7.5: The Fundamental Theorem of Calculus
- Page ID
- 22682
(Fundamental Theorem of Calculus)
Suppose \(f\) is integrable on \([a, b] .\) If \(F\) is continuous on \([a, b]\) and differentiable on \((a, b)\) with \(F^{\prime}(x)=f(x)\) for all \(x \in(a, b),\) then
\[\int_{a}^{b} f=F(b)-F(a).\]
- Proof
-
Given \(\epsilon>0,\) let \(P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\}\) be a partition of \([a, b]\) for which
\[U(f, P)-L(f, P)<\epsilon .\]
For \(i=1,2, \ldots, n,\) let \(t_{i} \in\left(x_{i-1}, x_{i}\right)\) be points for which
\[F\left(x_{i}\right)-F\left(x_{i-1}\right)=f\left(t_{i}\right)\left(x_{i}-x_{i-1}\right).\]
Then
\[\sum_{i=1}^{n} f\left(t_{i}\right)\left(x_{i}-x_{i-1}\right)=\sum_{i=1}^{n}\left(F\left(x_{i}\right)-F\left(x_{i-1}\right)\right)=F(b)-F(a).\]
But
\[L(f, P) \leq \sum_{i=1}^{n} f\left(t_{i}\right)\left(x_{i}-x_{i-1}\right) \leq U(f, P),\]
so
\[\left|F(b)-F(a)-\int_{a}^{b} f\right|<\epsilon .\]
Since \(\epsilon\) was arbitrary, we conclude that
\[\int_{a}^{b} f=F(b)-F(a).\]
Q.E.D.
(Integration by parts)
Suppose \(f\) and \(g\) are integrable
on \([a, b] .\) If \(F\) and \(G\) are continuous on \([a, b]\) and differentiable on \((a, b)\) with \(F^{\prime}(x)=f(x)\) and \(G^{\prime}(x)=g(x)\) for all \(x \in(a, b),\) then
\[\int_{a}^{b} F(x) g(x) d x=F(b) G(b)-F(a) G(a)-\int_{a}^{b} f(x) G(x) d x.\]
- Proof
-
By the Fundamental Theorem of Calculus,
\[\int_{a}^{b}(F(x) g(x)+f(x) G(x)) d x=F(b) G(b)-F(a) G(a).\]
Q.E.D.
7.5.1 The other Fundamental Theorem of Calculus
Suppose \(f\) is integrable on \([a, b]\) and \(F:[a, b] \rightarrow \mathbb{R}\) is defined by
\[F(x)=\int_{a}^{x} f(t) d t.\]
Then \(F\) is uniformly continuous on \([a, b] .\)
- Proof
-
Let \(\epsilon>0\) be given and let \(M>0\) be such that \(|f(x)| \leq M\) for all \(x \in[a, b] .\) Then for any \(x, y \in[a, b]\) with \(x<y\) and \(y-x<\frac{e}{M}\),
\[|F(y)-F(x)|=\left|\int_{x}^{y} f(t) d t\right| \leq M(y-x)<\epsilon .\]
Hence \(F\) is uniformly continuous on \([a, b] . \quad\) Q.E.D.
The following theorem is often considered to be part of the Fundamental Theorem of Calculus.
Suppose \(f\) is integrable on \([a, b]\) and continuous at \(u \in(a, b) .\) If \(F:[a, b] \rightarrow \mathbb{R}\) is defined by
\[F(x)=\int_{a}^{x} f(t) d t,\]
then \(F\) is differentiable at \(u\) and \(F^{\prime}(u)=f(u)\).
- Proof
-
Let \(\epsilon>0\) be given and choose \(\delta>0\) such that \(|f(x)-f(u)|<\epsilon\) whenever \(|x-u|<\delta .\) Then if \(0<h<\delta,\) we have
\[\begin{aligned}\left|\frac{F(u+h)-F(u)}{h}-f(u)\right| &=\left|\frac{1}{h} \int_{u}^{u+h} f(t) d t-f(u)\right| \\ &=\left|\frac{1}{h} \int_{u}^{u+h}(f(t)-f(u)) d t\right| \\ &<\epsilon . \end{aligned}\]
If \(-\delta<h<0,\) then
\[\begin{aligned}\left|\frac{F(u+h)-F(u)}{h}-f(u)\right| &=\left|-\frac{1}{h} \int_{u+h}^{u} f(t) d t-f(u)\right| \\ &=\left|\frac{1}{h} \int_{u+h}^{u} f(t) d t+f(u)\right| \\ &=\left|\frac{1}{h} \int_{u+h}^{u} f(t) d t-\frac{1}{h} \int_{u+h}^{u} f(u) d t\right| \\ &=\left|\frac{1}{h} \int_{u+h}^{u}(f(t)-f(u)) d t\right| \\ &<\epsilon . \end{aligned}\]
Hence
\[F^{\prime}(u)=\lim _{h \rightarrow 0} \frac{F(u+h)-F(u)}{h}=f(u) .\]
Q.E.D.
