11.1 Area Between Curvesap Calculus

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Pre-Calculus 11 Chapter 1 – Sequences and Series Created by Ms. Lee 4 of 19 Reference: McGraw-Hill Ryerson Pre-Calculus 11 1.2 – Arithmetic Series Carl Friedrich Gauss, mathematician born in 1977: When Gauss was 10, his math teacher challenged the class to find the sum of the numbers from 1 to 100, thinking it will take some time. 2.3: Solving Two-Step and Multi-Step Equations. 2.4: Solving Equations with Variables on Both Sides. 2.5: Solving for a Variable.

While the idea of a sequence of numbers, $a_1,a_2,a_3,ldots$ isstraightforward, it is useful to think of a sequence as a function. Wehave up until now dealt with functions whose domains are the realnumbers, or a subset of the real numbers, like $f(x)=sin x$. Asequence is a function with domain the natural numbers$N={1,2,3,ldots}$ or the non-negative integers,$ds Z^{ge0}={0,1,2,3,ldots}$. The range of the function is stillallowed to be the real numbers; in symbols, we say that a sequence isa function $fcolon NtoR$. Sequences are written in a few differentways, all equivalent; these all mean the same thing:$$ displaylines{a_1,a_2,a_3,ldotscr left{a_nright}_{n=1}^inftycr left{f(n)right}_{n=1}^inftycr}$$

  • Path between A and B Fig. 11.1 (a) The shortest time route between A and B Fig. 11.1 (b) f g f f f Length of the curve Area Surface Area Volume Fig. 11.3 A B A B Perimeter 2(l+b) Area (1/2)bh l b b h Surface Area πrl l r Volume (4 /3)πr3 r Unit11.indd 183 18:40:30.
  • Calculus 2 Lia Vas Areas between Curves If f(x) and g(x) are two continuous functions de ned on the interval a;b such that f(x) g(x) for all x in a;b; then the area between the graphs of f and g on a;b is A = Z b a (f(x) g(x)) dx: Note that in this consideration the position of f and g with respect to x-axis is not relevant.

As with functions on the real numbers,we will most often encounter sequences that can be expressed by aformula. We have already seen the sequence $ds a_i=f(i)=1-1/2^i$, and othersare easy to come by:$$eqalign{ f(i)&={iover i+1}cr f(n)&={1over2^n}cr f(n)&=sin(npi/6)cr f(i)&={(i-1)(i+2)over2^i}cr}$$Frequently these formulas will make sense if thought of either asfunctions with domain $R$ or $N$, though occasionally one will makesense only for integer values.

Faced with a sequence we are interested in the limit$$lim_{ito infty} f(i) = lim_{itoinfty} a_i.$$We already understand$$lim_{xtoinfty} f(x)$$when $x$ is a real valued variable; now we simply want to restrict the'input' values to be integers. No real difference is required in thedefinition of limit, except that we specify, perhaps implicitly, thatthe variable is an integer. Compare this definition todefinition 4.10.4.

Definition 11.1.1 Suppose that $dsleft{a_nright}_{n=1}^infty$ is a sequence.We say that $ds lim_{nto infty}a_n=L$ if for every $epsilon>0$there is an $N > 0$ so that whenever $n>N$, $ a_n-L < epsilon$. If$ds lim_{ntoinfty}a_n=L$ we say that the sequence converges,otherwise it diverges.

If $f(i)$ defines a sequence, and $f(x)$ makes sense, and $ds lim_{xtoinfty}f(x)=L$, then it is clear that$ds lim_{itoinfty}f(i)=L$ as well, but it is important to note thatthe converse of this statement is not true. For example, since$ds lim_{xtoinfty}(1/x)=0$, it is clear that also$ds lim_{itoinfty}(1/i)=0$, that is, the numbers$${1over1},{1over2},{1over3},{1over4},{1over5},{1over6},ldots$$get closer and closer to 0. Consider this, however: Let $f(n)=sin(npi)$. This is the sequence$$ sin(0pi), sin(1pi),sin(2pi),sin(3pi),ldots=0,0,0,0,ldots$$since $sin(npi)=0$ when $n$ is an integer. Thus$ds lim_{ntoinfty}f(n)=0$. But $ds lim_{xtoinfty}f(x)$, when $x$ isreal, does not exist: as $x$ gets bigger and bigger, the values$sin(xpi)$ do not get closer and closer to a single value, but takeon all values between $-1$ and $1$ over and over. In general, wheneveryou want to know $ds lim_{ntoinfty}f(n)$ you should first attempt tocompute $ds lim_{xtoinfty}f(x)$, since if the latter exists it is alsoequal to the first limit. But if for some reason$ds lim_{xtoinfty}f(x)$ does not exist, it may still be true that $ds lim_{ntoinfty}f(n)$ exists, but you'll have to figure outanother way to compute it.

