MATH 409 Lecture 21

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End Exam 2 content

Lecture Slides

Exam 2 Review

Topics

Derivatives

• Derivative of a function
• Differentiability Theorems
• derivative of inverse function
• Mean Value Theorem
  • Rolle's Theorem
  • Generalized Mean Value Theorem
• Taylor's Formula
• l'Hôpital's rule

Integrals

• Definitions
  • Darboux Sums
  • Riemann Sums
  • Riemann Integral
• Properties of Integrals
• Fundamental Theorem of Calculus (both parts)
• Integration by Parts
• Change of variable in an integral

Chapters 4.1–4.5, 5.1–5.3

Differentiability Theorems

• sum / difference rules
• product rule
• quotient rule
• chain rule
• differentiability implies continuity
• Rolle's Theorem: If a function $f$ is continuous on a closed interval $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then $f^{\prime }(c)=0$ for some $c\in (a,b)$
• Mean Value Theorem: If a function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in (a,b)$ such than $f(b)-f(a)=f^{\prime }(c)(b-a)$.
• $f$ is increasing on $[a,b]$ if and only if $f^{\prime }\ge 0$ on $(a,b)$
• $f$ is decreasing on $[a,b]$ if and only if $f^{\prime }\le 0$ on $(a,b)$
• $f$ is constant on $[a,b]$ if and only if $f^{\prime }=0$ on $(a,b)$

Properties of Integrals

• Linearity:
  • $\underset{a}{\overset{b}{\int }}(f(x)+g(x))\mathrm{d}x=\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x+\underset{a}{\overset{b}{\int }}g(x)\mathrm{d}x$
  • $\underset{a}{\overset{b}{\int }}(\alpha f(x))\mathrm{d}x=\alpha \underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x$
• Subinterval property: $\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x=\underset{a}{\overset{c}{\int }}f(x)\mathrm{d}x+\underset{c}{\overset{b}{\int }}f(x)\mathrm{d}x$
• Comparison Theorem: If $f(x)\le g(x)$ for all $x\in [a,b]$, then $\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x\le \underset{a}{\overset{b}{\int }}g(x)\mathrm{d}x$

Fundamental Theorem of Calculus

• Part 1: $F(x)=\underset{a}{\overset{x}{\int }}f(t)\mathrm{d}t$, $x\in [a,b]$ is continuously differentiable on $[a,b]$ and $F^{\prime }(x)=f(x)$ for all $x\in [a,b]$
• Part 2: 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 $x\in [a,b]$.

Sample Problems

Problem 1: Prove the Chain Rule

Theorem. If a function $f$...

Proved in class (except for one small part)

Problem 2: Find the Limits

$\underset{x\to 0}{\lim }{(1+x)}^{\frac{1}{x}}$

The function $f(x)={(1+x)}^{\frac{1}{x}}$ is well-defined on $(-1,\mathrm{\infty })$ except at $0$. Since $f(x)>0$ for all $x>1$, a function $g(x)=\log f(x)$ is well-defined on $(-1,\mathrm{\infty })$ except at $0$ as well. For any $x>1$, we have $g(x)=\log {(1+x)}^{\frac{1}{x}}=x^{-1}\log (1+x)$. Hence $g=\frac{h_1}{h_2}$, where $h_1(x)=\log (1+x)$ and $h_2(x)=x$ are continuously differentiable on $(-1,\mathrm{\infty })$. Since $h_1(0)=h_2(0)=0$, it follows that $\underset{x\to 0}{\lim }{h}_1(x)=\underset{x\to 0}{\lim }{h}_2(x)=0$. By l'Hôpital's rule, ${h_1}^{\prime }(0)=1$ and ${h_2}^{\prime }(0)=1$, so $\underset{x\to 0}{\lim }g(x)=\underset{x\to 0}{\lim }\frac{h_1(x)}{h_2(x)}=\frac{1}{1}=1$

Since $f={\mathrm{e}}^{g(x)}$, a composition of $g$ with a continuous function, it follows that $\underset{x\to 0}{\lim }f(x)={\mathrm{e}}^1=\mathrm{e}$

$\underset{x\to +\mathrm{\infty }}{\lim }{(1+x)}^{\frac{1}{x}}$ Similar to above, $\underset{x\to +\mathrm{\infty }}{\lim }{h}_1(x)=\underset{x\to +\mathrm{\infty }}{\lim }{h}_2(x)=+\mathrm{\infty }$. At the same time, ${h_1}^{\prime }(x)\to 0$ as $x\to +\mathrm{\infty }$, while ${h_2}^{\prime }$ is identically $1$. Using l'Hôpital's Rule, we obtain $\underset{x\to +\mathrm{\infty }}{\lim }g(x)=\underset{x\to +\mathrm{\infty }}{\lim }\frac{h_1(x)}{h_2(x)}=\frac{0}{1}=0$

Since $f={\mathrm{e}}^{g(x)}$, a composition of $g$ with a continuous function, it follows that $\underset{x\to 0}{\lim }f(x)={\mathrm{e}}^0=1$

Problem 3: Limit of a sequence

Find the limit of a sequence $x_n=\frac{1^k+2^k+\dots +n^k}{n^{k+1}}$ for $n\in \mathbb{\mathbb{N}}$, where $k$ is a natural number.

