5.3 The Definite Integral

OBJECTIVES

When you finish this section, you should be able to:

  1. Define a definite integral as the limit of Riemann sums (p. 353)
  2. Find a definite integral using the limit of Riemann sums (p. 356)

The area \(A \) under the graph of \(y = f(x)\) from \(a \) to \(b \) is obtained by finding\[\begin{equation}\boxed{\bbox[#FAF8ED,5pt]{A=\lim\limits_{n\rightarrow \infty}s_{n}=\lim\limits_{n\rightarrow \infty}{\sum\limits_{i=1}^{n}} {f}(c_{i}) \Delta x=\lim\limits_{n\rightarrow \infty}S_{n}=\lim\limits_{n\rightarrow \infty }{\sum\limits_{i=1}^{n}}{f}(C_{i}) \Delta x}}\end{equation}\]

where the following assumptions are made:

ORIGINS

Riemann sums are named after the German mathematician Georg Friedrich Bernhard Riemann (1826–1866). Early in his life, Riemann was homeschooled. At age 14, he was sent to a lyceum (high school) and then theUniversity of Göttingen to study theology. Once at Göttingen, he askedfor and received permission from his father to study mathematics. Hecompleted his PhD under Karl Friedrich Gauss (1777–1855). In histhesis, Riemann used topology to analyze complex functions. Later hedeveloped a theory of geometry to describe real space. His ideas were farahead of their time and were not truly appreciated until they provided themathematical framework for Einstein's Theory of Relativity.

In Section 5.1, we found the area \(A\) under the graph of \(f\) from\(a\) to \(b\) by choosing either the number \(c_{i}\), where \(f\) has an absoluteminimum on the \(i\)th subinterval, or the number \(C_{i},\) where\(f\) has an absolute maximum on the \(i\)th subinterval. Suppose wearbitrarily choose a number \(u_{i}\)in each subinterval \([x_{i-1},x_i]\), and draw rectangles of height\(f(u_{i})\) and width \(\Delta x.\) Then from thedefinitions of absolute minimum value and absolute maximum value\[f(c_{i}) \leq f (u_{i}) \leq f(C_{i})\]

and, since \(\Delta x > 0\),\[f(c_{i}) \Delta x\leq f (u_{i}) \Delta x\leq f(C_{i}) \Delta x\]

Then\[\begin{eqnarray*}\sum\limits_{i=1}^{n}f (c_{i}) \Delta x & \leq &\sum\limits_{i=1}^{n}f (u_{i}) \Delta x\leq\sum\limits_{i=1}^{n}f (C_{i}) \Delta x \\s_{n} &\leq &\sum\limits_{i=1}^{n}f(u_{i}) \Delta x\leq S_{n}\end{eqnarray*}\]

NEED TO REVIEW?

The Squeeze Theorem is discussed in Section 1.4, pp. xx-xx.

Since \(\lim\limits_{n\rightarrow \infty }s_{n}=\lim\limits_{n\rightarrow\infty }S_{n}=A,\) by the Squeeze Theorem, we have\[\lim\limits_{n\rightarrow \infty }\sum\limits_{i=1}^{n}f (u_{i}) \Delta x=A\]

In other words, we can use any number \(u_{i}\) in the \(i\)thsubinterval to find the area \(A.\)

5.3.1 Define a Definite Integral as the Limit of Riemann Sums

We now investigate sums of the form\[ \boxed{\bbox[#FAF8ED,5pt]{\sum\limits_{i=1}^{n}f (u_{i}) \Delta x_{i}}}\]

using the following more general assumptions:

354

The sums \({\sum\limits_{i=1}^{n}{f (u_{i}) }}\Delta x_{i},\) called Riemann sums for \(f\) on \([a, b]\), form the foundation of integralcalculus.

EXAMPLE 1 Forming Riemann Sums

For the function \(f(x) =x^{2}-3,\) \(0\leq x\leq 6,\) partitionthe interval \([0,6] \) into 4 subintervals \([0,1]\), \([1,2]\),\([2,4]\), \([4,6]\) and form the Riemann sum for which

  1. \(u_{i}\) is the left endpoint of each subinterval.
  2. \(u_{i}\) is the midpoint of each subinterval.