It is occasionally useful to think of the graph of a sequence. Sincethe function is defined only for integer values, the graph is just asequence of dots. In figure 11.1.1 we see thegraphs of two sequences and the graphs of the corresponding realfunctions.

11.1 Area Between Curvesap Calculus
Figure 11.1.1. Graphs of sequences and their corresponding real functions.

Not surprisingly, the properties of limits of real functions translateinto properties of sequences quite easily. Theorem 2.3.6 about limits becomes

Theorem 11.1.2 Suppose that $dslim_{ntoinfty}a_n=L$ and $dslim_{ntoinfty}b_n=M$ and$k$ is some constant. Then$$eqalign{&lim_{ntoinfty} ka_n = klim_{ntoinfty}a_n=kLcr&lim_{ntoinfty} (a_n+b_n) = lim_{ntoinfty}a_n+lim_{ntoinfty}b_n=L+Mcr&lim_{ntoinfty} (a_n-b_n) = lim_{ntoinfty}a_n-lim_{ntoinfty}b_n=L-Mcr&lim_{ntoinfty} (a_nb_n) = lim_{ntoinfty}a_ncdotlim_{ntoinfty}b_n=LMcr&lim_{ntoinfty} {a_nover b_n} = {lim_{ntoinfty}a_nover lim_{ntoinfty}b_n}={Lover M},hbox{ if $M$ is not 0}cr}$$

Likewise the Squeeze Theorem (4.3.1) becomes

Theorem 11.1.3 Suppose that $ds a_n le b_n le c_n$ for all $n>N$, for some $N$.If $dslim_{ntoinfty}a_n=dslim_{ntoinfty}c_n=L$, then $dslim_{ntoinfty}b_n=L$.

And a final useful fact:

Theorem 11.1.4 $dslim_{ntoinfty} a_n =0$ if and only if$dslim_{ntoinfty}a_n=0$.

This says simply that the size of $ds a_n$ gets close to zero if andonly if $ds a_n$ gets close to zero.

Example 11.1.5 Determine whether $dsleft{{nover n+1}right}_{n=0}^infty$ converges ordiverges. If it converges, compute the limit. Since this makes sensefor real numbers we consider$$lim_{xtoinfty}{xover x+1}=lim_{xtoinfty}1-{1over x+1}=1-0=1.$$Thus the sequence converges to 1.

Example 11.1.6 Determine whether $dsbigg{{ln nover n}bigg}_{n=1}^infty$ converges ordiverges. If it converges, compute the limit. We compute$$lim_{xtoinfty}{ln xover x}=lim_{xtoinfty}{1/xover 1}=0,$$using L'Hôpital's Rule. Thus the sequence converges to 0.

Example 11.1.7 Determine whether $ds{(-1)^n}_{n=0}^infty$ converges ordiverges. If it converges, compute the limit. This does not make sensefor all real exponents, but the sequence is easy to understand: it is$$1,-1,1,-1,1ldots$$and clearly diverges.

Example 11.1.8 Determine whether $ds{(-1/2)^n}_{n=0}^infty$ converges ordiverges. If it converges, compute the limit. We consider the sequence $ds{ (-1/2)^n }_{n=0}^infty={(1/2)^n}_{n=0}^infty$.Then$$ lim_{xtoinfty}left({1over2}right)^x=lim_{xtoinfty}{1over2^x}=0,$$so by theorem 11.1.4 the sequence converges to 0.

Example 11.1.9 Determine whether $ds{(sin n)/sqrt{n}}_{n=1}^infty$ converges ordiverges. If it converges, compute the limit. Since $ sin n le 1$, $ds 0le sin n/sqrt{n} le1/sqrt{n}$ and we can use theorem 11.1.3 with $ds a_n=0$ and $ds c_n=1/sqrt{n}$. Since$dslim_{ntoinfty} a_n=dslim_{ntoinfty} c_n=0$, $dslim_{ntoinfty}sin n/sqrt{n}=0$ and the sequence converges to 0.

Example 11.1.10 A particularly common and useful sequence is $ds {r^n}_{n=0}^infty$,for various values of $r$. Some are quite easy to understand: If $r=1$the sequence converges to 1 since every term is 1, and likewise if$r=0$ the sequence converges to 0. If $r=-1$ this isthe sequence of example 11.1.7 anddiverges. If $r>1$ or $r< -1$ the terms $ds r^n$ get large without limit,so the sequence diverges. If $0< r< 1$ then the sequence converges to0. If $-1< r< 0$ then $ds r^n = r ^n$ and $0< r < 1$, so the sequence$ds { r ^n}_{n=0}^infty$ converges to 0, so also $ds{r^n}_{n=0}^infty$ converges to 0.converges. In summary, $ds {r^n}$ converges precisely when$-1< rle1$ in which case$$ lim_{ntoinfty} r^n=cases{ 0& if $-1< r< 1$cr 1& if $r=1$cr}$$

Sometimes we will not be able to determine the limit of a sequence,but we still would like to know whether it converges. In some cases wecan determine this even without being able to compute the limit.