The general element of the sequence can be represented as

$x_n=\frac{1^k+2^k+\dots +n^k}{n^k}\cdot \frac{1}{n}={\left(\frac{1}{n}\right)}^k\frac{1}{n}+{\left(\frac{2}{n}\right)}^k\frac{1}{n}+\dots +{\left(\frac{n}{n}\right)}^k\frac{1}{n}$

which shows that $x_n$ is a Riemann sum of the function $f(x)=x^k$ on the interval $[0,1]$ that corresponds to the partition $P_n=\{0,\frac{1}{n},\frac{2}{n},\dots ,\frac{n-1}{n},1\}$ and samples $t_j=\frac{j}{n}$, $j=1,2,\dots ,n$. The norm of the partition is $\Vert P_n\Vert =\frac{1}{n}$. Since $\Vert P_n\Vert \to 0$ as $n\to \mathrm{\infty }$ and the function $f$ is integrable on $[0,1]$, the Riemann sums $x_n$ converge to the integral:

$\underset{n\to \mathrm{\infty }}{\lim }{x}_n=\underset{0}{\overset{1}{\int }}{x}^k\mathrm{d}x={\frac{x^{k+1}}{k+1}|}_{x=0}^1=\frac{1}{k+1}$

Problem 4: Find/evaluate indefinite/definite integrals

Subproblem 1

$\int \frac{x^2}{1-x}\mathrm{d}x$

A standard way to evaluate this type of function is to split it into the sum of a polynomial and a simple fraction:

$\frac{x^2}{1-x}=\frac{x^2-1+1}{1-x}=\frac{x^2-1}{1-x}+\frac{1}{1-x}=-x-1-\frac{1}{x-1}$

Since the domain of the function is $(-\mathrm{\infty },1)\cup (1\mathrm{\infty })$, the indefinite integral has differetn representations on the intervals $(-\mathrm{\infty },1)$ and $(1,\mathrm{\infty })$:

$\int \frac{x^2}{1-x}=\{\begin{array}{ll}-\frac{x^2}{2}-x-\log (1-x)+C_1& x<1\\ -\frac{x^2}{2}-x-\log (x-1)+C_2& x>1\end{array}$

Subproblem 2

$\underset{0}{\overset{\pi }{\int }}{\sin }^2(2x)\mathrm{d}x$

To integrate this function, we use a trigonometric formula $1-\cos 2\alpha =2{\sin }^{}\alpha$ and a new variable $u=4x$:

$\underset{0}{\overset{\pi }{\int }}{\sin }^2(2x)\mathrm{d}x=\underset{0}{\overset{\pi }{\int }}\frac{1-\cos (4x)}{2}\mathrm{d}x=\underset{0}{\overset{\pi }{\int }}\frac{1-\cos (4x)}{8}\mathrm{d}(4x)=\underset{0}{\overset{4\pi }{\int }}\frac{1-\cos u}{8}\mathrm{d}u={\frac{u-\sin u}{8}|}_{u=0}^{4\pi }=\frac{\pi }{2}$

Subproblem 3

$\int {\log }^{3}x\mathrm{d}x$

To find this indefinite integral, we integrate by parts:

$\begin{array}{rl}\int {\log }^{3}x\mathrm{d}x& =x{\log }^{3}x-\int x\mathrm{d}({\log }^{3}x)\\ & =x{\log }^{3}x-\int x\left({\log }^{3}x\right)\mathrm{d}x\\ & =x{\log }^{3}x-\int 3{\log }^{2}x\mathrm{d}x\\ & =x{\log }^{3}x-3x{\log }^{2}x+\int x\mathrm{d}\left(3{\log }^{2}x\right)\\ & =x{\log }^{3}x-3x{\log }^{2}x+\int 6\log x\mathrm{d}x\\ & =x{\log }^{3}x-3x{\log }^{2}x+6x\log x-\int x\mathrm{d}(6\log x)\\ & =x{\log }^{3}x-3x{\log }^{2}x+6x\log x-\int 6\mathrm{d}x\\ & =x{\log }^{3}x-3x{\log }^{2}x+6x\log x-6x+C\end{array}$

