Unit 2: Integration Theory
Preface
This is the note of the MIT Course 18.01.2 Calculus. For more details, refer to MIT 18.x Catalog, and it can also be found at MIT Open Learning Library.
The Definite Integral
Geometric Definition of the Definite Integral
The definite integral of $f$ from $a$ to $b$, denoted by
$$
\int_a^b f(x)\ dx
$$
which is the area of the region above the $x$-axis, below the curve $y=f(x)$, and in between the two vertical lines $x=a$ and $x=b$, as shown shaded in the figure below.
lower and upper limits of the integral. And remember that this is a different sense of the word “limit” from when we take the limit of a function.
The only difference between the notation for definite and indefinite integrals is that definite integrals have limits but indefinite integrals do not.
Summation Notation
The $\Sigma$ notation in a compact way to denote a sum in which each term is obtained from a formula: $$ \sum_{i=1}^n a_i=a_1+a_2+\cdots+a_{n-1}+a_n $$ where $i$ indexes the terms, and $a_i$ is a formula from the $i^{th}$ term of the sum.
The notation $\sum_{i=1}^n a_i$ reads “the sum of $a_i$ from $i=1$ to $i=n$”.
For example, if $a_i=i^2$, that is, the formula for the term indexed by $i$ is $i^2$, then $$ \sum_{i=1}^n i^2=1^2+2^2+3^2+\cdots+(n-1)^2+n^2 $$
Riemann Sums
Let us summarize in precise terms the steps for evaluating $\displaystyle\int_a^b f(x)\ dx$ using a Reimann Sum.
- Divide $[a,b]$ into $n$ equal subintervals.
Then each interval is of length $$ \Delta x=\frac{b-1}n $$ Let the $i^th$ subinterval be the base of $i^{th}$ rectangle.
- Choose a point $c_i$ within the $i^th$ subinterval. Choose $f(c_i)$ be the height of the $i^th$ rectangle.
- Add up the areas of the $n$ rectangles. The total area of $n$ rectangles is:
$$
\smash[b]{\underbrace{f(c_1)}_{\mathclap{\text{height}}}}
\smash[b]{\underbrace{\Delta x}_{\mathclap{\text{base}}}} +
\smash[b]{\underbrace{f(c_2)}_{\mathclap{\text{height}}}}
\smash[b]{\underbrace{\Delta x}_{\mathclap{\text{base}}}} + \cdots +
\smash[b]{\underbrace{f(c_n)}_{\mathclap{\text{height}}}}
\smash[b]{\underbrace{\Delta x}_{\mathclap{\text{base}}}}
=\sum_{i=1}^n{f(c_i)\Delta x}
$$
- Take the limit as the rectangles become infinitesimally thin, ($\Delta x\to 0$, or equivalent $n\to\infty$). This limit is the actual area under the curve between $a$ and $b$. $$ \lim_{n\to\infty}\sum_{i=1}^n f(c_i)\Delta x=\int_a^b f(x)\ dx $$
The sum of the areas of the $n$ rectangles, $\sum_{i=1}^n f(c_i)$, is called a Riemann Sum. If we pick $c_i$ to be the left endpoint of the $i^{th}$ subinterval, the Riemann Sum is called a left Riemann Sum. Similarly, if $c_i$ is the right endpoint of the $i^{th}$ interval, the Riemann Sum is called a right Riemann Sum.
Howerver, in the limit $n\to\infty$ (so that $\Delta x\to 0$), this distinction is no longger needed. The limit of any Reimann Sum, no matter what the $c_i$’s whtin the subinterval are, is equal to the exact area under the curve.
Cumulative Sums
We have been defining the definite integral $\displaystyle\int_a^b f(x)\ dx$ geometrically as the area under a curve. But $f(x)$ and $x$ can represent quantities other than length.
For instance, if $x$ represents time, with units of hours, and $f(x)$ represents velocity, with units of km/hour. So the units of the integral, $\displaystyle\int_a^b f(x)\ dx$, is km, which are the units of distance. But the integral $\displaystyle\int_a^b f(x)\ dx$ still represents the area under the curve, so the units of this area are km.