A sequence is called increasingor sometimes strictly increasing if $ds a_i< a_{i+1}$ for all$i$. It is called non-decreasing or sometimes(unfortunately) increasing if $ds a_ile a_{i+1}$ for all$i$. Similarly a sequence is decreasing if $ds a_i>a_{i+1}$ for all $i$and non-increasing if$ds a_ige a_{i+1}$ for all $i$. If a sequence has any of theseproperties it is called monotonic.

Example 11.1.11 The sequence$$ left{{2^i-1over2^i}right}_{i=1}^infty= {1over2},{3over4},{7over8},{15over16},ldots,$$is increasing, and$$ left{{n+1over n}right}_{i=1}^infty= {2over1},{3over2},{4over3},{5over4},ldots$$is decreasing.

A sequence is bounded aboveif there is some number $N$ such that $ds a_nle N$ for every $n$,and bounded below if there issome number $N$ such that $ds a_nge N$ for every $n$. If a sequenceis bounded above and bounded below it is bounded. If a sequence $ds{a_n}_{n=0}^infty$ is increasing or non-decreasing it is boundedbelow (by $ds a_0$), and if it is decreasing or non-increasing it isbounded above (by $ds a_0$). Finally, with all this new terminologywe can state an important theorem.

Theorem 11.1.12 If a sequence is bounded and monotonic then it converges.

We will not prove this; the proof appears in many calculus books. Itis not hard to believe: suppose that a sequence is increasing andbounded, so each term is larger than the one before, yet never largerthan some fixed value $N$. The terms must then get closer and closerto some value between $ds a_0$ and $N$. It need not be $N$, since $N$ maybe a 'too-generous' upper bound; the limit will be thesmallest number that is above all of the terms $ds a_i$.

Example 11.1.13 All of the terms $ds (2^i-1)/2^i$ are less than 2, and the sequence isincreasing. As we have seen, the limit of the sequence is 1—1 is thesmallest number that is bigger than all the terms in the sequence.Similarly, all of the terms $(n+1)/n$ are bigger than $1/2$, and thelimit is 1—1 is the largest number that is smaller than the terms ofthe sequence.

We don't actually need to know that a sequence is monotonic to applythis theorem—it is enough to know that the sequence is'eventually' monotonic, that is, that at some point it becomesincreasing or decreasing. For example, the sequence $10$, $9$, $8$,$15$, $3$, $21$, $4$, $3/4$, $7/8$, $15/16$, $31/32,ldots$ is notincreasing, because among the first few terms it is not. But startingwith the term $3/4$ it is increasing, so the theorem tells us that thesequence $3/4, 7/8, 15/16, 31/32,ldots$ converges. Since convergencedepends only on what happens as $n$ gets large, adding a fewterms at the beginning can't turn a convergent sequence into adivergent one.

Example 11.1.14 Show that $ds{n^{1/n}}$ converges.

We first show that this sequence is decreasing, that is, that $ds n^{1/n}>(n+1)^{1/(n+1)}$. Consider the real function $ds f(x)=x^{1/x}$ when$xge1$. We can compute the derivative, $ds f'(x)=x^{1/x}(1-ln x)/x^2$,and note that when $xge 3$ this is negative. Since the function hasnegative slope, $ds n^{1/n}>(n+1)^{1/(n+1)}$ when $nge 3$. Since all terms of the sequence arepositive, the sequence is decreasing and bounded when $nge3$, and sothe sequence converges. (As it happens, we can compute the limit inthis case, but we know it converges even without knowing the limit; seeexercise 1.)

Example 11.1.15 Show that $ds{n!/n^n}$ converges.

Again we show that the sequence is decreasing, and since each term ispositive the sequence converges. We can't take the derivative thistime, as $x!$ doesn't make sense for $x$ real. But we note that if $ds a_{n+1}/a_n < 1$ then $ds a_{n+1}< a_n$, which is what we want toknow. So we look at $ds a_{n+1}/a_n$:$$ {a_{n+1}over a_n} = {(n+1)!over (n+1)^{n+1}}{n^nover n!}= {(n+1)!over n!}{n^nover (n+1)^{n+1}}= {n+1over n+1}left({nover n+1}right)^n= left({nover n+1}right)^n < 1.$$(Again it is possible to compute the limit; seeexercise 2.)

Exercises 11.1

Ex 11.1.1Compute $dslim_{xtoinfty} x^{1/x}$.(answer)

Ex 11.1.2Use the squeeze theorem to show that $dslim_{ntoinfty} {n!over n^n}=0$.