Solution In forming Riemann sums\(\sum\limits_{i=1}^{n}f(u_{i}) \Delta x_{i}\), \(n\) is the numberof subintervals in the partition, \(f(u_{i}) \) is the value of \(f\)at the number \(u_{i}\) chosen in the \(i\)th subinterval, and \(\Delta x_{i}\) is the lengthof the \(i\)th subinterval.

(a) Figure 5.16 shows the graph of \(f,\) the partition of the interval \([0,6] \) into the 4 subintervals, and values of \(f(u_{i}) \) at the left endpoint of each subinterval, namely,\[\begin{array}{rcl@{\qquad}crcl}f(u_{1}) &=& f(0) =-3 & f(u_{2}) &=&f(1) =-2\\f(u_{3}) &=& f(2) =1 & f(u_{4}) &=& f(4) = 13\end{array}\]

Figure 5.16: \(f(x) = x^2-3, 0 \leq x \leq 6\)

The 4 subintervals have length\[\Delta x_{1}=1-0=1 \qquad \Delta x_{2}=2-1=1 \qquad \Delta x_{3}=4-2=2 \qquad \Delta x_{4}=6-4=2\]

The Riemann sum is formed by adding the products \(f(u_{i}) \Delta x_{i}\) for \(i=1,2,3,4\). \[\sum\limits_{i=1}^{4} f(u_{i}) \Delta x_{i}=-3\cdot 1 + (-2) \cdot 1+1\cdot 2 + 13 \cdot 2=23\]

(b) See Figure 5.17. If \(u_{i}\) is chosen as themidpoint of each subinterval, then the value of \(f(u_{i}) \) at the midpoint of eachsubinterval is \[\begin{array}{rcl@{\qquad}crcl}f(u_{1}) &=& f\left(\dfrac{1}{2}\right) =-\dfrac{11}{4} & f(u_{2}) &=&f\left( \dfrac{3}{2}\right) =-\dfrac{3}{4}\\f(u_{3}) &=&f (3) =6 & f(u_{4}) &=& f(5) = 22\end{array} \]

Figure 5.17: \(f(x) = x^2-3, 0 \leq x \leq 6\)

The Riemann sum formed by adding the products \(f(u_{i}) \Delta x_{i}\) for \(i=1,2,3,4\) is\[\sum\limits_{i=1}^{4}f(u_{i}) \Delta x_{i}=-\dfrac{11}{4}\cdot 1 + \left(-\dfrac{3}{4}\right) \cdot 1 + 6 \cdot 2 + 22 \cdot 2 =\dfrac{105}{2}=52.5 \]

NOW WORK

Problem 9.

Suppose a function \(f\) is defined on a closed interval \([a,b],\) and we partition the interval \([a,b]\) into \(n\) subintervals \[\lbrack x_{0}, x_{1}], \lbrack x_{1}, x_{2}], \lbrack x_{2}, x_{3}], \ldots , [x_{i-1},x_{i}], \ldots , \lbrack x_{n-1}, x_{n}]\]

where \[a=x_{0}\lt x_{1}\lt x_{2}\lt\cdots \lt x_{i-1} \lt x_{i}\lt\cdots \lt x_{n-1}\lt x_{n}=b\]

These subintervals are not necessarily of the same length. Denotethe length of the first interval by \(\Delta x_{1}=x_{1}-x_{0}\), the lengthof the second interval by \(\Delta x_{2}=x_{2}-x_{1}\), and so on. In general,the length of the \(i\)th subinterval is\[\boxed{\bbox[#FAF8ED,5pt]{\Delta x_{i}=x_{i}-x_{i-1}}}\]

355

for \( i=1, 2, \ldots , n\). This set of subintervals of the interval\([a,b]\) is called a partition of \([a,b]\). The length of the largest subinterval in a partition is called the norm ofthe partition and is denoted by \(\max \Delta x_{i}\).

