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I am having some trouble with this question out of Apostol's 'Calculus' Volume 1, Section 5.5 Question 19 (Page 209). This is not a homework question, I am doing this for personal study. The question is as follows:

Given a function $g$, continuous everywhere, such that $g(1) = 5$ and $\int_0^1g(t) dt = 2$. Let $f(x) = 0.5 \int_0^x (x-t)^2g(t) dt$. Prove $$ f'(x) = x\int_0^x g(t) dt - \int_0^x t g(t) dt $$ then compute $f''(1)$ and $f'''(1)$.

My thinking so far has been as follows: $$ f(x) = 0.5 \int_0^x (x-t)^2g(t) dt $$ which implies $$ f(x) = 0.5 \int_0^x (x^2-2tx +t^2)g(t) dt $$ which implies $$ 2f(x) = x^2 \int_0^x g(t) dt - 2x \int_0^x t g(t) dt + \int_0^xt^2 g(t) dt $$ Differentiating, we get: $$ 2f'(x) = 2x \int_0^x g(t) dt - 2 \int_0^x t g(t) dt $$ which implies $$ f'(x) = x\int_0^x g(t) dt - \int_0^x t g(t) dt $$ The problems I am having with this are as follows:

  1. I am not entirely convinced of the rigour of the argument defined above;
  2. I am stumped as to how to proceed with the computation of $f''(1)$ and $f'''(1)$. I think that the first fundamental theorem of calculus needs to be applied to the formula for $f'(x)$ , but for me this results in $f''(x) = 2 - xg(x)$ which would mean $f''(1) = -3$ (NB: The answers in the book are $f''(1) = 2$ and $f'''(1) = 5$). Clearly I'm missing the point somewhere.

Any suggestions and advice would be greatly appreciated.

Paramanand Singh
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emjay
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    When you wrote "Differentiating, we get", you computed the derivative of the first term wrong: it's a product of two things, $x^2$ and an integral. You have to apply the product rule, and the FTC shows the derivative of the integral gets you a $g(x)$. (Similar stuff applies to the other terms as well.) – John Hughes Mar 10 '14 at 03:50
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    Do you really mean an upper limit of $x$ in $\int_0^x g(t)\mathrm{d}t = 2$? – Eric Towers Mar 10 '14 at 03:52
  • Good point, Eric: the integral goes from 0 to 1: My bad :) – emjay Mar 10 '14 at 03:57
  • John: good point re: the product rule, however when I apply it I still get the same result: there are a bunch of $x^2g(x)$ that occur in the computation, but they cancel each other out as well. – emjay Mar 10 '14 at 04:07

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