5.5 缺陷综合:里曼苏姆的局限性

章节大纲

• 选择节常规

  折叠 展开

  常规

  全部折叠 展开全部

  • 选择活动新闻通告

    

    新闻通告 讨论区
• 选择节缺陷综合:里曼山的局限性

  折叠 展开

  缺陷综合:里曼山的局限性

  播放段落

  Approximating the area under a function curve by summing a finite number of rectangles in a Riemann Sum can yield very accurate results. Intuitively, we know, however, that the more sub-intervals we have the better the result. Taking the limit of the Riemann Sum as the subintervals get smaller (number of rectangles gets larger) should asymptotically give the true area. For some function curves, the Riemann limit can be evaluated algebraically; for complex curves, the area can only be determined by brute force numerical computations of Riemann Sums.
  ::通过在 Riemann Sum 中混合一定数量的矩形来将区域置于函数曲线之下, 从而接近此区域, 可以得出非常准确的结果 。 但是,我们知道, 直观地说, 越是次间距的结果越好。 当次间距越小( 矩形数目越大) 时, 使用Riemann Sum 的限度, 就会给真实区域带来非抽象的表达。 对于某些函数曲线, Riemann 的界限可以进行代数评估; 对于复杂的曲线, 区域只能由 Riemann Sum 的粗力数字计算来确定 。

  播放段落

  Limits and Reimann Sums
  ::限额和 Reimann 金额

  播放段落

  Earlier, the area under a curve was defined in terms of a limit of sums:
  ::此前,曲线下的区域按金额限额界定:

  播放段落

  $$\begin{align*}A=\lim_{n \to + \infty}S(P)=\lim_{n \to + \infty}T(P)\end{align*}$$
  ::A=limnS( P)=limnT( P)

  播放段落

  where
  ::何 地

  播放段落

  $$\begin{align*}S(P) &=\sum_{1}^n m_i (x_i-x_{i-1})=m_1(x_1-x_0)+m_2(x_2-x_1)+ \ldots+m_n(x_n-x_{n-1}), \\ T(P) &=\sum_{1}^n M_i(x_i-x_{i-1})=M_1(x_1-x_0)+M_2(x_2-x_1)+\ldots+M_n(x_n-x_{n-1}),\end{align*}$$
  ::S(P) = n1mi(xI-xI-1) = m1(x1-x0) +m2(x2-x1) +m2(x2-x1) +...+mn(xn-xn-1)、T(P) = n1MI(xI-xI-1) = M1(x1-x0) +M2(x2-x1) +...+mn(xn-xn-1) +...(xn-xn-1) +mn(xn-xn-1)

  播放段落

  $\begin{align*}S(P)\end{align*}$  and $\begin{align*}T(P)\end{align*}$  are examples of Riemann Sums .
  ::S(P)和T(P)是Riemann Summs的例子。

  播放段落

  In general, Riemann Sums are of form $\begin{align*}\sum\limits_{i=1}^n f(x_i^*) \triangle x\end{align*}$  where each $\begin{align*}x_i^*\end{align*}$  is the value we use to find the length of the rectangle in the  $\begin{align*}i^{th}\end{align*}$ sub-interval. For example, the maximum function value in each sub-interval to find the upper sums and the minimum function in each sub-interval to find the lower sums. But since the function is continuous, we could have used any points within the sub-intervals to find the limit.
  ::一般来说, Riemann Sums 是形式 ni=1f(x*i) {xx, 其中每个 x*i 是我们用来在 Ish 子interval 中查找矩形长度的值。 例如, 每个子interval 中的最大函数值, 以查找每个子interval 中的最大数值和最小函数, 以查找较低数额 。 但是, 由于函数是连续的, 我们本可以在子interval 中使用任何点来找到限制值 。