DEFINITION Definite Integral

Let \(f\) be a function defined on the closed interval \([a,b]\).Partition \([a,b] \) into \(n\) subintervals of length \(\Deltax_{i}\,{=}\,x_{i}\,{-}\,x_{i-1},\) \(i\,{=}\,1,2, \ldots , n.\)Choose a number \(u_{i}\) in each subinterval, evaluate \(f(u_{i}),\) and form the Riemann sums \({\sum\limits_{i=1}^{n}}{{f(u_{i})}}\Delta x_{i}.\) If \(\lim\limits_{{\max \Delta x}_{i}\rightarrow 0}{\sum\limits_{i=1}^{n}} {{f(u_{i})}}\Delta x_{i}=I\)exists and does not depend on the choice of the partition or on thechoice of \(u_{i}\), then the number \(I\) is called theRiemann integral or definiteintegral of \(f\) from \(a\) to \(b\) and is denoted by the symbol \(\int_{a}^{b} f (x)\,dx.\) That is,\[\boxed{\bbox[5pt]{\int_{a}^{b} f (x) dx=\lim\limits_{\max \Delta x_{i}\rightarrow 0}{\sum\limits_{i=1}^{n}} f(u_{i}) \Delta x_{i}}}\]

When the above limit exists, then we say that \(f\) isintegrable over \([a,b].\)

For the definite integral \({\int_{a}^{b}{f(x)\,dx}}\), the number \(a\) is called the lower limit of integration, the number \(b\) is called the upper limit of integration, the symbol \(\int\) (anelongated \(S\) to remind you of summation) is called the integral sign,\(f(x)\) is called the integrand, and \(dx\) is the differential of theindependent variable \(x\). The variable used in the definite integral is an artificial or a dummy variable because it may be replacedby any other symbol. For example,\[\int_{a}^{b} {f(x)\,dx} \qquad \int_{a}^{b} {f(t)\,dt} \qquad \int_{a}^{b}{f(s)\,ds}\qquad\int_{a}^{b} {f(\theta)\,d\theta}\]

all denote the definite integral of \(f\) from \(a\) to \(b\), andif any of them exist, they are all equal to the same number.

EXAMPLE 2 Expressing the Limit of Riemann Sums as a Definite Integral

  1. Find the Riemann sums for \(f(x) =x^{2}-3\) on the closedinterval \([0,6] \) for a partition of \([0,6] \) into \(n\)subintervals of length \(\Delta x_{i}=x_{i}-x_{i-1},\) \(i=1,2,\ldots , n.\)
  2. Assuming that the limit of the Riemann sums exists as \(\max\Delta x_{i}\rightarrow 0\), express the limit as a definiteintegral.

Solution (a) The Riemann sums for \(f(x)=x^{2}-3\)on the closed interval \([0,6] \) are\[\sum\limits_{i=1}^{n}f(u_{i}) \Delta x_{i}=\sum\limits_{i=1}^{n}\big( u_{i}^{2}-3\big) \Delta x_{i}\]

where \([0,6]\) is partitioned into \(n\) subintervals \([x_{i-1},x_{i}] \), and \(u_{i}\) is some number in the subinterval\([x_{i-1}, x_{i}]\), \(i=1,2, \ldots, n\).

(b) Since \(\lim\limits_{\max \Delta x_{i}\rightarrow0}\sum\limits_{i=1}^{n}\big( u_{i}^{2}-3\big) \Delta x_{i}\) exists,then \[\lim\limits_{\max \Delta x_i \rightarrow 0}\sum\limits_{i=1}^{n}\big(u_{i}^{2}-3\big) \Delta x_{i}=\int_{0}^{6}( x^{2}-3) dx \]

NOW WORK

Problem 15.

