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都怕香港六合彩的嘴。似乎嘴皮薄的人才擅长能说会道，王秀的厚嘴却也擅长，可见是对人们观念的一种颠覆。呵呵。周晨晨笑着碰了一下落魄的王秀，这一碰，香港六合彩差点从凳子上掉下。香港六合彩……香港六合彩，夫唱妇随！王秀从大脑深处搜出这么一个词，听得相关的两个人脸红了半天也没人表示要为这句话追讨名誉损失。嘴里刚咬进一大口汉堡包的王峰，可谓真正的瞠目结舌。这话太刺激人，香港六合彩王秀的嘴也太拙不择言了。但汉堡包毕竟有个男子汉的汉字，又有堡垒的堡字，王峰撑开的嗓子眼开始反抗，话可不能这么说，夫唱妇随是贬义词，王秀你这是侮辱人，知道吗？就是。子允附和道，还是哥们说话在理。应该是嫁(又鸟)随(又鸟)，嫁狗随狗。子允料不到王峰会给自己留这么一手，毫无戒备地吃了个闷棍。王秀大笑，王峰却心有余悸地看着周晨晨，说完后香港六合彩才发现这话冲着香港六合彩的。其实香港六合彩根本没有瞄准目标是谁，只是想勾动板机一下，也有插句话的本意。周晨晨笑中带羞，与王峰略带歉意的目光交错。王峰见不是担心的表情，才放下悬着的心。见王峰这样开自己玩笑，子允好气又好笑，王峰用词跟王秀不同，意思却差不多。香港六合彩得报复一下，也好把王峰的话更好地融入玩笑的气氛，否则这话会未免显得太那个。香港六合彩想从王峰和王秀都姓王入手，再从秀和峰上深入，但香港六合彩两个人条件相差太多，效果上难以精彩。子允准备另辟蹊径，又苦于拘谨的王峰没留下什么破绽，香港六合彩想引诱香港六合彩，人得意忘形时才最容易出纰漏。沉默让饥饿的香港六合彩得以专心享受饱食的乐趣，子允的乐趣还在于咀嚼王秀和王峰说的话，一种朦胧的关系如果被人歪来正去地当众说开，很有种三级片女主角脱而不脱的感觉。子允喜欢这种感觉，这种情形在影视中常会出现，往往起着推波助澜作用。子允曾笑过香港六合彩傻，现在发现所谓的傻有时也有傻的妙处。不知王秀因为什么也闭上了嘴，在子允现在的印象中，好像王秀的嘴要么是说，要么是吃，反正不能又吃又说，或者吃着说着，所以香港六合彩的沉默跟赶紧吃盘子里还没吃掉多少的东西有关。王秀说话速度快，吃的速度也快。香港六合彩举起左手一看，眼皮立即抽过脂似的抬高许多，但直到咽干净嘴里的东西才说，看，十二点半了。时间真快，还有十分钟就要关校门了。子允更慌，香港六合彩还处在马老师的热点察看期。显然你是个乐观主义者，我敢说，老班已在班上点名了。王峰早可以吃完的，但越吃越慢，香港六合彩不愿吃完后看别人吃，也不愿给女生留下狼吞虎咽的印象。说过刚才那句话后，香港六合彩越想越觉得不合适，应该再说句什么缓和一下的，但当时没太意识到，想到后又不好再拿过来说。刚才的话虽说是当玩笑说的，但有失平时风格，这让香港六合彩懊恼不已。所以王秀一开口，香港六合彩便笑着接子允的茬，虽不太合时适，总算结合紧密。子允不知道王峰为什么会为自己不值得一笑的话配上这么多笑容，但也笑了笑。香港六合彩站起来看着周晨晨，香港六合彩也一笑，好像无所谓迟不迟到。这让子允安定不少，也让子允感动不小，因为这时香港六合彩越无所谓，就说明越对子允好感，越有同甘共苦的决心。那怎么办？香港六合彩怎么解释？
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Index FAQ
Areas of Domains and Definite Integrals
Area under a ParabolaDefinition of the Area of Certain DomainsArea under a Parabola RevisitedIntegrals and Antiderivatives
Mika Seppälä: Areas and Definite
IntegralsIndex FAQ
Estimate AreasConsider the problem of determining the area of the domain bounded by the graph of the function x2, the x-axis, and the lines x=0 and x=1.
As the number n of the approximating rectangles grows, the approximation gets better.
We determine the area by approximating the domain with thin rectangles for which the area can be directly computed. Letting these rectangles get thinner, the approximation gets better and, at the limit, we get the area of the domain in question.