  播放段落

  To make use of the , we make the width of each rectangle approach 0, which is equivalent to making the number of rectangles, $\begin{align*}n\end{align*}$ , approach infinity. By doing so, we find the exact ,
  ::要使用 , 我们使每个矩形方向 0 的宽度, 相当于 矩形, n 的宽度, 无穷无尽。 这样, 我们发现准确的,

  播放段落

  $$\begin{align*}\lim_{n \to \infty} A_n=\lim_{n \to \infty} \sum_{i=1}^n f(x_i)\triangle x.\end{align*}$$
  ::

  播放段落

  We now define the most general situation as follows:
  ::我们现在将最普遍的情况确定如下:

  播放段落

  If $\begin{align*}f\end{align*}$  is continuous on $\begin{align*}[a,b]\end{align*}$ , and:
  ::如果 f 是 [a,b] 和:

  播放段落
  1. The interval $\begin{align*}[a,b]\end{align*}$  is divided into  $\begin{align*}n\end{align*}$ sub-intervals of equal width $\begin{align*}\triangle x\end{align*}$ , with $\begin{align*}\triangle x=\frac{b-a}{n}\end{align*}$ , and
     ::间距 [a,b] 分为相等宽度的 n 次隔热量 x, 与 x=b-an, 和
  播放段落
  2. The endpoints of these sub-intervals are $\begin{align*}x_0=a,x_1,x_2,..., x_n=b\end{align*}$ , and
     ::这些次迭代的终点是 x0=a,x1,x2,...,xn=b,以及
  播放段落
  3. $\begin{align*}x_1^*,x_2^*, \ldots ,x_n^*\end{align*}$ are any sample points in these sub-intervals, then the definite integral of  $\begin{align*}f\end{align*}$ from $\begin{align*}x=a\end{align*}$  to $\begin{align*}x=b\end{align*}$  is
     ::x*1, x**2,..., x*n 是这些次间隔中的任何样本点, 那么 f 从 x=a 到 x=b 的确定组成部分就是

  播放段落

  $$\begin{align*}\int\limits_{a}^{b}f(x)dx=\lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \triangle x.\end{align*}$$
  :x)dx=limnni=1f(x*i)x。

  播放段落

  provided the limit exists.
  ::前提是存在限制。

  播放段落

  If the above limit exists, $\begin{align*}f\end{align*}$  is said to be integrable on the closed interval  $\begin{align*}[a,b]\end{align*}$ and the definite integral exists.
  ::如果存在上述限制,f据说在封闭间隔[a,b]上是无法加固的,而且确定的整体部分是存在的。

  播放段落

  Note that the sample point $\begin{align*}x_i^*\end{align*}$  can be any sample point in the $\begin{align*}i\end{align*}$ -th sub-interval, with common choices being right, or mid-point, or left.
  ::请注意,样本点 x**i 可以是 i-th 子interval 中的任何样本点, 共同的选择是右, 或中点, 或左。

  播放段落

  For example, evaluate the Riemann Sum for $\begin{align*}f(x)=x^3\end{align*}$  from $\begin{align*}x=0\end{align*}$  to $\begin{align*}x=3\end{align*}$  using $\begin{align*}n=6\end{align*}$  sub-intervals, and take the sample points to be the midpoints of the sub-intervals.
  ::例如,使用 n=6 亚intervals, 评估 f(x)=x3 f(x)=x3, 从 x=0 到 x=3, 并使用 n=6 亚intervals, 将抽样点作为次intervals 的中点。

  播放段落

  If we partition the interval [0, 3] into $\begin{align*}n=6\end{align*}$  equal sub-intervals, then each sub-interval will have length $\begin{align*}\frac{3-0}{6}=\frac{1}{2}\end{align*}$ . So we have $\begin{align*}\triangle x=\frac{1}{2}\end{align*}$  and
  ::如果我们将间隔 [0, 3] 分隔为 n=6 相等的次间隔, 那么每个次间隔的长度 3-06=12。 所以我们有 \ x=12 和

