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 $f'(c)=0$ 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 $f(b)-f(a)=f'(c)\,(b-a)$.
• $f$ is increasing on $[a,b]$ if and only if $f'\geq 0$ on $(a,b)$
• $f$ is decreasing on $[a,b]$ if and only if $f'\leq 0$ on $(a,b)$
• $f$ is constant on $[a,b]$ if and only if $f'=0$ on $(a,b)$

Properties of Integrals

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

Fundamental Theorem of Calculus

• Part 1: $F(x)=\int _{a}^{x}f(t)\,\mathrm {d} t$, $x\in [a,b]$ is continuously differentiable on $[a,b]$ and $F'(x)=f(x)$ for all $x\in [a,b]$
• Part 2: 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 $x\in [a,b]$.

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

$\lim _{x\to 0}\left(1+x\right)^{\frac {1}{x}}$

The function $f(x)=\left(1+x\right)^{\frac {1}{x}}$ is well-defined on $(-1,\infty )$ except at $0$. Since $f(x)>0$ for all $x>1$, a function $g(x)=\log {f(x)}$ is well-defined on $(-1,\infty )$ except at $0$ as well. For any $x>1$, we have $g(x)=\log {\left(1+x\right)^{\frac {1}{x}}}=x^{-1}\,\log {\left(1+x\right)}$. Hence $g={\frac {h_{1}}{h_{2}}}$, where $h_{1}(x)=\log {\left(1+x\right)}$ and $h_{2}(x)=x$ are continuously differentiable on $\left(-1,\infty \right)$. Since $h_{1}(0)=h_{2}(0)=0$, it follows that $\lim _{x\to 0}h_{1}(x)=\lim _{x\to 0}h_{2}(x)=0$. By l'Hôpital's rule, $h_{1}'(0)=1$ and $h_{2}'(0)=1$, so $\lim _{x\to 0}g(x)=\lim _{x\to 0}{\frac {h_{1}(x)}{h_{2}(x)}}={\frac {1}{1}}=1$

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

$\lim _{x\to +\infty }\left(1+x\right)^{\frac {1}{x}}$ Similar to above, $\lim _{x\to +\infty }h_{1}(x)=\lim _{x\to +\infty }h_{2}(x)=+\infty$. At the same time, $h_{1}'(x)\to 0$ as $x\to +\infty$, while $h_{2}'$ is identically $1$. Using l'Hôpital's Rule, we obtain $\lim _{x\to +\infty }g(x)=\lim _{x\to +\infty }{\frac {h_{1}(x)}{h_{2}(x)}}={\frac {0}{1}}=0$

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

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 $n\in \mathbb {N}$, 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 $P_{n}=\left\{0,{\frac {1}{n}},{\frac {2}{n}},\ldots ,{\frac {n-1}{n}},1\right\}$ and samples $t_{j}={\frac {j}{n}}$, $j=1,2,\ldots ,n$. The norm of the partition is $\left\|P_{n}\right\|={\frac {1}{n}}$. Since $\left\|P_{n}\right\|\to 0$ as $n\to \infty$ and the function $f$ is integrable on $[0,1]$, the Riemann sums $x_{n}$ converge to the integral:

$\lim _{n\to \infty }x_{n}=\int _{0}^{1}x^{k}\,\mathrm {d} x=\left.{\frac {x^{k+1}}{k+1}}\right|_{x=0}^{1}={\frac {1}{k+1}}$

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 $(-\infty ,1)\cup (1\infty )$, the indefinite integral has differetn representations on the intervals $(-\infty ,1)$ and $(1,\infty )$:

$\int {\frac {x^{2}}{1-x}}={\begin{cases}-{\frac {x^{2}}{2}}-x-\log {\left(1-x\right)}+C_{1}&x<1\\-{\frac {x^{2}}{2}}-x-\log {\left(x-1\right)}+C_{2}&x>1\end{cases}}$

Subproblem 2

$\int _{0}^{\pi }\sin ^{2}{(2x)}\,\mathrm {d} x$

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

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

Subproblem 3

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

To find this indefinite integral, we integrate by parts:

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

Subproblem 4

$\int _{0}^{\frac {1}{2}}{\frac {x}{\sqrt {1-x^{2}}}}\,\mathrm {d} x$

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

${\begin{aligned}\int _{0}^{\frac {1}{2}}{\frac {x}{\sqrt {1-x^{2}}}}\,\mathrm {d} x&=-{\frac {1}{2}}\,\int _{0}^{\frac {1}{2}}{\frac {\left(1-x^{2}\right)'}{\sqrt {1-x^{2}}}}\,\mathrm {d} x\\&=-{\frac {1}{2}}\,\int _{0}^{\frac {1}{2}}{\frac {1}{\sqrt {1-x^{2}}}}\,\mathrm {d} \left(1-x^{2}\right)\\&=-{\frac {1}{2}}\,\int _{1}^{\frac {3}{4}}{\frac {1}{\sqrt {u}}}\,\mathrm {d} u\\&=\ldots \\&=1-{\frac {\sqrt {3}}{2}}\end{aligned}}$

Subproblem 5

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

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 ${\frac {\pi }{6}}$):

${\begin{aligned}\int _{0}^{1}{\frac {1}{\sqrt {4-x^{2}}}}\,\mathrm {d} x&=\int _{0}^{\frac {\pi }{6}}{\frac {1}{\sqrt {4-(2\sin {t})^{2}}}}\,\mathrm {d} (2\sin {t})\\&=\int _{0}^{\frac {\pi }{6}}{\frac {\left(2\sin {t}\right)}{\sqrt {4-4\sin ^{2}{t}}}}\,\mathrm {d} t\\&=\int _{0}^{\frac {\pi }{6}}{\frac {2\cos {t}}{\sqrt {4\cos ^{2}{t}}}}\,\mathrm {d} t\\&=\int _{0}^{\frac {\pi }{6}}1\,\mathrm {d} x\\&={\frac {\pi }{6}}\end{aligned}}$

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 $\epsilon >0$ such that $p$ coincides with a polynomial on the interval $\left(c-\epsilon ,c+\epsilon \right)$. Prove that $p$ is a polynomial.

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

${\begin{aligned}E_{+}&=\left\{x>0~\mid ~p(x)\neq p_{0}(x)\right\}\\E_{-}&=\left\{x<0~\mid ~p(x)\neq p_{0}(x)\right\}\end{aligned}}$

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

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 $E_{+}\subset \left[\epsilon _{0},\infty \right)$ Therefore $d\geq \epsilon _{0}>0$.

Observe that $p(x)=p_{0}(x)$ for $x\in (0,d)$ and $p(x)=p_{d}(x)$ for $x\in \left(d-\epsilon _{d},d+\epsilon _{d}\right)$. The interval $(0,d)$ overlaps with the interval $\left(d-\epsilon _{d},d+\epsilon _{d}\right)$. Hence $p_{d}$ coincides with $p_{0}$ on the intersection $(0,d)\cap (d-\epsilon _{d},d+\epsilon _{d})$. Equivalently, the difference $p_{d}-p_{0}$ is zero on $(0,d)\cap (d-\epsilon _{d},d+\epsilon _{d})$. 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 $x\in (0,d+\epsilon _{d})$, so $d\neq \inf {E_{+}}$, a contradiction.

Thus $E_{+}=\emptyset$. Similarly, we prove that the set $E_{-}$ is empty as well. Since $E_{+}=E_{-}=\emptyset$, the function $p$ coincides with the polynomial $p_{0}$ everywhere.

Bonus Problem 6

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

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.