356

If \(f\) is integrable over [\(a,b\)], then \(\lim\limits_{{\max \Delta x}_{i} \rightarrow 0}{\sum\limits_{i=1}^{n}} f (u_{i}) \Delta x_{i}\) exists for anychoice of \(u_{i}\) in the \(i\)th subinterval, so we are free tochoose the \(u_{i}\) any way we please. The choices could be the left endpointof each subinterval, or the right endpoint, or the midpoint, or any othernumber in each subinterval. Also, \(\lim\limits_{{\max \Delta x}_{i} \rightarrow 0}{\sum\limits_{i=1}^{n}} f(u_{i})\Delta x_{i}\) is independent of the partition of the closed interval \([a,b]\),provided \({\max \Delta x}_{i}\) can be made as close as we please to\(0\). It is these flexibilities that make the definite integral so important inengineering, physics, chemistry, geometry, and economics.

In defining the definite integral \({\int_{a}^{b} {f(x)\,dx,}}\) we assumedthat \(a\lt b\). To remove this restriction, we give the followingdefinitions.

DEFINITION

  • If \(f(a)\) is defined, then\[\boxed{\bbox[5pt]{\int_{a}^{a}f (x)\,dx=0 }}\]
  • If \(a > b\) and if \(\int_{b}^{a}f (x)\,dx\) exists, then \[\boxed{\bbox[5pt]{\int_{a}^{b}f (x) dx=-\int_{b}^{a}f (x) dx }}\]

IN WORDS

Interchanging the limits of integration reverses the sign of thedefinite integral.

For example, \[\int_{1}^{1} x^{2}\,dx=0 \qquad \text{ and }\qquad \int_{3}^{2} x^{2}\,dx=-\int_{2}^{3} x^{2}\,dx\]

Next we give a condition on the function \(f\) that guarantees\(f\) is integrable. The proof of this result may be found in advanced calculus texts.

THEOREM Existence of the Definite Integral

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

IN WORDS

If a function \(f\) is continuous on \([a,b]\), then it isintegrable over \([a,b]\). But if \(f\) is not continuous on\([a,b]\), then \(f\) may or may not be integrable over \([a,b].\)

The two conditions of the theorem deserve special attention. First,\(f\) is defined on a closed interval, and second, \(f\) iscontinuous on that interval. There are some functions that are continuous on an openinterval (or even a half-open interval) for which the integral does notexist. For example, although \(f(x)=\dfrac{1}{x^{2}}\) is continuouson \((0,1)\) (and on \((0,1]\)), the definite integral\({\int_{0}^{1}}\dfrac{{1}}{x^{2}} dx\) does not exist. Also, thereare many examples of discontinuous functions for which the integralexists. (See Problems 64 and 65.)

5.3.2 Find a Definite Integral Using the Limit of Riemann Sums

Suppose \(f\) is integrable over the closed interval [\(a,b\)]. To find\[\int\nolimits_{a}^{b}f(x)\, dx=\lim\limits_{\max\Delta x_{i}\rightarrow 0}\ \sum\limits_{i=1}^{n} f(u_{i})\Delta x_{i} \]

using Riemann sums, we usually partition \([a,b]\) into \(n\)subintervals, each of the same length \(\Delta x=\dfrac{b-a}{n}\). Such a partition is called a regular partition. For aregular partition, the norm of the partition is\[\boxed{\bbox[#FAF8ED,5pt]{{\max \Delta x}_{i}={\dfrac{b-a}{n}}}}\]

357

Since \(\lim\limits_{n\rightarrow \infty }\dfrac{b-a}{n}=0,\) it follows thatfor a regular partition, the two statements\[{\max \Delta x}_{i} {\rightarrow 0}\qquad \text{ and }\qquad {n\rightarrow \infty }\]

are interchangeable. As a result, for regular partitions \(\Delta x= \dfrac{b-a}{n}\),\[\boxed{\bbox[#FAF8ED,5pt]{\int_{a}^{b}{f(x)\,dx=}\lim\limits_{{\max \Delta x}_{i} \rightarrow0}{\sum\limits_{i=1}^{n}} {f(u_{i})} \Delta x_i=\lim\limits_{n\rightarrow \infty}{\sum\limits_{i=1}^{n}} {f(u_{i})}\Delta x }}\]

The next result uses Riemann sums to establish a formula to find thedefinite integral of a constant function.