  

  
  

  播放段落

  R 6 = 6 ∑ 1 f ( x ∗ i ) △ x = f ( 0.25 ) △ x + f ( 0.75 ) △ x + f ( 1.25 ) △ x + f ( 1.75 ) △ x + f ( 2.25 ) △ x + f ( 2.75 ) △ x = ( 1 64 ) ( 1 2 ) + ( 27 64 ) ( 1 2 ) + ( 125 64 ) ( 1 2 ) + ( 343 64 ) ( 1 2 ) + ( 729 64 ) ( 1 2 ) + ( 1331 64 ) ( 1 2 ) = 2556 64 = 39.93.

  $$\begin{align*}R_6 &=\sum_{1}^6 f(x_i^*)\triangle x=f(0.25)\triangle x+f(0.75)\triangle x+f(1.25)\triangle x+f(1.75)\triangle x+f(2.25)\triangle x+f(2.75)\triangle x \\ &= \left(\frac{1}{64} \right) \left(\frac{1}{2} \right) + \left (\frac{27}{64} \right) \left (\frac{1}{2} \right) + \left (\frac{125}{64} \right) \left (\frac{1}{2} \right) + \left (\frac{343}{64} \right) \left (\frac{1}{2} \right) + \left (\frac{729}{64} \right) \left (\frac{1}{2} \right) + \left (\frac{1331}{64} \right) \left (\frac{1}{2} \right) \\ &= \frac{2556}{64}=39.93.\end{align*}$$
  ::R6=61f(x*i) x=f(0.25) x+f(0.75) x+f(1.75)x+f(1.25) x+f(2.25) x+f(2.75) x+f(1.64(12) +(2764) (12)+(12564) (12)+(34364) +(12)+(72964) (12)+(133164)(12)=255664=39.93)。

  播放段落

  Now let’s compute the definite integral using the definition. We want to evaluate $\begin{align*}\int\limits_{0}^{3} x^3 dx\end{align*}$ .
  ::现在让我们使用定义来计算确定的整体。 我们要评估 30x3dx 。

  播放段落

  and show that   $\begin{align*}\int\limits_0^3 x^3 dx=\lim\limits_{n \to \infty} \sum\limits_{i=1}^n f(x_i^*) \triangle x\end{align*}$ .
  ::并显示 30x3dx=limnnni=1f(x*i)x。

  播放段落

  We will use right endpoints to compute the integral. First need to divide [0, 3] into  $\begin{align*}n\end{align*}$ sub-intervals of length $\begin{align*}\triangle x=\frac{3-0}{n}=\frac{3}{n}\end{align*}$ . Since we are using right endpoints,
  ::我们将使用右端点来计算元件。 首先需要将 [ 0, 3] 分隔为 n 长度的 n 子端点 {x=3- 0n=3n。 既然我们使用正确的端点,

  播放段落

  $\begin{align*}x_0=0,x_1=\frac{3}{n}, x_2=\frac{6}{n}, \ldots , x_i=\frac{3i}{n}.\end{align*}$
  ::x0=0,x1=3n,x2=6n,...xI=3in。

  播放段落

  The definite integral is then evaluated as follows:
  ::然后对确定的整体性作如下评估:

