A tale of three integrals

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\tkzTabSlope }[1]{\foreach \x /\y /\z in {#1}{\node [left = 3pt] at (Z\x 1) {\scriptsize $\y $}; \node [right = 3pt] at (Z\x 1) {\scriptsize $\z $}; }} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ddt }[0]{\frac {d}{\d t}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{\vec {proj}} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

At this point we have three “different” integrals.

At this point we have three different “integrals.” Let’s see if we can sort out the differences.

Indefinite integrals

An indefinite integral, also called an antiderivative computes classes of functions: $\int f(x) \d x = \text {``a class of functions whose derivative is $f$''}$ Here there are no limits of integration, and your answer will have a “ $+C$ ” at the end. Pay attention to the notation:

Where $F'(x) = f(x)$ .

Indefinite integrals havedo not have limits of integration, and they compute signed areaan antiderivativea class of antiderivatives .

Two students, say Devyn and Riley, are working with the following indefinite integral: $\int \frac {2}{x\ln (x^2)}\d x$ Devyn computes the integral as $\int \frac {2}{x\ln (x^2)}\d x = \ln |\ln |x^2|| + C$ and Riley computes the integral as $\int \frac {2}{x\ln (x^2)}\d x = \ln |\ln |x|| + C.$ Which student is correct?

Devyn is correct Riley is correct Both students are correct Neither student is correct

Both students are correct! The seeming discrepancy arises from the fact that the “+C” in each case is different!

Accumulation functions

An accumulation function, also called an area function computes accumulated area: $\int _a^x f(t) \d t = \text {``a function $F$ whose derivative is $f$''}$ This is a function of $x$ whose derivative is $f$ , with the additional property that $F(a)=0$ . Pay attention to the notation:

Where $F'(x) = f(x)$ .

Accumulation functions havedo not have limits of integration, and they compute signed areaan antiderivativea class of antiderivatives .

True or false: There exists a function $f$ such that $\int _0^x f(t) \d t = e^x$

true false

Let $F(x) = \int _0^x f(t) \d t,$ this is an accumulation function and $F(0) = 0$ , since no area is accumulated yet. However, $e^0 =1$ . Hence there can be no such function $f$ . On the other hand, there is a function $g$ with $\int _0^x g(t) \d t = e^x-1$ namely, $g(x) = e^x$ . This subtlety arises from the fact that an accumulation function $F(x) = \int _a^x f(t) \d t$ gives a specific antiderivative of $f$ , the one that when evaluated at $x=a$ is zero.

Definite integrals

A definite integral computes signed area: $\int _a^b f(x) \d x = \text {``the signed area between the $x$-axis and $f$''}$ Here we always have limits of integration, both of which are numbers. Moreover, definite integrals have definite values, the signed area between $f$ and the $x$ -axis. Pay attention to the notation:

Where $F'(x) = f(x)$ .

Definite integrals havedo not have limits of integration, and they compute signed areaan antiderivativea class of antiderivatives .

Consider $f(x) = \begin {cases} -2 &\text {if $x< 1$,}\\ 2 &\text {if $x\ge 1$.} \end {cases}$ If we compute an antiderivative of $f$ , we find $F(x) = \begin {cases} -2x &\text {if $x< 1$,}\\ 2x &\text {if $x\ge 1$.} \end {cases}$ Is it correct to say $\begin{align*} \int_0^1 f(x) \d x &= \eval{F(x)}_0^1 \\ &=F(1) - F(0)\\ &=2 ? \end{align*}$

yes no

Perhaps the first thing to do would be to attempt to analyze this geometrically. Here we see our function and the signed area computed by the integral:

From the graph above, we can see that $\int _0^1 f(x) \d x =-2.$ So now the question is, “what went wrong” above? In this case our function $f$ is not continuous! For The Fundamental Theorem of Calculus to apply, the integrand must be continuous on the interval that one is integrating on. If this is not the case, the fundamental theorem may or may not yield valid results.