First Fundamental Theorem of Calculus
The First Fundamental Theorem of Calculus states that:
If $F$ is differentiable, and $F^{\prime}=f$ is continuous, then
$$
\int_a^b f(x)\ dx = F(b)-F(a)=\left.F(x)\right\vert_a^b
$$
In other words, the definite integral of a function is the difference between the values of its antiderivative at the limits of the definite integral.
We will abbreviate the First Fundamental Theorem of Calculus as FTC1.
The FTC1 connects the definite integral to the antiderivative. With this connectionk, we can now compute definite integrals using antiderivatives, and dispense with Reimann sums.
The Definite Integral of Any Continuous Function
Since the equation in the statement of FTC1 makes sense of general functions, we can use the FTC1 to extend the definition of the definite integral to functions that are not necessarily non-negative.
That is,
For any continuous function $f$ with an antiderivative $F$,
$$
\int_a^b f(x)\ dx=F(b)-F(a)\quad\text{(for any continuous $f$)}
$$
It turns out that to be consistent with FTC1, the Riemann sum formula for definite integrals does not need to change. Consequently, the geometric definition of $\displaystyle\int_a^b f(x)\ dx$ that is consistent with FTC1 is as follows.
signed area bounded by the curve $y=f(x)$.
We will continue to use FTC1, not Riemann sums, to evaluate definite integrals. We will also often use the area interpretation to dedeuce properties of definite integrals, for example when looking for symmetry.
Velocity and Speed, Displacement and Distance Travelled
Suppose you are travelling between time $a$ and time $b$, and your velocity is given by $v(t)$ and your position by $x(t)$ at time $t$.
Recall the Speed is $\vert v(t)\vert$, the absolute value of the velocity function.
Then,
$$
\int_a^b v(t)\ dt=x(b)-x(a)\quad\text{(Displacement)}
$$
That is, the integral of the velocity function is the change if position between time $a$ and $b$, also called the displacement.
On the other hand, $\displaystyle\int_a^b\vert v(t)\vert\ dt$ gives the total distance travelled between time $a$ and $b$.
In the case when $v(t)>0$ for all $t$, $\vert v(t)\vert=v(t)$, and therefore $\displaystyle\int_a^b v(t)\ dt=\displaystyle\int_a^b\vert v(t)\vert\ dt$. In other words, when you are always travelling in the same direction, then speed is equal to velocity, and total distance traveled is equal to displacement.
Properties of Definite Integrals
Here are some properties of definite integrals. We have discussed and used most of these properties for $f>0$. The properties below are true for definite integrals for functions which can be negative as well.
Sums: $$ \int_a^b(f(x)+g(x))\ dx=\int_a^bf(x)\ dx+\int_a^bg(x)\ dx $$
Constant Multiples: $$ \int_a^bc\cdot f(x)\ dx=c\cdot\int_a^b f(x)\ dx\quad\text{for any constant $c$} $$
Same Upper and Lower Limits: $$ \int_a^af(x)\ dx=0 $$
Reversing Limits of Integrals: $$ \int_b^af(x)\ dx=-\int_a^bf(x)\ dx\quad\text{for any $a$,$b$} $$
This is the definition of a definite integral with its lower limit greater than its upper limit. It is defined this way to be consistent with FTC1.
Combining Integrals: $$ \int_a^cf(x)\ dx=\int_a^bf(x)\ dx+\int_b^cf(x)\ dx\quad\text{for any $a$,$b$,$c$} $$
We have seen this property in the last section with $a<b<c$. With the property above on reversing the limits of a definite integral, we can now use this property with $a$,$b$,$c$ in any order.\
Estimation
If $f(x)\le g(x)$, and $a\le b$, then $$ \int_a^bf(x)\ dx\le\int_a^b g(x)\ dx\quad \text(for $a\le b$) $$ Notice that the order of $a$ and $b$ matters for this inequality. If instead of $a\le b$ we have $b\le a$, then $$ \int_a^bf(x)\ dx\ge \int_a^bg(x)\ dx\quad(b\le a) $$ In other words, if the limits of the integrals are reversed, the inequality is also reversed.
