MATH 409 Lecture 20

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Lecture Slides

Exam next Thursday

Homework problems just require definition of improper Riemann definition; look at hint for problem 5.3.12

Integral with Variable Limit

(See Integral with a variable limit→)

Theorem. If $f$ is continuous at a point $x\in [a,b]$, then $F$ is differentiable at $x$ and $F'(x)=f(x)$

Proof. For any $x,y\in [a,b]$ such that $x<y$, we have

$\int _{a}^{y}f(t)\,\mathrm {d} t=\int _{a}^{x}f(t)\,\mathrm {d} t+\int _{x}^{y}f(t)\,\mathrm {d} t\,\!$

Then

$F(y)-F(x)-f(x)(y-x)=\int _{x}^{y}f(t)\,\mathrm {d} t-\int _{x}^{y}f(x)\,\mathrm {d} t$

So that

$\left|F(y)-F(x)-f(x)(y-x)\right|=\left|\int _{x}^{y}\left(f(t)-f(x)\right)\,\mathrm {d} t\right|\leq \int _{x}^{y}\left|f(t)-f(x)\right|\,\mathrm {d} t\leq \sup _{t\in [x,y]}\left|f(t)-f(x)\right|\,(y-x)$

Dividing both sides by $(y-x)$ gives

${\frac {F(y)-F(x)}{y-x}}-f(x)\leq \sup _{t\in [x,y]}\left|f(t)-f(x)\right|$

If the function $f$ is right continuous at $x$ (i.e. $f(y)\to f(x)$ as $y\to x^{+}$), then $\sup _{t\in [x,y]}\left|f(t)-f(x)\right|\to 0$ as $y\to x^{+}$. It follows that $f(x)$ is the right-hand derivative of $F$ at $x$. Likewise, one can prove that left continuity of f at x implies $f(x)$ is the left derivative at $x$.

quod erat demonstrandum

The Fundamental Theorem of Calculus

Theorem. [Part I]. If a function $f$ is continuous on an interval $[a,b]$, then the function

$F(x)=\int _{a}^{x}f(t)\,\mathrm {d} t\qquad x\in [a,b]$

is continuously differentiable on $[a,b]$. Moreover, $F'(x)=f(x)$ for all $x\in [a,b]$.

Proof. Since $f$ is continuous, it is also integrable on $[a,b]$. As proved earlier, integrability of $f$ implies that the function $F$ is well-defined on $[a,b]$. Moreover, $F'(x)=f(x)$ whenever $f$ is continuous at the point $x$. Therefore, the continuoity of $f$ on $[a,b]$ implies that $F'(x)=f(x)$ for all $x\in [a,b]$. In particular, $F$ is continuously differentiable on $[a,b]$.

quod erat demonstrandum

Theorem. [Part II]. If a function $F$ is differentiable on $[a,b]$ and the derivative $F'$ is integrable on $[a,b]$, then

$\int _{a}^{x}F'(t)\,\mathrm {d} t=F(x)-F(a)$ for all $x\in [a,b]$

Proof. The case $x=a$ is trivial: $0=0$. Assume $x\in (a,b]$. Since $F'$ is integrable on $[a,b]$, it is also integrable on any subinterval $[a,x]$, $x\in (a,b)$. Therefore, it is no loss to assume that $x=b$.

Consider an arbitrary partition $P=\left\{x_{0},x_{1},\ldots ,x_{n}\right\}$ of $[a,b]$. Let us choose samples $t_{j}\in [x_{j-1},x_{j}]$ for the Riemann sum ${\mathcal {S}}(F',P,t_{j})$ so that $F(x_{j})-F(x_{j-1})=F'(t_{j})\,(x_{j}-x_{j-1})$ (this is possible due to the Mean Value Theorem). Then

${\begin{aligned}{\mathcal {S}}(F',P,t_{j})&=\sum _{j=1}^{n}F'(t_{j})\,(x_{j}-x_{j-1})\\&=\sum _{j=1}^{n}\left(F(x_{j})-F(x_{j-1})\right)\\&=F(x_{n})-F(x_{0})\\&=F(b)-F(a)\end{aligned}}$

Since the sums ${\mathcal {S}}(F',P,t_{j})$ converge to $\int _{a}^{b}F'(t)\,\mathrm {d} t$ as $\left\|P\right\|\to 0$, the theorem follows: $\int _{a}^{b}F'(t)\,\mathrm {d} t=F(b)-F(a)$.

quod erat demonstrandum

The Indefinite Integral

Given a function $f:[a,b]\to \mathbb {R}$, a function $F:[a,b]\to \mathbb {R}$ is called the indefinite integral (or antiderivative, primitive integral, or the primitive) of $f$ if $F'(x)=f(x)$ for all $x\in [a,b]$. Notation:

$\int f(x)\,\mathrm {d} x$

If the function $f$ is continuous on $[a,b]$, then the function $F(x)=\int _{a}^{x}f(t)\,\mathrm {d} t$ for $x\in [a,b]$ is an indefinite integral of $f$ due to the Fundamental Theorem of Calculus.

Suppose $F$ is an antiderivative of $f$. If $G$ is another antiderivative of $f$, then $G'=F'$ on $[a,b]$. Hence $(G-F)'=G'-F'=0$ on $[a,b]$. It follows that $G-F$ is a constant function. Conversely, for any constant $C$, the function $G(x)=F(x)+C$ is also an antiderivative of $f$.

$\int f(x)\,\mathrm {d} x=F(x)+C$

Where $C$ is an arbitrary constant.