Ex 11.1.3Determine whether $ds{sqrt{n+47}-sqrt{n}}_{n=0}^infty$ converges or diverges. If it converges, compute the limit.(answer)

Ex 11.1.4Determine whether $dsleft{{n^2+1over (n+1)^2}right}_{n=0}^infty$ converges or diverges. If it converges, compute the limit.(answer)

Ex 11.1.5Determine whether $dsleft{{n+47oversqrt{n^2+3n}}right}_{n=1}^infty$ converges or diverges. If it converges, compute the limit.(answer)

11.1 Area Between Curves Ap Calculus Algebra

Ex 11.1.6Determine whether $dsleft{{2^nover n!}right}_{n=0}^infty$ converges or diverges. (answer)

Recall that the area under a curve and above thex-axis can be computed by the definite integral. If we have two curves

( y = f(x) ) and ( y=g(x) )

such that

[ f(x) > g(x) nonumber]

then the area between them bounded by the horizontal lines (x = a) and (x = b) is

[ text{Area}=int_{c}^{b} left [ f(x) - g(x) right ] ;dx. nonumber]

To remember this formula we write

[ text{Area}=int_{a}^{b}text{(Top-Bottom)};dx nonumber]

Example 1

Find the area between the curves ( y=x^2) and (y=x^3).

Solution

First we note that the curves intersect at the points ((0,0)) and ((1,1)). Then we see that

[ x^3 < x^2 nonumber]

in this interval. Hence the area is given by

[begin{align*} int_{0}^{1} left( x^2 - x^3 right) dx &= {left[ frac{1}{3}x^3 - frac{1}{4}x^4 right]}_0^1 &= dfrac{1}{3} - dfrac{1}{4} &= dfrac{1}{12}. end{align*}]

Area Bounded by Two Functions of (y)

Example 2

Find the area between the curves ( x = 1 - y^2 ) and ( x = y^2-1 ).

Solution

Here the curves bound the region from the left and the right.

We use the formula

[ text{Area}=int_{c}^{b}text{(Right-Left)};dy. nonumber]

For our example:

[begin{align*} int_{-1}^{1}big[ (1-y^2)-(y^2-1) big] dy &= int_{-1}^{1}(2-y^2) dy &= left(2y-dfrac{2}{3}y^3right]_{-1}^1 &=big(2-dfrac{2}{3}big)-big(-2-dfrac{2}{3} big) &= dfrac{8}{3}. end{align*}]

Example 3

Find the area between the curves ( y =0 ) and (y = 3 left( x^3-x right) ).

Solution

When we graph the region, we see that the curves cross each other so that the top and bottom switch. Hence we split the integral into two integrals:

[begin{align*} int_{-1}^{0}big[ 3(x^3-x)-0big] dx +int_{0}^{1}big[0-3(x^3-x) big] dx &= left(dfrac{3}{4}x^4-dfrac{3x^2}{2}right]_{-1}^0 - left(dfrac{3}{4}x^4-dfrac{3x^2}{2}right]_0^1 &=big(-dfrac{3}{4}+dfrac{3}{2} big) - big(dfrac{3}{4}-dfrac{3}{2} big) &=dfrac{3}{2} end{align*}.]

Application

Let (y = f(x)) be the demand function for a product and (y = g(x)) be the supply function. Then we define the equilibrium point to be the intersection of the two curves. The consumer surplus is defined by the area above the equilibrium value and below the demand curve, while the producer surplus is defined by the area below the equilibrium value and above the supply curve.

11.1 Area Between Curves Ap Calculus Calculator

Example 4

Find the producer surplus for the demand curve

[ f(x) = 1000 - 0.4x^2 nonumber]

and the supply curve of

[ g(x) = 42x. nonumber]

Solution

We first find the equilibrium point:

11.1 Area Between Curves Ap Calculus Formulas

We set

[ 1000 - 0.4x^2 = 42x nonumber ]

or

[ 0.4x^2 + 42x - 1000 = 0. nonumber]

We get

[x=20 nonumber]

hence

[y=42(20)=840. nonumber]

We integrate

[ begin{align*} int_{0}^{20} left ( 840 - 42x right ) dx &= {left[ 840x-21x^2 right] }_0^{20} [4pt] &= 8400. end{align*}]

11.1 Area Between Curves Ap Calculus 14th Edition

Exercises

  1. Find the area between the curves ( y = x^2 ) and ( y =sqrt{x} ).
  2. Find the area between the curves ( y = x^2 - 4) and ( y = -2x ).
  3. Find the area between the curves ( y = 2/x ) and ( y = -x + 3 ).
  4. Find the area between the curves ( y = x3^x ) and ( y = 2x +1 ).

Contributors and Attributions

  • Larry Green (Lake Tahoe Community College)
  • Integrated by Justin Marshall.