  播放段落

  $$\begin{align*}\int\limits_{0}^{3} x^3 dx&= \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \triangle x. \qquad \ \ldots \text{The definite integral as the limit of Riemann sum.} \\ &= \lim_{n \to \infty} \sum_{i=1}^n f \left(\frac{3i}{n} \right) \left(\frac{3}{n} \right) \quad \ldots \text{Use sub-intervals of equal length and right side endpoints.} \\ &= \lim_{n \to \infty} \left(\frac{3}{n} \right) \sum_{i=1}^n \left(\frac{3i}{n} \right)^3 \\ &= \lim_{n \to \infty} \left(\frac{3}{n} \right)^4 \sum_{i=1}^n i^3 \qquad \quad \ldots \text{Makes use of the following:} \\ &= \lim_{n \to \infty} \left(\frac{3}{n} \right)^4 \left(\frac{n(n+1)}{2} \right )^2 \sum_{i=1}^n i^3=\left(\frac{n(n+1)}{2} \right)^2 \\ &= \left (\frac{3^4}{4} \right) \cdot \lim_{n \to \infty} \left(\frac{1}{n^4} \right) \left(\frac{n(n+1)}{1} \right)^2 \\ &= \left (\frac{3^4}{4} \right) \cdot \lim_{n \to \infty} \left(1+ \frac{1}{n} \right)^2 \\ \int\limits_{0}^{3} x^3 dx &=\left (\frac{3^4}{4} \right) \end{align*}$$
  ::=0x3dx=limnnnn}i=1f(x*i)x. 确定的整体性为 Riemann sum. =limnnnnni=1(3in)3=1(3in)3=limn(3n)4ni=1i3... 使用以下值:=limn(3n)44(n(+1)2) 2ni=1i3=(n+1)2) 2=344{(1n)(n+1)1}2=(344)_(1+1)2=(344)1}=(344)\limn(1+1n)23_0x3dx=(344)

  播放段落

  Hence  $\begin{align*}\int\limits_0^3 x^3 dx=\frac{81}{4}\end{align*}$ .
  ::因此, 30x3dx=814。

  播放段落

  Notice that the value $\begin{align*}x=3\end{align*}$ , could have be some $\begin{align*}x=c\end{align*}$ , where  $\begin{align*}c\end{align*}$ is a constant  $\begin{align*}> 0\end{align*}$ , so that we would have $\begin{align*}\int\limits_0^c x^3 dx=\frac{c^4}{4}\end{align*}$ .
  ::注意值 x=3, 可能是某种 x=c, c 是常数 > 0, 这样我们就会有 c0x3dx=c44 。

  播放段落

  The following additional definitions support the definition of the definite integral:
  ::以下补充定义支持确定的整体性定义:

  播放段落

  If $\begin{align*}f(x)\end{align*}$  is an integrable function on the closed interval $\begin{align*}[a,b]\end{align*}$ , then:
  ::如果 f(x) 是封闭间隔[a,b] 上的不可调试函数,则:

  播放段落
  1. Definition:  $\begin{align*}\int\limits_a^a f(x)dx=0\end{align*}$ if $\begin{align*}f(a)\end{align*}$  exists.
     ::定义: aaf(x)dx=0,如果f(a) 存在的话。
  播放段落
  2. Definition: If $\begin{align*}b > a\end{align*}$ , then $\begin{align*}\int\limits_a^a f(x)dx=- \int\limits_a^b f(x)dx\end{align*}$
     ::定义: 如果 b>a, 那么 a\\\ f( x) dx\\\ b\\\ b\\ f( x) dx

  播放段落

  In the discussion up to this point, the Riemann sums have been tied to the area under a curve. But, the idea of definite integrals as giving the area under a curve can be a bit confusing since sometimes the results do not make sense when interpreted as areas.
  ::在讨论过程中,Riemann的金额一直与曲线下的区域挂钩。 但是,确定的整体体将区域置于曲线下的想法可能有点混乱,因为有时当被解读为区域时结果不合理。