If \(a<b\) and \(f\) is continuous on \([a, b],\) then there exists a function \(F:[a, b] \rightarrow \mathbb{R}\) which is continuous on \([a, b]\) with \(F^{\prime}(x)=f(x)\) for all \(x \in(a, b) .\)
- Proof
-
Let
\[F(x)=\int_{a}^{x} f(t) d t .\]
Q.E.D.
If
\[g(x)=\int_{0}^{x} \sqrt{1+t^{4}} d t,\]
then \(g^{\prime}(x)=\sqrt{1+x^{4}}\).
(Integration by substitution)
Suppose \(I\) is an open interval, \(\varphi: I \rightarrow \mathbb{R}, a<b,[a, b] \subset I,\) and \(\varphi^{\prime}\) is continuous on \([a, b] .\) If \(f: \varphi([a, b]) \rightarrow \mathbb{R}\) is continuous, then
\[\int_{\varphi(a)}^{\varphi(b)} f(u) d u=\int_{a}^{b} f(\varphi(x)) \varphi^{\prime}(x) d x.\]
- Proof
-
If \(m\) and \(M\) are the minimum and maximum values, respectively, of \(\varphi\) on \([a, b],\) then \(\varphi([a, b])=[m, M] .\) If \(m=M,\) then \(\varphi(x)=m\) for all \(x \in[a, b],\) and both sides of \((7.5 .17)\) are \(0 .\) So we may assume \(m<M .\) Let \(F\) be a function which is continuous on \([m, M]\) with \(F^{\prime}(u)=f(u)\) for every \(u \in(m, M) .\) Let \(g=F \circ \varphi .\) Then
\[g^{\prime}(x)=F^{\prime}(\varphi(x)) \varphi^{\prime}(x)=f(\varphi(x)) \varphi^{\prime}(x).\]
So if \(\varphi(a) \leq \varphi(b)\),
\[\begin{aligned} \int_{a}^{b} f(\varphi(x)) \varphi^{\prime}(x) d x &=g(b)-g(a) \\ &=F(\varphi(b))-F(\varphi(a)) \\ &=\int_{\varphi(a)}^{\varphi(b)} f(u) d u. \end{aligned}\]
If \(\varphi(a)>\varphi(b),\) then
\[\begin{aligned} \int_{a}^{b} f(\varphi(x)) \varphi^{\prime}(x) d x &=g(b)-g(a) \\ &=F(\varphi(b))-F(\varphi(a)) \\ &=-(F(\varphi(a))-F(\varphi(b))) \\ &=-\int_{\varphi(b)}^{\varphi(a)} f(u) d u \\ &=\int_{\varphi(a)}^{\varphi(b)} f(u) d u. \end{aligned}\]
Q.E.D.
Evaluate
\[\int_{0}^{1} u \sqrt{u+1} d u\]
using (a) integration by parts and (b) substitution.
Suppose \(\varphi: \mathbb{R} \rightarrow \mathbb{R}\) is differentiable on \(\mathbb{R}\) and periodic with period \(1(\text { that is, } \varphi(x+1)=\varphi(x) \text { for every } x \in \mathbb{R}) .\) Show that for any continuous function \(f: \mathbb{R} \rightarrow \mathbb{R}\),
\[\int_{0}^{1} f(\varphi(x)) \varphi^{\prime}(x) d x=0.\]
(Integral Mean Value Theorem)
If \(f\) is continuous on \([a, b],\) then there exists \(c \in[a, b]\) such that
\[\int_{a}^{b} f=f(c)(b-a).\]
Prove the Integral Mean Value Theorem.
(Generalized Integral Mean Value Theorem)
If \(f\) and \(g\) are continuous on \([a, b]\) and \(g(x)>0\) for all \(x \in[a, b],\) then there exists \(c \in[a, b]\) such that
\[\int_{a}^{b} f g=f(c) \int_{a}^{b} g.\]
Prove the Generalized Integral Mean Value Theorem.