Change of Variables for Definite Integrals
When we make a change of variables of an integral in order to evaluate it, we are using the method of substitution. The method of substitution for definite integrals is exactly analogous to the method of substitution for indefinite integrals, except we now need to pay attention to the limits of the integrals.
If $$ \int_a^bf(x)\ dx=\int_a^bg(u(x))u^{\prime}(x)\ dx $$ and $u^{\prime}$ does not change sign between $a$ and $b$, then $$ \int_a^bf(x)\ dx=\int_a^bg(u(x))u^{\prime}(x)\ dx=\int_{u(a)}^{u(b)}g(u)\ du $$ That is, the limits of the integral over $u$ are the values of $u$ corresponding to the limits of the integral over $x$.
Caution
When using the method of substituion, we need to be very careful about when $u^{\prime}$ or $du$ changes sign. If $u^{\prime}$ changes sign within the integration interval $[a,b]$, the method of substition may give the wrong answer. In this case, we need to first rewrite the integral as a sum of two integrals such that within the limits of each integral $u^{\prime}$ does not change sign, and then use the method of substitution on each integral separately.
Comparing FTC1 and MVT
If $F(x)$ is differentiable and $F^{\prime}$ is continuous on $[a,b]$. And let $$ \begin{aligned} \Delta F =& F(b)-F(a)\\ \Delta x =& b-a \end{aligned} $$ Then, the MVT states that $$ \frac{\Delta F}{\Delta x}=F^{\prime}(c)\quad\text{for some }c,\ a<c<b $$ On the other hand, the FTC1 gives $$ \frac{\Delta F}{\Delta x}=\frac 1 {b-a}\int_a^b f^{\prime}\ dx $$ We see that the FTC1 gives a specific value for $\frac{\Delta F}{\Delta x}$, the average rate of change of $F$ over $[a,b]$, but the MVT does not, sinve it does not tell us where $c$ is.
Therefore, the First Fundamental Theorem much more useful than the Mean Value Theorem. Once we have FTC1 at our disposal, we do not need to use MVT anymore. Nonetheless, the Mean Value Theorem is important as the basis of calculus. We needed it to establish the fact that two antiderivatives of the same function can only differ by a constant. We will need this fact again in order to prove FTC1 in the next section.
Second Fundamental Theorem of Calculus
The Second Fundemental Theorem of Calculus states the following.
Given a continuous function $f(x)$, if
$$
G(x)=\int_a^x f(t)\ dt\quad\text{($t$ between $a$ and $x$)}
$$
then
$$
G^{\prime}(x)=f(x)
$$
Geometrically, $G(x)$ is the area between $a$ and $x$. This area varies as $x$ varies.
We will abbreviate this theorem by FTC2.
In terms of differential equations, FTC2 says that $G(x)$ is the solution to the following differential equation and initial condition: $$ \begin{aligned} y^{\prime}\quad=&\quad f\quad\text{(differential equation)}\\ y(a)\quad=&\quad 0\quad\text{(initial condition)} \end{aligned} $$ The initial condition $y(a)=0$ is satisified because $G(a)=\displaystyle\int_a^a f(t)\ dt=0$.
Recall that any function $G$ such that $G^{\prime}=f$ is an antiderivative of $f$. Hence, FTC2 gives us a formula for an antiderivative of $f(x)$. This formula is different from the formulas you have seen. It is terms of a definite integral, and is called an integral formula.
Note: The integrand $f(x)$ can be nay continuous function, not just the one whose antiderivative we know how to find. These integral formulas are still useful there are numerical methods that allow us to compute them.