  播放段落

  For example, if we were to compute the definite integral $\begin{align*}\int\limits_{-3}^{3}x^3 dx\end{align*}$ , then due to the symmetry of $\begin{align*}f(x)=x^3\end{align*}$ about the origin, we would find that $\begin{align*}\int\limits_{-3}^{3}x^3 dx=0\end{align*}$ . This is because for every sample point  $\begin{align*}x_j^*\end{align*}$ , we also have  $\begin{align*}-x_j^*\end{align*}$ as a sample point with $\begin{align*}f(-x_j^*)=-f(x_j^*)\end{align*}$ . Hence, it is more accurate to say that  $\begin{align*}\int\limits_{-3}^{3}x^3 dx\end{align*}$ gives us the net area between $\begin{align*}x=-3\end{align*}$  and  $\begin{align*}x=3\end{align*}$ . If we wanted the total area bounded by the graph and the  $\begin{align*}x\end{align*}$ -axis, then we would compute $\begin{align*}2 \int\limits_0^3 x^3 dx=\frac{81}{2}\end{align*}$ .
  ::例如,如果我们计算33x3dx,那么由于对准了来源的 f(x) =x3,我们就会发现 33x3dx=0。这是因为对于每个样本点 x*j,我们也有 -x*j 作为样本点,有 f(-x*j)\f(x*j)\f(x*j)。因此,更准确地说, 33x3dx给了我们x*3 和 x=3 之间的净区域。如果我们想要图表和x轴所约束的总面积,那么我们就会计算出 230x3dx=812。

  播放段落

  As you might suspect from the problems shown so far in this concept, algebraically finding the limit at infinity of a Riemann Sum can get complicated very quickly. The solutions in the examples we used were possible because of a few summation identities. But most problems become intractable if we attempt to use limits. This is why the numerical approach using a calculator (computer) is better in most cases. However, it’s satisfying to be able to show that the Riemann sum approach can yield an algebraic solution.
  ::正如你可能从这个概念中迄今所显示的问题中猜想的那样,代数地发现里曼山无限的极限会很快变得复杂。 我们使用的例子中的解决方案之所以可能,是因为几个相加特征。 但是,如果我们试图使用限制,大多数问题就会变得棘手。 这就是为什么在多数情况下使用计算器(计算机)的数字方法会更好。然而,能够证明里曼之和方法能够产生代数解决方案是令人满意的。

  播放段落

  Examples
  ::实例

  播放段落

  Example 1
  ::例1

  播放段落

  Use a Riemann Sum to determine the area under the curve $\begin{align*}f(x)=3x^2-x+7\end{align*}$  over the interval [0, 1].
  ::使用 riemann um 来确定在 [0, 1] 间隔范围内的曲线 f(x) = 3x2 - x+7 下的面积。

  播放段落

  We need to find  $\begin{align*}\int\limits_0^1 (3x^2-x+7)dx=\lim\limits_{n \to \infty} \sum\limits_{i=1}^n f(x_i^*) \triangle x\end{align*}$
  ::我们需要找到 10( 3x2 - x+7) dx=limnni=1f( x*i) x

  播放段落

  We will use right endpoints to compute the integral. First divide [0, 1] into $\begin{align*}n\end{align*}$  sub-intervals of length $\begin{align*}\triangle x=\frac{1-0}{n}=\frac{1}{n}\end{align*}$ . Since we are using right endpoints, $\begin{align*}x_0=0,x_1=\frac{1}{n},x_2=\frac{2}{n},...,x_i=\frac{i}{n}\end{align*}$ . 
  ::我们将使用右端点来计算积分。 第一次将 [ 0, 1] 分隔为长度为 {x=1- 0n=1n的 n次间距。 既然我们使用右端点, x0=0, x1=1n, x2=2n,..., xi=in 。

  播放段落

  The definite integral is then evaluated as follows:
  ::然后对确定的整体性作如下评估:

