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 $F^{\prime }(x)=f(x)$

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

$\underset{a}{\overset{y}{\int }}f(t)\mathrm{d}t=\underset{a}{\overset{x}{\int }}f(t)\mathrm{d}t+\underset{x}{\overset{y}{\int }}f(t)\mathrm{d}t$

Then

$F(y)-F(x)-f(x)(y-x)=\underset{x}{\overset{y}{\int }}f(t)\mathrm{d}t-\underset{x}{\overset{y}{\int }}f(x)\mathrm{d}t$

So that

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

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

$\frac{F(y)-F(x)}{y-x}-f(x)\le \underset{t\in [x,y]}{\sup }|f(t)-f(x)|$

If the function $f$ is right continuous at $x$ (i.e. $f(y)\to f(x)$ as $y\to x^{+}$), then $\underset{t\in [x,y]}{\sup }|f(t)-f(x)|\to 0$ 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

$F(x)=\underset{a}{\overset{x}{\int }}f(t)\mathrm{d}tx\in [a,b]$

is continuously differentiable on $[a,b]$. Moreover, $F^{\prime }(x)=f(x)$ 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, $F^{\prime }(x)=f(x)$ whenever $f$ is continuous at the point $x$. Therefore, the continuoity of $f$ on $[a,b]$ implies that $F^{\prime }(x)=f(x)$ 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 $F^{\prime }$ is integrable on $[a,b]$, then

$\underset{a}{\overset{x}{\int }}{F}^{\prime }(t)\mathrm{d}t=F(x)-F(a)$ for all $x\in [a,b]$

Proof. The case $x=a$ is trivial: $0=0$. Assume $x\in (a,b]$. Since $F^{\prime }$ 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 $P=\{x_0,x_1,\dots ,x_n\}$ of $[a,b]$. Let us choose samples $t_j\in [x_{j-1},x_j]$ for the Riemann sum ${𝒮}(F^{\prime },P,t_j)$ so that $F(x_j)-F(x_{j-1})=F^{\prime }(t_j)(x_j-x_{j-1})$ (this is possible due to the Mean Value Theorem). Then

$\begin{array}{rl}{𝒮}(F^{\prime },P,t_j)& =\sum _{j=1}^n{F}^{\prime }(t_j)(x_j-x_{j-1})\\ & =\sum _{j=1}^n(F(x_j)-F(x_{j-1}))\\ & =F(x_n)-F(x_0)\\ & =F(b)-F(a)\end{array}$

Since the sums ${𝒮}(F^{\prime },P,t_j)$ converge to $\underset{a}{\overset{b}{\int }}{F}^{\prime }(t)\mathrm{d}t$ as $\Vert P\Vert \to 0$, the theorem follows: $\underset{a}{\overset{b}{\int }}{F}^{\prime }(t)\mathrm{d}t=F(b)-F(a)$.

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 $F^{\prime }(x)=f(x)$ 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 $F(x)=\underset{a}{\overset{x}{\int }}f(t)\mathrm{d}t$ 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 $G^{\prime }=F^{\prime }$ on $[a,b]$. Hence $(G-F{)}^{\prime }=G^{\prime }-F^{\prime }=0$ 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 $(0,\mathrm{\infty })$ for $\alpha \ne 1$.
• $\int \frac{1}{x}\mathrm{d}x=\log x+C$ on $(0,\mathrm{\infty })$.
• $\int \sin x=-\cos x+C$
• $\int \cos x=\sin x+C$

Integration by Parts

Theorem. Suppose that functions $f$ and $g$ are differentiable on $[a,b]$ with derivatives $f^{\prime }$ and $g^{\prime }$ integrable on $[a,b]$. Then

$\underset{a}{\overset{b}{\int }}f(x)g^{\prime }(x)\mathrm{d}x=f(b)g(b)-f(a)g(a)-\underset{a}{\overset{b}{\int }}{f}^{\prime }(x)g(x)\mathrm{d}x$

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

$\begin{array}{rl}f(b)g(b)-f(a)g(a)& =\underset{a}{\overset{b}{\int }}(fg{)}^{\prime }(x)\mathrm{d}x\\ & =\underset{a}{\overset{b}{\int }}{f}^{\prime }(x)g(x)\mathrm{d}x+\underset{a}{\overset{b}{\int }}f(x)g^{\prime }(x)\mathrm{d}x\end{array}$

quod erat demonstrandum

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

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

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

Now the formula of integration by parts can be rewritten as

$\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}g(x)={f(x)g(x)|}_a^b-\underset{a}{\overset{b}{\int }}g(x)\mathrm{d}f(x)$

for definite integrals, and as

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

for indefinite integrals

Examples

• $\int \log x\mathrm{d}x=x\log x-\int x\mathrm{d}(\log x)=x\log x-\int \mathrm{d}x=x\log x-x+C$
• $\underset{0}{\overset{\frac{\pi }{2}}{\int }}x\sin x\mathrm{d}x={-x\cos x|}_0^{\frac{\pi }{2}}-\underset{0}{\overset{\frac{\pi }{2}}{\int }}(-\cos x)\mathrm{d}x=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos x\mathrm{d}x={\sin x|}_0^{\frac{\pi }{2}}=1$.

Change of the variable in an integral

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

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

$\underset{\varphi (a)}{\overset{\varphi (b)}{\int }}f(t)\mathrm{d}t=\underset{a}{\overset{b}{\int }}f(\varphi (x)){\varphi }^{\prime }(x)\mathrm{d}x=\underset{a}{\overset{b}{\int }}f(\varphi (x))\mathrm{d}\varphi (x)$

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

Proof. Let us define two functions

$\begin{array}{rl}F(u)& =\underset{\varphi (a)}{\overset{u}{\int }}f(t)\mathrm{d}tu\in \varphi ([a,b])\\ G(u)& =\underset{a}{\overset{x}{\int }}f(\varphi (s)){\varphi }^{\prime }(s)\mathrm{d}sx\in [a,b]\end{array}$

It follows from the Fundamental Theorem of Calculus that $F^{\prime }(u)=f(u)$ and $G^{\prime }(x)=f(\varphi (x)){\varphi }^{\prime }(x)$. By the Chain Rule, $(F\circ \varphi {)}^{\prime }(x)=F^{\prime }(\varphi (x)){\varphi }^{\prime }(x)=f(\varphi (x)){\varphi }^{\prime }(x)=G^{\prime }(x)$.

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

quod erat demonstrandum

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