1. 11.1 Area Between Curves Ap Calculus 14th Edition
2. 11.1 Area Between Curves Ap Calculus Formulas
3. 11.1 Area Between Curves Ap Calculus Algebra
4. 11.1 Area Between Curves Ap Calculus Calculator

While much more can be said about sequences, we now turn to ourprincipal interest, series. Recall that a series, roughly speaking, isthe sum of a sequence: if $ds{a_n}_{n=0}^infty$ is a sequence then theassociated series is$$sum_{i=0}^infty a_n=a_0+a_1+a_2+cdots$$Associated with a series is a second sequence, called the sequence of partial sums $ds{s_n}_{n=0}^infty$:$$s_n=sum_{i=0}^n a_i.$$So$$s_0=a_0,quad s_1=a_0+a_1,quad s_2=a_0+a_1+a_2,quad ldots$$A series converges if the sequence of partial sums converges, and otherwise the series diverges.

11.1 Area Between Curves Ap Calculus 14th Edition

Example 11.2.1 If $ds a_n=kx^n$, $dssum_{n=0}^infty a_n$ is called a geometric series.A typical partial sum is$$s_n=k+kx+kx^2+kx^3+cdots+kx^n=k(1+x+x^2+x^3+cdots+x^n).$$We note that$$eqalign{ s_n(1-x)&=k(1+x+x^2+x^3+cdots+x^n)(1-x)cr &=k(1+x+x^2+x^3+cdots+x^n)1-k(1+x+x^2+x^3+cdots+x^{n-1}+x^n)xcr &=k(1+x+x^2+x^3+cdots+x^n-x-x^2-x^3-cdots-x^n-x^{n+1})cr &=k(1-x^{n+1})cr}$$so$$eqalign{ s_n(1-x)&=k(1-x^{n+1})cr s_n&=k{1-x^{n+1}over 1-x}.cr}$$If $|x|< 1$, $dslim_{ntoinfty}x^n=0$ so$$ lim_{ntoinfty}s_n=lim_{ntoinfty}k{1-x^{n+1}over 1-x}= k{1over 1-x}.$$ Thus, when $|x|< 1$ the geometric series converges to $k/(1-x)$. When, for example, $k=1$ and $x=1/2$:$$ s_n={1-(1/2)^{n+1}over 1-1/2}={2^{n+1}-1over 2^n}=2-{1over 2^n} quadhbox{and}quad sum_{n=0}^infty {1over 2^n} = {1over 1-1/2} = 2.$$We began the chapter with the series$$sum_{n=1}^infty {1over 2^n},$$namely, the geometric series without the first term $1$. Each partialsum of this series is 1 less than the corresponding partial sum for the geometric series, so of course the limit is also one less than thevalue of the geometric series, that is,$$sum_{n=1}^infty {1over 2^n}=1.$$

It is not hard to see that the following theorem follows fromtheorem 11.1.2.

(a) Compute the area enclosed within the ice cream cone (including the scoop por-tion). (b) After a bite is taken from the top, the remaining area is enclosed by y= 6jxj, y= 16 x2, and y= x2 + 12. Compute the area of the remaining portion. Consider the region R shown below: The area of the region Ris equivalent to Z 1 1 1 1 + x2 dx. The area under a general curve, you need calculus. Each of these situations involves the same general strategy—the reformulation of precalculus mathematics through the use of a limit process. So, one way to answer the question “What is calculus?” is to say that calculus is a “limit machine” that involves three stages.

Area between Curves Notesheet 01 Completed Notes World Bank Data for Income Inequality Investigation 01 N/A Area between Curves Homework 01 - HW Solutions Integrals as Net Change Practice 02 Solutions Area between Curves Practice 02 Solutions INC and Area Review Homework 02 - HW Solutions VSoR Disk Method Notesheet 03. Lesson 1: Area Between Curves. And (a) Plot both functions on the same axes. (b) Find the area of the region enclosed between the curves from x = 0 to x = 6. 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.

Theorem 11.2.2 Suppose that $sum a_n$ and $sum b_n$ are convergent series,and $c$ is a constant. Then

1. $dssum ca_n$ is convergent and $dssum ca_n=csum a_n$

2. $dssum (a_n+b_n)$ is convergent and $dssum (a_n+b_n)=sum a_n+sum b_n$.

The two parts of this theorem are subtly different. Suppose that $suma_n$ diverges; does $sum ca_n$ also diverge if $c$ is non-zero? Yes:suppose instead that $sum ca_n$ converges; then by the theorem, $sum(1/c)ca_n$ converges, but this is the same as $sum a_n$, which byassumption diverges. Hence $sum ca_n$ also diverges. Note that we areapplying the theorem with $a_n$ replaced by $ca_n$ and $c$ replaced by$(1/c)$.

Now suppose that $sum a_n$ and $sum b_n$ diverge; does$sum (a_n+b_n)$ also diverge? Now the answer is no: Let $a_n=1$ and$b_n=-1$, so certainly $sum a_n$ and $sum b_n$ diverge. But$sum (a_n+b_n)=sum(1+-1)=sum 0 = 0$. Of course, sometimes $sum (a_n+b_n)$ will also diverge, for example, if $a_n=b_n=1$, then$sum (a_n+b_n)=sum(1+1)=sum 2$ diverges.

