If an interval $[a,b]$ is partitioned into $n$ subintervals $[x_{0},\,x_{1}],$ $[x_{1},\,x_{2}],$ $[x_{2},\,x_{3}],$ $\ldots,$ $[x_{n-1},\,x_{n}]$, where $a=x_{0}\lt x_{1}\lt x_{2}\lt\cdots \lt x_{n-1}\lt x_{n}=b,$ then the set of subintervals of the interval $[a,b]$ is called a(n) ________ of $[a,b]$.

Multiple Choice  In a regular partition of $[0,40]$ into $20$ subintervals, $\Delta x=$ [(a)  20 (b)  40  (c)  2  (d)  4].

True or False  A function $f$ defined on the closed interval $[a,b]$ has an infinite number of Riemann sums.

In the notation for a definite integral $\int_{a}^{b}f(x)\, dx$, $a$ is called the _______ _______; $b$ is called the _______ _______; $\int$ is called the _________ _________; and $f(x)$ is called the_________.

If $f(a)$ is defined, $\int_{a}^{a}f(x)\, dx=$ _________.

True or False  If a function $f$ is integrable over a closed interval $[a,b]$, then $\int_{a}^{b}f(x) dx=\int_{b}^{a}f(x) dx$.

True or False  If a function $f$ is continuous on a closed interval $[a,b]$, then the definite integral $\int_{a}^{b}f(x)\, dx$ exists.

Multiple Choice  Since $\int_{0}^{2}(3x-8) dx=-10$, then $\int_{2}^{0}(3x-8)\, dx=$ [(a) $-$2  (b)  10  (c)  5  (d)  0].

${f(x)=x}$, $0\leq x \leq 2$. Partition the interval $[0,2]$ as follows: $$\begin{eqnarray*} &&x_{0}=0, x_{1}=\dfrac{1}{4}, x_{2}=\dfrac{1}{2}, x_{3}=\dfrac{3}{4}, x_{4}=1, x_{5}=2;\\ &&\left[0,\dfrac{1}{4}\right], \left[\dfrac{1}{4},\dfrac{1}{2}\right], \left[\dfrac{1}{2},\dfrac{3}{4}\right], \left[\dfrac{3}{4},1\right], [1,2] \end{eqnarray*}$$

and choose $$u_{1}=\dfrac{1}{8}, u_{2}=\dfrac{3}{8}, u_{3}=\dfrac{5}{8}, u_{4}=\dfrac{7}{8}, u_{5}=\dfrac{9}{8}.$$

${f(x)=x}$, $0\leq x \leq 2$. Partition the interval $[0,2]$ as follows: $\left[0,\dfrac{1}{2}\right]$, $\left[\dfrac{1}{2},1\right]$, $\left[1,\dfrac{3}{2}\right]$, $\left[\dfrac{3}{2},2\right]$, and choose $u_{1}=\dfrac{1}{2}$, $u_{2}=1$, $u_{3}=\dfrac{3}{2}$, $u_{4}=2$.

$f(x)=x^{2}$, $-2\leq x \leq 1$. Partition the interval $[-2,1]$ as follows: $[-2,-1]$, $[-1,0]$, $[0,1]$ and choose $u_{1}={-}{\dfrac{3}{2}}$, $u_{2}={-}{\dfrac{1}{2}}$, $u_{3}={\dfrac{1}{2}}$.

${f(x)=x^{2}}$, $1\leq x\leq 2$. Partition the interval $[1,2]$ as follows: $\left[1,\dfrac{5}{4}\right]$, $\left[\dfrac{5}{4},\dfrac{3}{2}\right]$, $\left[\dfrac{3}{2},\dfrac{7}{4}\right]$, $\left[\dfrac{7}{4},2\right]$ and choose $u_{1}=\dfrac{5}{4}$, $u_{2}=\dfrac{3}{2}$, $u_{3}=\dfrac{7}{4}$, $u_{4}=2$.

$\lim\limits_{{\max \Delta x}_{i} \rightarrow 0}\sum\limits_{i=1}^{n}\left(e^{u_{i}}+2\right) \Delta x_{i}$ on $[0,2]$

$\lim\limits_{{\max \Delta x}_{i}\rightarrow 0}\sum\limits_{i=1}^{n}\ln u_{i}\Delta x_{i}$ on $[1,8]$

$\lim\limits_{{\max \Delta x}_{i} \rightarrow 0}\sum\limits_{i=1}^{n}\cos u_{i}\Delta x_{i}$ on $[0,2\pi]$

$\lim\limits_{{\max \Delta x}_{i} \rightarrow 0}\sum\limits_{i=1}^{n}\left( \cos u_{i}+\sin u_{i}\right) \Delta x_{i}$ on $[0,\pi]$

$\lim\limits_{{\max \Delta x}_{i} \rightarrow 0}\sum\limits_{i=1}^{n}{\dfrac{{2}}{{u^{2}_i}}}\Delta x_{i}$ on $[1,4]$

$\lim\limits_{{\max \Delta x}_{i} \rightarrow 0}\sum\limits_{i=1}^{n}u^{1/3}_i\Delta x_{i}$ on $[0,8]$

