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# 7.5: The Fundamental Theorem of Calculus

• • Dan Sloughter
• Professor (Mathematics) at Furman University
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##### Theorem $$\PageIndex{1}$$

(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.

##### Proposition $$\PageIndex{2}$$

(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

##### Proposition $$\PageIndex{3}$$

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.

##### Theorem $$\PageIndex{4}$$

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.

##### Proposition $$\PageIndex{5}$$

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.

##### Example $$\PageIndex{1}$$

If

$g(x)=\int_{0}^{x} \sqrt{1+t^{4}} d t,$

then $$g^{\prime}(x)=\sqrt{1+x^{4}}$$.

##### Proposition $$\PageIndex{6}$$

(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.

##### Exercise $$\PageIndex{1}$$

Evaluate

$\int_{0}^{1} u \sqrt{u+1} d u$

using (a) integration by parts and (b) substitution.

##### Exercise $$\PageIndex{2}$$

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 $$f: \mathbb{R} \rightarrow \mathbb{R}$$,

$\int_{0}^{1} f(\varphi(x)) \varphi^{\prime}(x) d x=0.$

##### Theorem $$\PageIndex{7}$$

(Integral Mean Value Theorem)

If $$f$$ is continuous on $$[a, b],$$ then there exists $$c \in[a, b]$$ such that

$\int_{a}^{b} f=f(c)(b-a).$

##### Exercise $$\PageIndex{3}$$

Prove the Integral Mean Value Theorem.

##### Theorem $$\PageIndex{8}$$

(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

$\int_{a}^{b} f g=f(c) \int_{a}^{b} g.$

##### Exercise $$\PageIndex{4}$$

Prove the Generalized Integral Mean Value Theorem.