FTC2 and The Chain Rule
For any continuous function $f$, and differentiable function $u(x)$, $$ \frac d {dx}\int_a^{u(x)}f(t)\ dt\quad=\quad f(u(x))\cdot u^{\prime}(x) $$ This is because $$ \int_a^{u(x)}f(t)\ dt\ =\ G(u(x))\quad\text{where}\quad G(y)\ =\ \int_a^y f(t)\ dt $$ In other words, the integral $\displaystyle\int_a^{u(a)}f(t)\ dt$ can be written as a composition of two function $G$ and $u$. We can then apply the chain rule and FTC2 to get the derivative.
The Two Fundamental Theorems of Calculus
Here is another point of view of the two fundamental theorems of calculus together.
FTC1:
Given a differentiable function $F$ with continuous derivative $F^{\prime}$, $$ \int_a^x F^{\prime}(t)\ dt=F(x)-F(a) $$
FTC2:
Given a continuous function $f$, $$ \frac d{dx}\int_a^x f(t)\ dt=f(x) $$ So, the two fundamental theorems together say that differentiation and integration (from $a$ to $x$) are “inverse” operation of on another, up to a constant.
Constructing Functions Using Integrals
The FTC2 gives the integral formula $$ G(x)=\int_a^xf(t)\ dt $$ as the solution to the following differential equation and initial condition: $$ \begin{aligned} y^{\prime}\quad=&\quad f\quad\text{(differential equation)}\\ y(a)\quad=&\quad 0\quad\text{(initial condition)} \end{aligned} $$ Depending on what the differential equation is, $G(x)$ can be a function we already know, or onw that cannot be expressed without an integral in terms of any function we already know. There are many functions of either kind.
For instance, the logarithmic function can be defined using an integral:
$$
L(x)\ =\ \int_1^x\frac{dt}t\ =\ \ln(x)\quad\text{where $x>0$}
$$
On the other hand, the integral of $e^{-t^2}$, the bell curve, is a function defined by an integral and cannot be expressed in terms of functions we already know:
$$
F(x)\ =\ \int_0^xe^{-t^2}\ dt
$$
This function is central in probability and will be discussed more in the next unit.
Finally, even though we may not have explicit formulas for the functions designed by integral formulas, we can apply FTC2 to find their derivatives. Therefore, we have all of our usual tools at our disposal. For example, we can graph them, or approximate them by linear and quadratic approximations.
Summary of The Logarithm
In Calculus IA, we defined the exponential function $E(y)$ to be the solution to the differential equation and initial condition. $$ \begin{aligned} E^{\prime}\quad=&\quad E(y)\quad\text{(differential equation)}\\ E(0)\quad=&\quad 1\quad\text{(initial condition)} \end{aligned} $$ The Properties of the logarithm were derived from the properties of the exponential.
Product: $$ \begin{array}{lcl} \text{Product:}&\qquad e^a\cdot e^b\ =\ e^{a+b}\qquad&\Longrightarrow\qquad&\ln(A\cdot B)\ =\ \ln(A)+\ln(B)\\ \text{Reciprocal:}&\qquad\frac{1}{e^a}\ =\ e^{-a}\qquad&\Longrightarrow\qquad& \ln(\frac 1 A)\ =\ -\ln(A)\\ \end{array} $$ On the other hand, we first define the logarithm to be the solution to the differential equation and initial condition below: $$ \begin{aligned} L^{\prime}(x)\ =&\ \frac 1 x\quad\text{(differential equation)}\\ L(1)\ =&\ 0\quad\text{(initial condition)} \end{aligned} $$ The FTC2 gives the integral formula for $L$. $$ \text{Definition of the logarithm: }\quad L(x)\ =\ \int_1^x\frac{dt}t\\ $$ We then define $e$ to be the number such that $L(e)=1$, and the exponential function to be the inverse of $L$. $$ \text{Definition of the exponential:}\quad e^y\ =\ L^{-1}(y) $$ To derive the properties of logarithm, we perform different changes of variable to the integral formula $L(x)=\displaystyle\int_1^x\frac{dt}t$.