$\lim\limits_{{\max \Delta x}_{i} \rightarrow 0}\sum\limits_{i=1}^{n}u_{i}\ln u_{i}$ $\Delta x_{i}$ on $[1, e]$

$\lim\limits_{{\max \Delta x}_{i} \rightarrow 0} \sum\limits_{i=1}^{n}\ln (u_{i}+1)\Delta x_{i}$ on $[0, e]$

$\int_{-3}^{4}e\,dx$

$\int_{0}^{3}(-\pi) \,dx$

$\int_{3}^{0}(-\pi)\, dt$

$\int_{7}^{2}2\, ds$

$\int_{4}^{4}2\, \theta\ d\theta$

$\int_{-1}^{-1}8\, dr$

$\int_{0}^{\pi }\sin x\,dx$

$\int_{-\pi /4}^{\pi /4}\tan x\,dx$

$\int_{1}^{4}(x-2) ^{1/3}dx$

$\int_{1}^{4}(x+2) ^{1/3}dx$

$\int_{1}^{4}(\vert x \vert \,-2 ) \,dx$

$\int_{-2}^{4}\vert x \vert \,\,dx$

$f(x) =x^{2}-1$ on $[0,2]$

$f(x) =x^{3}-2$ on $[0,5]$

$f(x) =\sqrt{x+1}$ on $[0,3]$

$f(x) =\sin x$ on $[0, \pi]$

$f(x) =e^{x}$ on $[0, 2]$

$f(x) =e^{-x}$ on $[0,1]$

${\int_{0}^{1}(x - 4)dx}$

$\int_{0}^{3}{(3x - 1)dx}$

$f(x) =2+\sqrt{x}$ on $[1,5]$

$f(x) =e^{x}+e^{-x}$ on $[-1,3]$

$f(x) =\dfrac{3}{1+x^{2}}$ on $[-1,1]$

$f(x) = \dfrac{1}{\sqrt{x^2+4}}$ on $[0,2]$

Find an approximate value of ${\int_{1}^{2}}\dfrac{1}{x}\,{dx}$ by finding Riemann sums corresponding to a partition of $[1,2]$ into four subintervals, each of the same length, and evaluating the integrand at the midpoint of each subinterval. Compare your answer with the true value, $0.6931\ldots .$

Units of an Integral  In the definite integral $\int_{0}^{5}F(x)\, dx$, $F$ represents a force measured in newtons and $x,$ $0\leq x\leq 5,$ is measured in meters. What are the units of $\int_{0}^{5}F(x)\, dx?$

Units of an Integral  In the definite integral $\int_{0}^{50}C(x)\, dx$, $C$ represents the concentration of a drug in grams per liter and $x,$ $0\leq x\leq 50,$ is measured in liters of alcohol. What are the units of $\int_{0}^{50}C(x)\, dx?$

Units of an Integral  In the definite integral $\int_{a}^{b}v(t)\, dt,$ $v$ represents velocity measured in meters per second and time $t$ is measured in seconds. What are the units of $\int_{a}^{b}v(t)\, dt?$

Units of an Integral  In the definite integral $\int_{a}^{b}S(t)\, dt$, $S$ represents the rate of sales of a corporation measured in millions of dollars per year and time $t$ is measured in years. What are the units of $\int_{a}^{b}S(t)\, dt?$

Area

Area

The interval $[1,5]$ is partitioned into eight subintervals each of the same length.

The floor function $f(x) =$ $\lfloor x\rfloor$ is not continuous on $[0,4].$ Show that ${\int_{0}^{4}{f(x)\,dx}}$ exists.

Consider the Dirichlet function $f,$ where $$f(x)=\left\{ \begin{array}{@{}l@{ }l@{ }l} {1} & \hbox{if} & {x}~\hbox{is rational} \\ {0} & \hbox{if} & {x}~\hbox{is irrational} \end{array} \right.$$ Show that ${\int_{0}^{1}{f(x)\,dx}}$ does not exist. (Hint: Evaluate the Riemann sums in two different ways: first by using rational numbers for ${ u}_{i}$ and then by using irrational number seak for ${ u}_{i}$.)

It can be shown (with a certain amount of work) that if $f(x)$ is integrable on the interval $[a,b],$ then so is $\vert f(x) \vert$. Is the converse true?

If only regular partitions are allowed, then we could not always partition an interval $[a,b]$ in a way that automatically partitions subintervals $[a,c]$ and $[c,b]$ for $a\lt c\lt\,b$. Why not?

If $f$ is a function that is continuous on a closed interval $[a,b]$, except at $x_{1},$ $x_{2},$ $\ldots$, $x_{n},$ $n\geq 1$ an integer, where it has a jump discontinuity, show that $f$ is integrable on $[a,b] .$

If $f$ is a function that is continuous on a closed interval $[a,b]$, except at $x_{1},$ $x_{2},$ $\ldots$, $x_{n},$ $n\geq 1$ an integer, where it has a removable discontinuity, show that $f$ is integrableon $[a,b] .$