Examples

• $\int x^{\alpha }\,\mathrm {d} x={\frac {x^{\alpha +1}}{\alpha +1}}+C$ on $(0,\infty )$ for $\alpha \neq 1$.
• $\int {\frac {1}{x}}\,\mathrm {d} x=\log {x}+C$ on $(0,\infty )$.
• $\int \sin {x}=-\cos {x}+C$
• $\int \cos {x}=\sin {x}+C$

Integration by Parts

Theorem. Suppose that functions $f$ and $g$ are differentiable on $[a,b]$ with derivatives $f'$ and $g'$ integrable on $[a,b]$. Then

$\int _{a}^{b}f(x)\,g'(x)\,\mathrm {d} x=f(b)\,g(b)-f(a)\,g(a)-\int _{a}^{b}f'(x)\,g(x)\,\mathrm {d} x$

Proof. By the product rule, $(f\,g)'=f'\,g+f\,g'$ on $[a,b]$. Since $f,f',g,g'$ are integrable on $[a,b]$ by the hypothesis, so are the products $f'\,g$ and $f\,g'$. Then $(f\,g)'$ is integrable on $[a,b]$ as well. By the Fundamental Theorem of Calculus,

${\begin{aligned}f(b)\,g(b)-f(a)\,g(a)&=\int _{a}^{b}(f\,g)'(x)\,\mathrm {d} x\\&=\int _{a}^{b}f'(x)\,g(x)\,\mathrm {d} x+\int _{a}^{b}f(x)\,g'(x)\,\mathrm {d} x\end{aligned}}$

quod erat demonstrandum

Corollary. Suppose that functions $f,g$ are continuously differentiable on $[a,b]$. Then

$\int f(x)\,g'(x)\,\mathrm {d} x=f(x)\,g(x)-\int f'(x)\,g(x)\,\mathrm {d} x$ on $[a,b]$

To simplify notation, it is convenient to use the Leibniz differential $\mathrm {d} f$ of a function $f$ defined by $\mathrm {d} f(x)=f'(x)\,\mathrm {d} x={\frac {\mathrm {d} f}{\mathrm {d} x}}\,\mathrm {d} x$. Another convenient notation is $\left.f(x)\right|_{x=a}^{b}$ or simply $\left.f(x)\right|_{a}^{b}$, which denotes the difference $f(b)-f(a)$.

Now the formula of integration by parts can be rewritten as

$\int _{a}^{b}f(x)\,\mathrm {d} g(x)=\left.f(x)\,g(x)\right|_{a}^{b}-\int _{a}^{b}g(x)\,\mathrm {d} f(x)$

for definite integrals, and as

$\int f\,\mathrm {d} g=f\,g-\int g\,\mathrm {d} f$

for indefinite integrals

Examples

• $\int \log {x}\,\mathrm {d} x=x\,\log {x}-\int x\,\mathrm {d} (\log {x})=x\,\log {x}-\int \mathrm {d} x=x\,\log {x}-x+C$
• $\int _{0}^{\frac {\pi }{2}}x\,\sin {x}\,\mathrm {d} x=\left.-x\,\cos {x}\right|_{0}^{\frac {\pi }{2}}-\int _{0}^{\frac {\pi }{2}}(-\cos {x})\,\mathrm {d} x=\int _{0}^{\frac {\pi }{2}}\cos {x}\,\mathrm {d} x=\left.\sin {x}\right|_{0}^{\frac {\pi }{2}}=1$.

Change of the variable in an integral

(commonly called u-substitution as $\phi$ is often replaced by $u$ in practice.)

Theorem. if $\phi$ is continuously differentiable on a closed, nondegenerate interval $[a,b]$ and $f$ is continuous on $\phi ([a,b])$, then

$\int _{\phi (a)}^{\phi (b)}f(t)\,\mathrm {d} t=\int _{a}^{b}f(\phi (x))\,\phi '(x)\,\mathrm {d} x=\int _{a}^{b}f(\phi (x))\,\mathrm {d} \phi (x)$

Be aware that $t=\phi (x)$ is a proper change of variable only if the function $\phi$ is strictly monotone. However, the theorem holds even without this assumption.

Proof. Let us define two functions

${\begin{aligned}F(u)&=\int _{\phi (a)}^{u}f(t)\,\mathrm {d} t\qquad u\in \phi ([a,b])\\G(u)&=\int _{a}^{x}f(\phi (s))\,\phi '(s)\,\mathrm {d} s\qquad x\in [a,b]\end{aligned}}$

It follows from the Fundamental Theorem of Calculus that $F'(u)=f(u)$ and $G'(x)=f(\phi (x))\,\phi '(x)$. By the Chain Rule, $(F\circ \phi )'(x)=F'(\phi (x))\,\phi '(x)=f(\phi (x))\,\phi '(x)=G'(x)$.

Therefore, $\left(F(\phi (x))-G(x)\right)'=0$ for all $x\in [a,b]$. It follows that the function $F(\phi (x))-G(x)$ is constant on $[a,b]$. In particular, $F(\phi (b))-G(b)=F(\phi (a))-G(a)=0-0=0$. Hence $F(\phi (b))=G(b)$.

quod erat demonstrandum

Note: It is possible that $\phi (a)\geq \phi (b)$. To make sense of this case, we set $\int _{c}^{d}f(t)\,\mathrm {d} t{=}-\int _{d}^{c}f(t)\,\mathrm {d} t$ if $c>d$. Also we set the integral to be 0 if $c=d$.