10
y=x2
Mika Seppälä: Areas and Definite
IntegralsIndex FAQ
Estimate Areas (2)2
1
1 1The total area of the rectangles is .
n
nk
ks
n n
Height of the kth rectangle.
Length of the bottom.
Let A denote the actual area of the domain in question. Clearly sn<A for all n.
2
2 2
The approximation uses rectangles of
1 1height over the interval , .
Replace these rectangles with rectangles of height
1 to get the upper estimate
n
nk
s
k k k
n n n
k kS
n n n
1
.n
Lower est. sn
Upper est. Sn
Mika Seppälä: Areas and Definite
IntegralsIndex FAQ
Estimate Areas (3)
2 2
1 1
To compute the area we have now the estimates ,
1 1 1i.e., .
n n
n n
n nk k
A s A S
k ks A S
n n n n
1
Observe that, in this case, 0.
Hence lim lim .
n n n
n nn n
S sn
S s A
2 32
1 1
To compute the limit observe that
1 1 .
n n
nk k
kS k
n n n
This can be computed directly using a previously derived formula for the sum of squares. Solution follows.
Mika Seppälä: Areas and Definite
IntegralsIndex FAQ
Estimate Areas (4)3 2
2
1
Recall that .3 2 6
n
k
n n nk
32
21
Hence
1 1 1 1 1.
3 2 6 3
n
n nk
S kn n n
The blue area under the curve y=x2 over the interval [0,1] equals 1/3.
10
y=x2
Conclude
Mika Seppälä: Areas and Definite
IntegralsIndex FAQ
Definition of the Area of a Domain under the Graph of a Function
Theorem
1
1
Assume that f is continuous and f 0 for . Then
1 lim min f
1 lim max f
n
nk
n
nk
x a x b
k b a k b a b ax a x a
n n n
k b a k b a b ax a x a
n n n
The previous considerations were based on some intuitive idea about areas of domain. We make that precise in the statement of the following result.
Definition
The common value of these limits is the area of the domain under the graph of the function f and over the interval [a,b].
Mika Seppälä: Areas and Definite
IntegralsIndex FAQ
The Integral (1)Theorem
1
1
Assume that f is continuous on the interval , . Then
1 lim min f
1 lim max f
n
nk
n
nk
a b
k b a k b a b ax a x a
n n n
k b a k b a b ax a x a
n n n
Definition
Notation fb
a
x dx
The common value of these limits is the integral of the function f over the interval [a,b].
Mika Seppälä: Areas and Definite
IntegralsIndex FAQ
The Integral (2)Remark
1
1
Observe that, by the previous Theorem, for a continuous function f
1lim min f
lim f f ,
where the points satisfy
n
nk
bn
knk a
k
k b a k b ax a x a
n n n
b ax x dx
n
x a
1
.k
k b a k b ax a
n n
This means that the points xk may be freely selected from the intervals [a + (k-1)/n, a + k/n]. The limit of the sum does not depend on the choice of the points xk.
Mika Seppälä: Areas and Definite
IntegralsIndex FAQ
The Integral (3)
Remark
1
Using the notation , and choosing
-- -- the points from
1the intervals
it does not matt
,
we get
er how
lim f f .
k
bn
knk a
b ax
nx
k b a k b aa a
n n
x x x dx
Mika Seppälä: Areas and Definite
IntegralsIndex FAQ
Examples (1)Example 1
2
0
Compute .b
x dx
Solution
22
10
limb n
nk
bk bx dx
n n
By the Definition
3
23
1
limn
nk
bk
n3 3 2 3 3 3 3
3 2lim lim .
3 2 6 3 2 6 3n n
b n n n b b b b
n n n
Conclude
32
0
.3
b bx dx
Now use the formula for the sum of squares
Mika Seppälä: Areas and Definite
IntegralsIndex FAQ
Examples (2)Example 2
2Compute .b
a
x dx
Solution
2
2 2 2
0 0
Interpreting the integral as the area
under the graph of the function
we observe that
.b b a
a
y x
x dx x dx x dx
Conclude
3 32 2 2
0 0
.3 3
b b a
a
b ax dx x dx x dx
a0 b
The red area under the graph of x2 over [a,b] equals the area over [0,b] minus the area over [0,a].
Mika Seppälä: Areas and Definite
IntegralsIndex FAQ
Integrals and Antiderivatives
3 32
3 3
x
a
x at dt
Rewriting the previous result in the following form
we observe that the integral defines the function
3 3
2F .3 3
x
a
x ax t dt
Direct differentiation yields F’(x) = x2, i.e., the function F is an antiderivative of the function f(x) = x2.
Theorem
Let f be a continuous function. The function
F f
is an antiderivative of the function f, i.e., F f .
x
a
x t dt
x x
A proof of this result will be presented later.