THEOREM

If \(f(x) =h,\) where \(h\) is some constant, then\[\boxed{\bbox[5pt]{\int_{a}^{b}f(x)\,dx=\int_{a}^{b}h\,dx=h(b-a)} }\]

Proof

The constant function \(f(x) =h\) is continuous on the set of realnumbers and so is integrable. We form the Riemann sums for \(f\) onthe closed interval \([a,b] \) using a regular partition. Then\(\Delta x_{i}=\Delta x=\dfrac{b-a}{n}\), \(i=1,2, \ldots , n.\) The Riemann sums of \(f\) on the interval \([a,b] \) are\[\begin{eqnarray*}\sum\limits_{i=1}^{n}f(u_{i}) \Delta x_{i} &=& \sum\limits_{i=1}^{n}f(u_{i}) \Delta x \underset{\underset{\color{#0066A7}{\hbox{\(f(x) =h\)}}}{\color{#0066A7}{\uparrow }}} {=} \sum\limits_{i=1}^{n}h\,\Deltax=\sum\limits_{i=1}^{n}h\left( \dfrac{b-a}{n}\right) \\& =& h\left(\dfrac{b-a}{n}\right) \sum\limits_{i=1}^{n}1=h\left( \dfrac{b-a}{n}\right) \cdot n=h(b-a)\end{eqnarray*}\]

Then\[\lim\limits_{n\rightarrow \infty }\sum\limits_{i=1}^{n}f(u_{i}) \Delta x=\lim\limits_{n\rightarrow \infty} [h(b-a)] =h(b-a)\]

So,\[\int_{a}^{b}h\,dx=h(b-a)\]

For example, \[\begin{eqnarray*}\begin{array}{@{\hspace*{-4pc}}l}\int_{1}^{2}3\,dx=3(2-1) =3 \qquad \int_{2}^{6}dx=1(6-2) =4 \qquad \int_{-3}^{4} (-2)\, dx=(-2) [4- (-3 )] =-14\end{array}\end{eqnarray*} \]

NOW WORK

Problem 23.

EXAMPLE 3 Finding a Definite Integral Using the Limit of Riemann Sums

Find \({\int_{0}^{3}{(3x -8)}}\,dx\).

Solution Since the integrand \(f(x) = 3x - 8\) is continuous onthe closed interval \([0, 3]\), the function \(f\) is integrable over\([0, 3].\) Although we can use any partition of \([0,3] \)whose norm can be made as close to \(0\) as we please, and we canchoose any \(u_{i}\) in each subinterval\(,\) we use a regularpartition and choose \(u_{i}\)as the right endpoint of each subinterval. This will result in a simpleexpression for the Riemann sums.

358

We partition \([0,3]\) into \(n\) subintervals, each of length\(\Delta x=\dfrac{3-0}{n}=\dfrac{3}{n}\). The endpoints of each subinterval of the partition,written in terms of \(n\), are \[\begin{eqnarray*}x_{0}&=&0,\ x_{1}=\dfrac{3}{n},\ {x_{2}=2}\left({\dfrac{3}{n}}\right), \ldots , x_{i-1}=(i-1)\left(\dfrac{3}{n}\right),\\x_{i} &=& i\left(\dfrac{3}{n}\right) ,\ldots , x_{n}=n\left(\dfrac{3}{n}\right)=3\end{eqnarray*}\]

