}f''(b)(x-b)^2+\cdots)dx=\\ \frac{f(b)}{1!}(b-a)-\frac{f'(b)}{2!}(b-a)^2+\frac{f''(b)}{3!}(b-a)^3+\cdots$$

Therefore

$$\int_a^bf(x)dx=\frac{1}{1!}\frac{f(a)+f(b)}{2}(b-a)+\frac{1}{2!}\frac{f'(a)-f'(b)}{2}(b-a)^2+\frac{1}{3!}\frac{f''(a)+f''(b)}{2}(b-a)^3+\cdots$$

However, one can also consider $$\int_a^bf(x)dx=\int_a^df(x)dx+\int_d^bf(x)dx$$

where $d=\frac{a+b}{2}$

Then

$$\int_a^df(x)dx=\int_a^d(f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2+\cdots)dx=\\ \frac{1}{1!}\frac{f(a)}{2}(b-a)+\frac{1}{2!}\frac{f'(a)}{2^2}(b-a)^2+\frac{1}{3!}\frac{f''(a)}{2^3}(b-a)^3+\cdots$$

$$\int_d^bf(x)dx=\int_d^b(f(b)+f'(b)(x-b)+\frac{1}{2}f''(b)(x-b)^2+\cdots)dx=\\ \frac{1}{1!}\frac{f(b)}{2}(b-a)-\frac{1}{2!}\frac{f'(b)}{2^2}(b-a)^2+\frac{1}{3!}\frac{f''(b)}{2^3}(b-a)^3+\cdots$$

and so

$$\int_a^bf(x)dx=\frac{1}{1!}\frac{f(a)+f(b)}{2}(b-a)+\frac{1}{2!}\frac{f'(a)-f'(b)}{2^2}(b-a)^2+\frac{1}{3!}\frac{f''(a)+f''(b)}{2^3}(b-a)^3+\cdots$$

My question is, the two estimations are different. Which one is correct?

velut luna
  • 9,961
  • In the first two integrals, you're Taylor expanding about $a$ first and then expanding about $b$. Then, you're again saying that $\int^b_af(x)dx$ is the average of these two integrals - that part doesn't make sense to me. Could you justify? –  Apr 16 '14 at 00:05
  • After "Therefore" should it be $f'(a)+f'(b)$ in the numerator of the second term on the right of the equality? And similarly about the minus sign in front of the second term in the last line of the mass following "Then"? –  Apr 16 '14 at 00:10
  • Check the various methods to compute integrals numerically, there is quite a remarkable body of theory there. – vonbrand Apr 16 '14 at 00:24
  • Disregard my comment I was mistaken... I think I see what happened though I'll post an answer if I figure this out. –  Apr 16 '14 at 00:29
  • Wow this is undoubtedly the best question I have seen in a long time - wish I could upvote. –  Apr 16 '14 at 00:37
  • @user139981 Would you be ok with me posting a question asking if those two series could be equal? Or would you like to do that yourself? I think it would get more responses that way... great question btw!! –  Apr 16 '14 at 02:25
  • Dear @user142299, please feel free to post that question. And btw, thanks for your answer! – velut luna Apr 16 '14 at 02:49

1 Answers1

1

Remarkably, they appear to both be correct. I have checked the following integrals using both formulas: $$\int_0^{\pi/2}\sin x\,\mathrm{d}x=1,\;\;\int_0^{\pi/2}\cosh x\,\mathrm{d}x\approx 2.301,\;\;\int_a^{b}x^2\,\mathrm{d}x=\frac{b^3-a^3}{3},\;\;\int_a^{b}x^3\,\mathrm{d}x=\frac{b^4-a^4}{4}$$ See for example Here and Here

(scroll down to "Alternate forms")

The first two were performed by a program in Python.

There does not appear to be any mistake in your argument, so I can only conclude that in fact they are equal to each other. Note that the second series converged much more rapidly in my approximations than did the first. Hopefully someone can provide a proof.

  • "Hopefully someone can provide a proof." - the OP's derivation is a proof. What are you looking for from another proof? – user2357112 Apr 21 '14 at 09:34