Chapter 5 The Definite Integral

5.1 Introduction to Area

The problem of finding the tangent line led us to the derivative. The problem of finding area will lead us to the definite integral.

Sigma Notation \(\Sigma\) and its Properties

Consider, for example, the following sums \[1^2+2^2+3^2+4^2+\ldots+100^2\] or \[a_1+a_2+a_3+a_4+\ldots+a_{100}\]

To indicate these sums in a compact way, we write these sums as \[\sum_{i=1}^{100} i^2 \quad\text{ and }\quad \sum_{i=1}^{n}a_i\] respectively.

Thought of as an operator, \(\Sigma\) operates on sequences, and it does so in a linear way.

N.B. The symbol used for the index does not matter.It is just a dummy index.

Some Special Sum Formulas

Proofs.

The proof of Formula 1 is in textbook. To prove Formulae 2-4, see Problems 29-31, page 282.

Area by Inscribed Polygons

Consider the region \(R\) bounded by the parabola \(y=f(x)=x^2\), the \(x\)-axis, and the vertical line \(x=2\). Our aim is to calculate its area \(A(R)\). Partition the interval \([0, 2]\) into \(n\) subintervals, each of length \(\Delta x=2/n\), by means of the \(n+1\) points.

Consider a typical rectangle with base \([x_{i-1}, x_i]\) and height \(f(x_{i-1})=x^2_{i-1}\). Its area is \(f(x_{i-1})\Delta x\). The union \(R_n\) of all such rectangles forms the inscribed polygon shown in the lower-right part of Figure 7.

N.B. The diagrams in Figure 8 should help you to visualize what is happening as n gets larger and larger.

Area by Circumscribed Polygons

Another Problem—Same Theme

N.B. The distance traveled is the area of the region under the velocity curve.

Approximate the distance by the sum of circumscribed polygons,

5.2 The Definite Integral

Riemann Sums

In formulating the definition of the definite integral, we are guided by the ideas we discussed in the previous section. The first notion is that of a Riemann sum.

Example 1, page 285. Equally spaced points lead to intervals of equal length, and the sample points are the midpoint of each subinterval.

N.B. When \(f\) is negative, sample point \(\bar x_i\) with the property \(f(\bar x_i)<0\) that will lead to a rectangle that is entirely below the \(x\)-axis, and the product \(f(\bar x_i)\Delta \bar x_i\) will be negative. This means that the contribution of such a rectangle to the Riemann sum is negative.

Example 2, page 285. Evaluate the function at each sample point, and sum up all the areas of the rectangle. Negative sign is included in the Riemann sum for the negative function.

Definition of the Definite Integral

\(\int_a^b f(x)dx\) gives the signed area of the region trapped between the curve \(y=f(x)\) and the \(x\)-axis on the interval \([a, b]\), meaning that a positive sign is attached to areas of parts above the \(x\)-axis, and a negative sign is attached to areas of parts below the \(x\)-axis. In symbols, \[\int_a^b f(x)dx=A_{\text{up}}-A_{\text{down}}\]

In the symbol \(\int_a^b f(x)dx\) that implicitly assumed \(a<b\), we might call \(a\) the lower end point and \(b\) the upper end point for the integral. However, most authors use the terminology lower limit of integration and upper limit of integration. \(x\) in the symbol \(\int_a^b f(x)dx\) is a dummy variable.

We remove the restriction \(a<b\) with the following definitions.

What Functions Are Integrable?

Not every function is integrable on a closed interval \([a, b]\). For example, the unbounded function \[f(x)=\left\{\begin{aligned}\frac{1}{x^2}&\quad \text{if } x\neq 0\\1&\quad\text{if }x=0\end{aligned}\right.\]

Calculating Definite Integrals

Knowing that a function is integrable allows us to calculate its integral by using a regular partition (i.e., a partition with equal-length subintervals) and by picking the sample points \(\bar x_i\) in any way that is convenient for us.

Examples 3 & 4, page 288, evaluate the definite integral of polynomials.

The Interval Additive Property

Velocity and Position

In general, the position/displacement (which could be positive or negative) is equal to the definite integral of the velocity function (which could be positive or negative).

More specifically, if \(v(t)\) is the velocity of an object at time \(t\), where \(t\geq 0\), and if the object is at position \(0\) at time \(0\), then the position of the object at time \(a\) is \(\int_0^av(t)dt\).

Comments: Evaluating using the formulae for the area of a triangle and a rectangle is unnecessary after we learn the integration techniques.

5.3 The First Fundamental Theorem of Calculus

What is the connection between these two kinds of limits – antiderivatives and definite integrals? This important relationship is called the First Fundamental Theorem of Calculus.