In general, the sequence of partial sums $ds s_n$ is harder to understandand analyze than the sequence of terms $ds a_n$, and it is difficultto determine whether series converge and if so to what. Sometimesthings are relatively simple, starting with the following.

Theorem 11.2.3 If $sum a_n$ converges then $dslim_{ntoinfty}a_n=0$.

Proof.
Since $sum a_n$ converges, $dslim_{ntoinfty}s_n=L$ and $dslim_{ntoinfty}s_{n-1}=L$, because this really says the samething but 'renumbers' the terms. Bytheorem 11.1.2, $$ lim_{ntoinfty} (s_{n}-s_{n-1})= lim_{ntoinfty} s_{n}-lim_{ntoinfty}s_{n-1}=L-L=0.$$But$$ s_{n}-s_{n-1}=(a_0+a_1+a_2+cdots+a_n)-(a_0+a_1+a_2+cdots+a_{n-1}) =a_n,$$so as desired $dslim_{ntoinfty}a_n=0$.

This theorem presents an easy divergence test: if given a series $suma_n$ the limit $dslim_{ntoinfty}a_n$ does not exist or has a valueother than zero, the series diverges. Note well that the converse isnot true: If $dslim_{ntoinfty}a_n=0$ then the series doesnot necessarily converge.

Example 11.2.4 Show that $dssum_{n=1}^infty {nover n+1}$ diverges.

We compute the limit:$$lim _{ntoinfty}{nover n+1}=1not=0.$$Looking at the first few terms perhaps makes it clear that the serieshas no chance of converging:$${1over2}+{2over3}+{3over4}+{4over5}+cdots$$will just get larger and larger; indeed, after a bit longer the seriesstarts to look very much like $cdots+1+1+1+1+cdots$, and of courseif we add up enough 1's we can make the sum as large as we desire.

Example 11.2.5 Show that $dssum_{n=1}^infty {1over n}$ diverges.

Here the theorem does not apply: $dslim _{ntoinfty} 1/n=0$, so itlooks like perhaps the series converges. Indeed, if you have thefortitude (or the software) to add up the first 1000 terms you will find that$$sum_{n=1}^{1000} {1over n}approx 7.49,$$so it might be reasonable to speculate that the series converges tosomething in the neighborhood of 10. But in fact the partial sums do goto infinity; they just get big very, very slowly. Consider thefollowing:

$ds 1+{1over 2}+{1over 3}+{1over 4} > 1+{1over 2}+{1over 4}+{1over 4} = 1+{1over 2}+{1over 2}$

$ds 1+{1over 2}+{1over 3}+{1over 4}+{1over 5}+{1over 6}+{1over 7}+{1over 8} > 1+{1over 2}+{1over 4}+{1over 4}+{1over 8}+{1over 8}+{1over 8}+{1over 8} = 1+{1over 2}+{1over 2}+{1over 2}$

$ds 1+{1over 2}+{1over 3}+cdots+{1over16}>1+{1over 2}+{1over 4}+{1over 4}+{1over 8}+cdots+{1over 8}+{1over16}+cdots +{1over16} =1+{1over 2}+{1over 2}+{1over 2}+{1over 2}$

11.1 Area Between Curves Ap Calculus Formulas

and so on. By swallowing up more and more terms we can always manageto add at least another $1/2$ to the sum, and by adding enough ofthese we can make the partial sums as big as we like. In fact, it'snot hard to see from this pattern that$$1+{1over 2}+{1over 3}+cdots+{1over 2^n} > 1+{nover 2},$$so to make sure the sum is over 100, for example, we'd addup terms until we get to around $ds 1/2^{198}$, that is,about $ds 4cdot 10^{59}$ terms. This series, $sum (1/n)$, is called theharmonic series.

Exercises 11.2

Ex 11.2.1Explain why $dssum_{n=1}^infty {n^2over 2n^2+1}$diverges.(answer)

Ex 11.2.2Explain why $dssum_{n=1}^infty {5over 2^{1/n}+14}$diverges.(answer)

Ex 11.2.3Explain why $dssum_{n=1}^infty {3over n}$diverges.(answer)

11.1 Area Between Curves Ap Calculus Algebra

Ex 11.2.4Compute $dssum_{n=0}^infty {4over (-3)^n}- {3over 3^n}$. (answer)

11.1 Area Between Curves Ap Calculus Calculator

Ex 11.2.5Compute $dssum_{n=0}^infty {3over 2^n}+ {4over 5^n}$. (answer)

Ex 11.2.6Compute $dssum_{n=0}^infty {4^{n+1}over 5^n}$.(answer)

Ex 11.2.7Compute $dssum_{n=0}^infty {3^{n+1}over 7^{n+1}}$.(answer)

Ex 11.2.8Compute $dssum_{n=1}^infty left({3over 5}right)^n$.(answer)

Ex 11.2.9Compute $dssum_{n=1}^infty {3^nover 5^{n+1}}$.(answer)