ProofOfTheSecondFundamentalTheoremOfCalculus 2003-07-17 05:35:18 2006-09-01 06:26:25 Proof fundamental theorem of calculus 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here %We prove the following result. % %Let $f\colon[a,b]\to \mathbf R$ be a continuous function, let $c\in [a,b]$ and %consider the integral function %\[ % F(x)=\int_c^x f(t) \, dt. %\] %Then $F$ is differentiable and %\[ % F'(x)=f(x)\quad \forall x\in[a,b] %\] % %\emph{Proof}. Recall that a continuous function is Riemann integrable, so the integral \[ F(x) = \int_c^x f(t)\, dt \] is well defined. Consider the increment of $F$: \[ F(x+h)-F(x) = \int_c^{x+h} f(t)\, dt - \int_c^x f(t)\, dt = \int_x^{x+h} f(t)\, dt \] (we have used the linearity of the integral with respect to the function and the additivity with respect to the domain). Now let $M$ be the maximum of $f$ on $[x,x+h]$ and $m$ be the minimum. Clearly we have \[ m h \le \int_x^{x+h} f(t)\, dt \le M h \] (this is due to the monotonicity of the integral with respect to the integrand) which can be written as \[ \frac{F(x+h)-F(x)}{h} = \frac{\int_{x}^{x+h}f(t)\, dt}{h} \in [m,M] \] Since $f$ is continuous, by the mean-value theorem, there exists $\xi_h\in [x,x+h]$ such that $f(\xi_h) = \frac{F(x+h)-F(x)}{h}$ so that \[ F'(x)= \lim_{h\to 0} \frac{ F(x+h)-F(x)}{h} = \lim_{h\to 0} f(\xi_h) = f(x) \] since $\xi_h\to x$ as $h\to 0$. This proves the first part of the theorem. For the second part suppose that $G$ is any antiderivative of $f$, i.e.\ $G'=f$. Let $F$ be the integral function \[ F(x)= \int_a^x f(t) \, dt. \] We have just proven that $F'=f$. So $F'(x)=G'(x)$ for all $x\in [a,b]$ or, which is the same, $(G-F)'=0$. This means that $G-F$ is constant on $[a,b]$ that is, there exists $k$ such that $G(x)=F(x)+k$. Since $F(a)=0$ we have $G(a)=k$ and hence $G(x)=F(x)+G(a)$ for all $x\in[a,b]$. Thus \[ \int_a^b f(t)\, dt = F(b) = G(b) - G(a). \]