  播放段落

  $$\begin{align*}\int\limits_0^1 (3x^2-x+7)dx &=\lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \triangle x. \quad \ldots \text{The definite integral as the limit of Riemann sum}. \\ \int\limits_0^1 (3x^2-x+7)dx &= \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \triangle x. \quad \ldots \text{The definite integral as the limit of Riemann sum.} \\ &=\lim_{n \to \infty} \sum_{i=1}^n f \left(\frac{i}{n} \right) \frac{1}{n} \quad \ \ \ldots \text{Use sub-intervals of equal length and right side endpoints.} \\ &=\lim_{n \to \infty} \left(\frac{1}{n} \right) \sum_{i=1}^n \left[3 \left(\frac{i}{n} \right)^2- \left(\frac{i}{n} \right)+7 \right] \\ &=\lim_{n \to \infty} \left[3 \left(\frac{1}{n} \right)^3 \sum_{i=1}^n i^2 - \left(\frac{1}{n} \right)^2 \sum_{i=1}^n i+ \left(\frac{7}{n} \right) \sum_{i=1}^n 1 \right] \quad \ldots \text{Makes use of the following:} \\ &=\lim_{n \to \infty} \left[\frac{3}{n^3} \left(\frac{n^3}{3}+ \frac{n^2}{2}+ \frac{n}{6} \right)- \frac{1}{n^2} \left(\frac{n^2}{2}+ \frac{n}{2} \right)+ \left(\frac{7}{n} \right) n \right] \quad \quad \sum_{i=1}^n i=\frac{n(n+1)}{2} \ \text{and} \\ &= \lim_{n \to \infty} \left[1+ \frac{3}{2n}+ \frac{1}{2n^2}- \frac{1}{2}- \frac{1}{2n}+7 \right] \quad \quad \sum_{i=1}^n i^2=\left(\frac{n(n+1)(2n+1)}{6} \right) \\ &= \lim_{n \to \infty}\left[7.5+ \frac{1}{n}+ \frac{1}{2n^2} \right] \\ \int\limits_{0}^{1}(3x^2-x+7)dx &= 7.5\end{align*}$$
  ::=0( 3x-2- x+7)\\xxxxxxxxxxxxxx... 确定的整体性为 riemann sum. = limn 尼尼尼 i= 1f(x)xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx...确定的整体性为 riemann sum.= limnn{= limnn{{{{尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼, 2 , 和尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼尼,

  播放段落

  Hence, the limit of the Riemann Sum is $\begin{align*}\int\limits_0^1 (3x^2-x+7)dx=7.5\end{align*}$ .
  ::因此,Riemann Sum的限额为10(3x2-x+7)dx=7.5。

  播放段落

  Review
  ::回顾

  播放段落

  For  #1-6, use the definition of the definite integral as the limit of a Riemann Sum to compute the areas under the curves.
  ::对于#1-6,使用确定的整体体的定义作为Riemann Sum的限度来计算曲线下的区域。

  播放段落
  1. $\begin{align*}\int\limits_3^7 4 dx\end{align*}$ .
     ::734dx 734dx.
  播放段落
  2. $\begin{align*}\int\limits_0^5 2x dx\end{align*}$ .
     ::502xxxxx.
  播放段落
  3. $\begin{align*}\int\limits_0^1 3x^2 dx\end{align*}$ .
     ::103x2dxx.
  播放段落
  4. $\begin{align*}\int\limits_0^2 3x^3 dx\end{align*}$ .
     ::203x3dx.
  播放段落
  5. $\begin{align*}\int\limits_2^4 (3x^2+5x+1)dx\end{align*}$ .
     ::*================================================================================================================================================ ============================================================================================================================================================================================================================================================================================================================================================================
  播放段落
  6. $\begin{align*}\int\limits_0^2 3x^4 dx\end{align*}$ (Hint: Use the relationship $\begin{align*}\sum\limits_{i=1}^n i^4=\frac{n(n+1)(6n^3+9n^2+n-1)}{2 \cdot 15}\end{align*}$  in your limit).
     ::03x4dx (提示: 在您的限制范围内使用 ni=1i4=n( n+1)( 6n3+9n2+n- 1) 215 关系) 。

  播放段落

  For  #7-12, write each of the following limits as a definite integral over the given interval where  $\begin{align*}x_i\end{align*}$ is a point in the $\begin{align*}i\end{align*}$ -th subinterval:
  ::对于 # 7-12, 请写下以下的每个限制, 作为给定间隔中xi 是 i-th 次隔热点的一个点的确定内含值 :