The First Fundamental Theorem

Heuristic Example: velocity and displacement

Suppose the velocity of an object at time \(t\) is given by \(v=f(t)\). Using the terminology from Section 5.2, we now see that the distance traveled from time \(t=0\) to time \(t=3\) is equal to the definite integral \[\int_0^3f(t)dt=\lim_{n\to \infty}\sum_{i=1}^nf(t_i)\Delta t\] In general, the distance \(s\) traveled from time \(t=0\) to time \(t=x\) is \[s(x)=\int_0^xf(t)dt\] The question we now pose is this: What is the derivative of s?

Since the derivative of distance traveled (as long as the velocity is always positive) is the velocity, we have \[s^\prime(x)=v=f(x)\] In other words, \[\frac{d}{dx}s(x)=\frac{d}{dx}\int_0^xf(t)dt=f(x)\]

Heuristic Example: area under the graph and the accumulation function

N.B. To find \(\frac{d}{dx}A(x)\), just evaluate the integrand of \(\int_1^x \left(\frac{2}{3}+\frac{1}{2}t\right)dt\) at \(x\).

Another example of accumulation function \(B\) as the area under the curve \(y=t^2\) is on page 294. The procedure is a bit more laborious, but the conclusion follows the same rule, i.e., substitute \(x\) for \(t\) of the integrand, we obtain the derivative of the accumulation function, \[\frac{d}{dx}\int_0^xt^2dt=x^2\]The results suggest that the derivative of an accumulation function is equal to the function being accumulated (i.e., the integrand).

The rate of accumulation at \(t=x\) is equal to the value of the function being accumulated at \(t=x\).

We will give a complete proof later in this section.

Comparison Properties

Note that \(m(b-a)\) is the area of the lower, small rectangle, \(M(b-a)\) is the area of the large rectangle, and \(\int_a^b f(x)\) is the area under the curve.

Let \(g(x)=M\) and \(h(x)=m\) for all \(x\) in \([a, b]\). Then, by Theorem B, \[\int_a^bf(x)dx\leq\int_a^bg(x)dx\] and \[\int_a^bh(x)dx\leq\int_a^bf(x)dx\] Using the geometric formula for the area of rectangle with width \(b-a\) and height \(M\) and \(m\), respectively, to calculate \(\int_a^bg(x)dx\) and \(\int_a^bh(x)dx\), we obtain \[\int_a^bf(x)dx\leq M(b-a)\] and \[m(b-a)\leq\int_a^bf(x)dx\]

Proof of the First Fundamental Theorem of Calculus

Now we are ready to prove the First Fundamental Theorem of Calculus. For \(x\) in \([a,b]\), define \(F(x)=\int_a^xf(t)dt\), and with the Interval Additive Property we established the fact that \[F(x+h)-F(x)=\int_a^{x+h}f(t)dt-\int_a^{x}f(t)dt=\int_x^{x+h}f(t)dt\] Assume for the moment that \(h>0\) and let \(m\) and \(M\) be the minimum value and maximum value, respectively, of \(f\) on the interval \([x,x+h]\) (Figure 7). By Theorem C Boundedness Property, \[mh\leq \int_x^{x+h}f(t)dt\leq Mh\] or \[mh\leq F(x+h)-F(x)\leq Mh\] Dividing by \(h\), we obtain \[m\leq \frac{F(x+h)-F(x)}{h}\leq M\] Since \(m\) and \(M\) depends on \(h\), as \(h\to 0\), both \(m\) and \(M\) approach \(f(x)\), thus by the Squeeze Theorem (Sandwich Theorem), \[\lim_{h\to 0} \frac{F(x+h)-F(x)}{h}=f(x)\] Note that the limit on the left-hand side (LHS) of the above equation is \(F^\prime (x)\), which is equal to the notation \(\frac{d}{dx}\int_a^x f(t)dt\) with the definition \(F(x)=\int_a^xf(t)dt\).

The Definite Integral Is a Linear Operator

Like \(D_x\), \(\int \ldots dx\), and \(\Sigma\), the definite integral \(\int_a^b\ldots dx\) is a linear operator.

Example 1-3, page 298. Just apply the theorem A. N.B. The lower limit does not matter, it is a general \(a\).

Position as Accumulated Velocity

Now let’s revisit the Example 5 of Section 5.2.

Then, we set \(F(a)=0\) and solve for \(a\).

5.4 The Second Fundamental Theorem of Calculus and the Method of Substitution

The First Fundamental Theorem of Calculus, given in the previous section, gives the inverse relationship between definite integrals and derivatives. The Second Fundamental Theorem of Calculus in this section gives us a powerful tool for evaluating definite integrals.

N.B. From the perspective of definition, the indefinite integral is an antiderivative, while the definite integral as the limit of a Riemann sum. The fundamental theorem shows how indefinite integration (antidifferentiation) and definite integration (signed area) are related.

Example 1-2, page 304. We need the antidifferentiation formulae in section 4.10 along with the second fundamental theorem to evaluate the definite integral.