The Riemann sums of \(f(x)=3x - 8\) from \(0\) to \(3\), using\(u_{i}= x_{i}= \dfrac{3i}{n}\) (the right endpoint) and \(\Deltax=\dfrac{3}{n},\) are 357\[\begin{eqnarray*}\sum\limits_{i=1}^{n}\,f(u_{i})\,\Delta x_{i}&=&\sum\limits_{i=1}^{n}\,f(x_{i})\,\Delta x \underset{\underset{{\color{#0066A7}{\hbox{\(\Delta x=\dfrac{3}{n}\)}}}}{\color{#0066A7}{\left\uparrow\right.}}} {=} \sum\limits_{i=1}^{n}(3x_{i}-8)\dfrac{3}{n} \underset{\underset{{\color{#0066A7}{\hbox{\(x_{i}=\dfrac{3i}{n}\)}}}}{\color{#0066A7}{\left\uparrow\right.}}} {=} \sum\limits_{i=1}^{n}\left[3\left( {\dfrac{{3i}}{{n}}}\right) -8\right] \dfrac{3}{n} \\&=& \sum \limits_{i=1}^{n}\left( \dfrac{27i}{n^{2}}-\dfrac{24}{n}\right) =\dfrac{27}{n^{2}}\sum\limits_{i=1}^{n}{i}-\dfrac{24}{n}\sum\limits_{i=1}^{n}1\\[5pt]&=&\dfrac{27}{n^{2}}\cdot \dfrac{n(n+1)}{{2}}-\dfrac{24}{{n}}\cdot n=\dfrac{27}{2}+\dfrac{27}{2n}-24 =-\dfrac{21}{2}+\dfrac{27}{2n}\end{eqnarray*}\]

Now \[{\int_{0}^{3}{(3x-8)\,dx=}}\lim\limits_{n\rightarrow \infty }\left( -\dfrac{21}{2}+\dfrac{27}{2n}\right) =-\dfrac{21}{2} \]

NOW WORK

Problem 45.

Figure 5.18 shows the graph of \(f(x)=3x - 8\) on \([0,3] \). Since\(f\) is not nonnegative on \([0,3] ,\) we cannot interpret\({\int_{0}^{3}{(3x - 8)\,dx}}\) as an area. The fact that the answeris negative is further evidence that this is not an area problem.When finding a definite integral, do not presume it represents area.As you will see in Section 5.7 and in Chapter 6, the definiteintegral has many interpretations. Interestingly enough, definiteintegrals are used to find the volume of a solid of revolution, thelength of a graph, the work done by a variable force, and otherquantities.

Figure 5.18: \(f(x) = 3x-8,0\leq x \leq 3\)

EXAMPLE 4 Interpreting a Definite Integral

Determine if each definite integral can be interpreted as an area. If itcan, describe the area; if it cannot, explain why.

  1. \(\int_{0}^{3\pi /4}\cos x\,dx\)
  2. \(\int_{2}^{10}\left \vert x-4\right\vert dx\)

Solution (a) See Figure 5.19 on page 359. Since\(\cos x\lt0\) on the interval \(\left(\dfrac{\pi}{2},\dfrac{3\pi}{4}\right],\) the integral \(\int_{0}^{3\pi /4}\cosx\,dx\) cannot be interpreted as area.

(b) See Figure 5.20. Since \(\vert x-4\vert \geq0\) on the interval \([2,10] \), the integral \(\int_{2}^{10}\vertx-4\vert \,dx\) can be interpreted as the area enclosed by the graph of \(y=\vert x-4\vert \), the \(x\)-axis, and the lines \(x=2\) and\(x=10.\)

359

Figure 5.19: \(f(x) = \cos x, 0 \leq x \leq \dfrac{3\pi}{4}\)
Figure 5.20: \(f(x) = \vert x- 4 \vert, 2 \leq x \leq 10\)

NOW WORK

Problem 33.

EXAMPLE 5 Finding a Definite Integral Using Technology

  1. Use a graphing utility to find \(\int_{1/2}^{4}\ln x\,dx\)
  2. Use a computer algebra system to find \(\int_{1/2}^{4}\lnx\,dx.\)

Solution (a) A graphing utility, such as a TI-84, provides only an approximate numerical answer to the integral \(\int_{1/2}^{4}\ln x\,dx.\) As shown in Figure 5.21,\(\int_{1/2}^{4}\ln\, x\,dx\approx 2.391751035\).

(b) Because a computer algebra system manipulatessymbolically, it can find an exact value of the definite integral. UsingMuPAD in Scientific WorkPlace, we get \[\int_{1/2}^{4}\ln x\,dx= \dfrac{17}{2}\ln 2-\dfrac{7}{2}\]

An approximate numerical value of the definite integral using MuPADis \(2.3918.\)

NOW WORK

Problem 43.