Subproblem 4

$\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{x}{\sqrt{1-x^2}}\mathrm{d}x$

To integrate this function, we introduce a new variable $u=1-x^2$:

$\begin{array}{rl}\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{x}{\sqrt{1-x^2}}\mathrm{d}x& =-\frac{1}{2}\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{(1-x^2)}{\sqrt{1-x^2}}\mathrm{d}x\\ & =-\frac{1}{2}\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{1}{\sqrt{1-x^2}}\mathrm{d}(1-x^2)\\ & =-\frac{1}{2}\underset{1}{\overset{\frac{3}{4}}{\int }}\frac{1}{\sqrt{u}}\mathrm{d}u\\ & =\dots \\ & =1-\frac{\sqrt{3}}{2}\end{array}$

Subproblem 5

$\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{4-x^2}}\mathrm{d}x$

To integrate this function, we use a substitution $x=2\sin t$. Observe that $x$ changes from $0$ to $1$ when $t$ changes from $0$ to $\frac{\pi }{6}$):

$\begin{array}{rl}\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{4-x^2}}\mathrm{d}x& =\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{1}{\sqrt{4-(2\sin t{)}^2}}\mathrm{d}(2\sin t)\\ & =\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{(2\sin t)}{\sqrt{4-4{\sin }^{2}t}}\mathrm{d}t\\ & =\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{2\cos t}{\sqrt{4{\cos }^{2}t}}\mathrm{d}t\\ & =\underset{0}{\overset{\frac{\pi }{6}}{\int }}1\mathrm{d}x\\ & =\frac{\pi }{6}\end{array}$

Bonus Problem 5

Suppose $p:\mathbb{ℝ}\to \mathbb{ℝ}$ is locally a polynomial, which means that for every $c\in \mathbb{ℝ}$, there exists $ϵ>0$ such that $p$ coincides with a polynomial on the interval $(c-ϵ,c+ϵ)$. Prove that $p$ is a polynomial.

Proof. For any $c\in \mathbb{ℝ}$ let $p_c$ denote a polynomial and ${ϵ}_c$ denote a positive number such that $p(x)=p_c(x)$ for all $x\in (c-{ϵ}_c,c+{ϵ}_c)$. Consider two sets:

$\begin{array}{rl}{E}_{+}& =\{x>0\mid p(x)\ne p_0(x)\}\\ E_{-}& =\{x<0\mid p(x)\ne p_0(x)\}\end{array}$

We are going to show that $E_{+}=E_{-}=\mathrm{\varnothing }$.

Assume that the set $E_{+}$ is not empty. Clearly, $E_{+}$ is bounded below, hence $d=\inf E_{+}$ is a well-defined real number. Note that $E_{+}\subset [{ϵ}_0,\mathrm{\infty })$ Therefore $d\ge {ϵ}_0>0$.

Observe that $p(x)=p_0(x)$ for $x\in (0,d)$ and $p(x)=p_d(x)$ for $x\in (d-{ϵ}_d,d+{ϵ}_d)$. The interval $(0,d)$ overlaps with the interval $(d-{ϵ}_d,d+{ϵ}_d)$. Hence $p_d$ coincides with $p_0$ on the intersection $(0,d)\cap (d-{ϵ}_d,d+{ϵ}_d)$. Equivalently, the difference $p_d-p_0$ is zero on $(0,d)\cap (d-{ϵ}_d,d+{ϵ}_d)$. Since $p_d-p_0$ is a polynomial and any nonzero polynomial has only finitely many roots, we conclude that $p_d-p_0$ is identically 0. Then the polynomials $p_d$ and $p_0$ are the same. It follows that $p(x)=p_0(x)$ for $x\in (0,d+{ϵ}_d)$, so $d\ne \inf E_{+}$, a contradiction.

Thus $E_{+}=\mathrm{\varnothing }$. Similarly, we prove that the set $E_{-}$ is empty as well. Since $E_{+}=E_{-}=\mathrm{\varnothing }$, the function $p$ coincides with the polynomial $p_0$ everywhere.

Bonus Problem 6

Show that a function $f(x)=\{\begin{array}{ll}{\mathrm{e}}^{-\frac{1}{1-x^2}}& \left|x\right|<1\\ 0& \left|x\right|\ge 1\end{array}$

is infinitely differentiable on $\mathbb{ℝ}$.

Observe that the integral of this function goes from 0 to 1 smoothly. Constant functions and ${\mathrm{e}}^{f(x)}$ are infinitely differentiable.