N.B. the Convenient Notation:

Example 4-6, page 304-305. Use the simple notation, the second fundamental theorem, and the linearity property.

The Method of Substitution

The nontrivial part of applying the theorem is always to find the indefinite integral \(\int f(x)dx\). One of the most powerful techniques for doing this is the method of substitution.

In Section 4.10, we introduced the method of substitution for the power rule, a.k.a. the generalized power rule. This rule can be extended to a more general case as the following theorem shows. Please be aware that the substitution rule is nothing more than the Chain Rule in reverse.

Example 7-9, page 306-307. Specify \(u\), and use the antiderivative formulae well.

N.B. No law says that you have to write out the \(u\)-substitution. If you can do the substitution mentally, that is fine. Illustration in Example 10.

Example 11-12, 307. In the definite integration, always choose \(C=0\) in applying the Second Fundamental Theorem.

N.B. If you want to use the upper and lower limits of the integral given in the question, be sure to express the indefinite integral in terms of \(x\) before applying the Second Fundamental Theorem. If the indefinite integral is in terms of \(u\), be sure to make the corresponding changes in the limits of integration to \(u\). That is, either of the following expression will lead you to the correct result, \[\int_0^{\frac{\pi}{4}}\sin^32x\cos2xdx=\left[\frac{\sin^42x}{8}\right]\Bigg|_0^{\frac{\pi}{4}}=\frac{1}{8}-0=\frac{1}{8}\] and equivalently, \[\int_0^{\frac{\pi}{4}}\sin^32x\cos2xdx=\left[\frac{1}{2}\frac{u^4}{4}\right]\Bigg|_0^{1}=\frac{1}{8}-0=\frac{1}{8}\]

Here is the general result, which lets us substitute the limits of integration, thereby producing a procedure with fewer steps.

Example 13-14, page 308.

Example 15, page 309. This example illustrates the remarkable property that to evaluate a definite integral all we need to know are the values of an antiderivative at the end points a and b.

Accumulated Rate of Change

The Second Fundamental Theorem of Calculus can be restated in this way: \[\int_a^b F^\prime(t)dt=F(b)-F(a)\]then the Second Fundamental Theorem of Calculus says that the accumulated rate of change from time \(t=a\) to time \(t=b\) is equal to the net change in that quantity over the interval \([a,b]\), that is, the amount present at time \(t=b\) minus the amount present at time \(t=a\).

Example 16, page 309. \(V^\prime(t)\) is the rate of the water leak-out. \(V(t)\) is the accumulated rate of change, i.e., the amount of leak-out through time \(t\).

5.5 The Mean Value Theorem for Integrals and the Use of Symmetry

This suggests the following definition.

Example 1 & 2, page 313. Apply the definition formula and evaluate the integral to find the average value.

Example 2 suggests that there should always be a value of \(x\) with the property that equals the average value of the function. This is true provided only that the function \(f\) is continuous.

The Mean Value Theorem for Integrals says that

there is some point \(c\) in the interval \([a, b]\) at which the average value of a function \(\frac{1}{b-a}\int_a^bf(t)dt\) is equal to the actual value of the function, \(f(c)\).

Recall the Mean Value Theorem for Derivatives, it says that

there is some point \(c\) in the interval \([a, b]\) at which the average rate of change of \(f\), equals the instantaneous rate of change, \(f^\prime(c)\).

Another way to express the Mean Value Theorem for Integrals: \[\int_a^bf(t)dt=(b-a)f(c)\]which says that there is some \(c\) in the interval \([a, b]\) such that the area of the rectangle with height \(f(c)\) and width \(b-a\) is equal to the area under the curve, as shown in Figure 3.

Example 3 & 4, page 315. Apply theorem A, evaluate the integral and solve for \(c\). Note the value of \(c\) should be in the given interval.

The Use of Symmetry in Evaluating Definite Integrals

If \(f\) is odd, \(f(u)=f(-x)=-f(x)\), \[\int_{-a}^0f(x)dx=\int_{-a}^0f(-x)(-dx)=\int_{a}^0f(u)du=-\int_{0}^af(u)du=-\int_{0}^af(x)dx\] Therefore, \[\int_{-a}^af(x)dx=-\int_{0}^af(x)dx+\int_{0}^af(x)dx=0\]

Example 4-7, page 316-317.

Example 8, page 317. N.B. Recall the properties of the odd and even functions: An odd function raised to an odd power is odd. An even function raised to any integer power is even, so is even. An odd function times an even function is odd.

Use of Periodicity

Recall that a function \(f\) is periodic if there is a number \(p\) such that \(f(x+p)=f(x)\) for all \(x\) in the domain of \(f\). If \(f\) is nonconstant, then the smallest such positive number \(p\) is called the period of \(f\). The trigonometric functions are examples of periodic functions.