  播放段落
  7. $\begin{align*}\lim\limits_{n \to \infty} \sum\limits_{i=1}^n 8x{_i}^2 \left(\frac{3}{n} \right)\end{align*}$  over [7, 10].
     ::超过[7, 10], =18xx2(3n)。
  播放段落
  8. $\begin{align*}\lim\limits_{n \to \infty} \sum\limits_{i=1}^n \sqrt[4]{x{_i}^3+2x{_i}^2+1\left(\frac{2}{n} \right)}\end{align*}$  over [3, 5].
     ::超过[3、5]的立方公尺=14xxx3+2xx2+1(2n)。
  播放段落
  9. $\begin{align*}\lim\limits_{n \to \infty} \sum\limits_{i=1}^n \frac{4}{(2x_i+1)^2} \left(\frac{6}{n} \right)\end{align*}$  over [1, 7].
     ::在 [1, 7] 以上, limnni=14( 2xi+1)2( 6n) 。
  播放段落
  10. $\begin{align*}\lim\limits_{n \to \infty} \sum\limits_{i=1}^n x{_i}^2 e^{-x_i} \left(\frac{4}{n} \right)\end{align*}$  over [0, 4].
      ::大于[0, 4] 的limnni=1xx2e-xxi(4n)
  播放段落
  11. $\begin{align*}\lim\limits_{n \to \infty} \sum\limits_{i=1}^n (x{_i}^3+3x{_i}^2+3x_i+1)\left(\frac{3}{n} \right)\end{align*}$  over [2, 5].
      ::在 [2, 5] 以上, limnni=1(x13+3xII+3xI2+3xI+1)(3n) 以上。
  播放段落
  12. $\begin{align*}\lim\limits_{n \to \infty} \sum\limits_{i=1}^n x{_i}^2 \cos x_i \left(\frac{\pi}{2n}\right)\end{align*}$  over $\begin{align*}\left[0, \frac{\pi}{2} \right]\end{align*}$ .
      ::大于[0,%2] 的立方公尺=1xx2cosxi(xxxxxxx(xxxxxxxxxxxx(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
  播放段落
  13. Consider  $\begin{align*}f(x)=x+3\end{align*}$ from  $\begin{align*}x=0\end{align*}$ to $\begin{align*}x=5\end{align*}$ . Use geometry to calculate the exact area of the region under the graph. (Hint: Sketch a graph of the region and compute its area using formulas from geometry.)
      ::考虑 f(x) = x+3 从 x=0 到 x=5 。 使用几何来计算图下区域的确切区域 。 (提示: 绘制一个区域图, 使用几何公式计算其区域 。 )
  播放段落
  14. Repeat problem #13 using the definition of the definite integral to calculate the exact area of the region under the graph of  $\begin{align*}f(x)=x+3\end{align*}$ from  $\begin{align*}x=0\end{align*}$ to $\begin{align*}x=5\end{align*}$ .
      ::重复问题 # 13 , 使用定义确定的组成部分来计算 f( x) =x+ 3 的图形下区域的确切面积, 从 x=0 到 x=5 。
  播放段落
  15. Use the definition of the definite integral and function symmetry to show that  $\begin{align*}\int\limits_{-2}^{3}2(x^2-x)dx=2 \int\limits_{\frac{1}{2}}^{3}2(x^2-x)dx\end{align*}$ for  $\begin{align*}f(x)=2(x^2-x)\end{align*}$ over the interval [-2, 3].
      ::使用确定的整体和函数对称定义来显示 322(x2-xx) dx=23122(x2-x) dx) 用于 f(x) =2(x2) (x2-x) 的间隔[- 3] 。

  播放段落

  Review (Answers)
  ::回顾(答复)

  播放段落

  Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。