the fundamental theorem of calculus

Theorem(the Fundamental Theorem of Calculus, Part 1) Let $f$ be a continuous real-valued function defined on a closed interval $[a,b]$. Let $F$ be the function defined by \(F(x)=\int_{a}^x f(t)dt\) Then, $F’=f$ on $(a,b)$.

proof. We will prove that the derivative of $F$ equals $f$ by calculating

For the right limit, we notice that

Since $f$ is a continuous function on $[x,x+h]$, there exist $u,v\in[x,x+h]$ such that $f(u)\leq f(t)\leq f(v)$ for all $t\in[x,x+h]$. By the definition of definite integrals, we have

which implies

Now, we let $h$ approach $0$, and so both $u$ and $v$ approach $x$. Hence, we conclude that $\displaystyle\lim_{h\to 0^+}\frac{F(x+h)-F(x)}{h} = f(x)$. A similar argument will be carried out for the left limit. Therefore, we can conclude that $F’=f$.

Theorem(the Fundamental Theorem of Calculus, Part 2) Let $f$ be a real-valued function on a closed interval $[a,b]$ and $F$ a continuous function on $[a,b]$ which is an antiderivative of $f$ in $(a,b)$: \(F'=f.\) If $f$ is Riemann integrable on $[a,b]$, then \(\int_a^bf(t)dt = F(b)-F(a)\)

proof(if we assume f is continuous) If $f$ is continuous, then by part 1 of the fundamental theorem of Calculus we know $G(x)=\int_0^xf(t)dt$ is another antiderivative of $f$. Thus, by a corollary of mean value theorem, $F-G$ is a constant, i.e. $F=G+c$ for some constant $c$. Note that $G(a)=0$ and $G(b)=\int_a^bf(t)dt$, so \(F(b)-F(a) = G(b)-G(a) = G(b)=\int_a^bf(t)dt\)

proof Since $F$ is continuous on $[a,b]$, we can apply the mean value theorem to $F$ on any subinterval in $[a,b]$, i.e. \(F'(c^\*)(x_2-x_1) = F(x_2)-F(x_1)\) for $[x_1,x_2]\subseteq [a,b]$ and for some $c^*\in (x_1,x_2)$. Given a partition $a\leq x_0<x_1<x_2<\cdots<x_n=b$, we can rewrite $F(b)-F(a)$ as \(\begin{align*} F(b)-F(a) &= F(x_n)-[-F(x_{n-1})+F(x_{n-1})]+\cdots [-F(x_1)+F(x_1)]-F(x_0)\\\\ &= [F(x_n)-F(x_{n-1})]+[F(x_{n-1})-F(x_{n-2})]+\cdots [F(x_1)-F(x_0)] \end{align*}\) Applying the mean value theorem to each subinterval $[x_{i},x_{i+1}]$, we further rewrite \(F(b)-F(a) = \sum_{i=0}^{n-1}F(x_{i+1})-F(x_{i}) = \sum_{i=0}^{n-1}F'(c_i^\*)(x_{i+1}-x_i)\) for some $c_i^*\in(x_{i},x_{i+1})$. The sum is a Riemann sum of $f$. Since we assume $f$ is integrable, as $\displaystyle\max_{i=0}^{n-1}\{x_{i+1}-x_i\}\to 0$, the Riemann sum will approach $\int_{a}^bf(t)dt$. Therefore, we can conclude that $\int_{a}^bf(t)dt=F(b)-F(a)$.

考型 Give you a function in terms of definite integral, and ask you to derive the function.

Example 1. Calculuate $\displaystyle \frac{d}{dx}\int_{g(x)}^{f(x)}h(t)dt$.

Sol)

STEP 1. Suppose $H(t)=\int h(t)dt$.

STEP 2. $\displaystyle\int_{g(x)}^{f(x)}h(t)dt = H(f(x))-H(g(x))$

STEP 3. $\displaystyle \frac{d}{dx}\int_{g(x)}^{f(x)}h(t)dt = (H(f(x))-H(g(x))) = H’(f(x))f’(x)-H’(g(x))g’(x)= h(f(x))f’(x)-h(g(x))g’(x).$