上图展示了当 x≥0x \ge 0 时 y=f(x)y = \text{f}(x) 的函数图像简图。在曲线下方,在区间 1≤x≤31 \le x \le 3 上画有 nn 个宽度均为 hh 的矩形。当每个矩形绕 xx 轴旋转 2π2\pi 弧度时,将得到一个圆柱形薄片。这 nn 个圆柱形薄片的总体积 V1V_1 可以用于估算体积 VV,其中 VV 是由该曲线、直线 x=1x = 1 和 x=3x = 3 以及 xx 轴所围成的区域绕 xx 轴旋转 2π2\pi 弧度时所产生的实际体积。

  1. (a)

    证明这 nn 个圆柱形薄片的总体积 V1V_1 由 V1=πh∑r=0n−1[f(1+rh)]2V_1 = \pi h \sum_{r=0}^{n-1} [\text{f}(1 + rh)]^2 给出。

    用 nn 表示 hh 的值。

  2. (b)

    从上图中可以观察到 V1<VV_1 < V。写出 V2V_2,一个类似于 V1V_1 且满足 V2>VV_2 > V 的表达式。

  3. (c)

    现在已知当 x≥0x \ge 0 时,f(x)=x21+ex\text{f}(x) = \dfrac{x^2}{\sqrt{1 + \text{e}^x}}。

    求 lim⁡n→∞V1\lim_{n \to \infty} V_1 的值。

  4. (d)

    求由 y=f(x)y = \text{f}(x) 与另一条方程为 (x−1)2+(y−3)2=9(x-1)^2 + (y-3)^2 = 9 的曲线(对于 y≤3y \le 3)围成的区域的面积。

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完整解法

10(a)

1. Express the volume of the cylindrical disc formed by the rectangle at index $r$.

For the rectangle corresponding to index rr (where r=0,1,2,…,n−1r = 0, 1, 2, \dots, n-1), its left edge is at x=1+rhx = 1 + rh.
The height of this rectangle is f(1+rh)\text{f}(1 + rh).
When rotated 2π2\pi radians about the xx-axis, it forms a cylindrical disc of radius f(1+rh)\text{f}(1 + rh) and thickness hh.
The volume of this single disc is π[f(1+rh)]2h\pi [\text{f}(1 + rh)]^2 h.

2. Sum the volumes of all $n$ discs to show the expression for $V_1$.

To find the total volume V1V_1, we sum the volumes of all nn discs from r=0r = 0 to r=n−1r = n-1:
V1=∑r=0n−1π[f(1+rh)]2hV_1 = \sum_{r=0}^{n-1} \pi [\text{f}(1 + rh)]^2 h
Factoring out the constants π\pi and hh gives:
V1=πh∑r=0n−1[f(1+rh)]2(Shown)V_1 = \pi h \sum_{r=0}^{n-1} [\text{f}(1 + rh)]^2 \quad \text{(Shown)}

3. Determine the value of $h$ in terms of $n$.

The region spans from x=1x = 1 to x=3x = 3, so the total width of the interval is 3−1=23 - 1 = 2.
Since there are nn rectangles of equal width hh, we have:
nh=2nh = 2
h=2nh = \frac{2}{n}

答案

h=2nh = \frac{2}{n}

10(b)

1. Identify the geometrical requirement for an overestimate

From the diagram, the curve y=f(x)y = \text{f}(x) is an increasing function. V1V_1 uses the left-hand endpoint of each subinterval (from r=0r=0 to n−1n-1), which leaves a gap under the curve, hence V1<VV_1 < V.
To create an overestimate (V2>VV_2 > V), we must use rectangles whose heights are determined by the right-hand endpoint of each subinterval.

2. Write the new expression using right-hand endpoints

The right-hand endpoints of the nn intervals occur at x=1+rhx = 1+rh for r=1,2,…,nr=1, 2, \dots, n.
Replacing the summation limits in V1V_1 to reflect these new evaluation points gives:
V2=πh∑r=1n[f(1+rh)]2V_2 = \pi h \sum_{r=1}^{n} [\text{f}(1 + rh)]^2

答案

V2=πh∑r=1n[f(1+rh)]2V_2 = \pi h \sum_{r=1}^{n} [\text{f}(1 + rh)]^2

10(c)

1. Relate the limit of the Riemann sum to the definite integral for the exact volume of revolution.

As n→∞n \to \infty, the width of each cylinder h→0h \to 0. The limit of the Riemann sum V1V_1 representing the total volume of the cylindrical discs corresponds to the exact volume VV of the solid of revolution.
lim⁡n→∞V1=V=π∫13[f(x)]2 dx\lim_{n \to \infty} V_1 = V = \pi \int_1^3 [\text{f}(x)]^2 \,dx

2. Substitute the given function and evaluate the integral using a graphing calculator.

Substitute f(x)=x21+ex\text{f}(x) = \frac{x^2}{\sqrt{1 + \text{e}^x}} into the integral:
lim⁡n→∞V1=π∫13(x21+ex)2 dx\lim_{n \to \infty} V_1 = \pi \int_1^3 \left( \frac{x^2}{\sqrt{1 + \text{e}^x}} \right)^2 \,dx
lim⁡n→∞V1=π∫13x41+ex dx\lim_{n \to \infty} V_1 = \pi \int_1^3 \frac{x^4}{1 + \text{e}^x} \,dx
Using a graphing calculator to evaluate the definite integral:
lim⁡n→∞V1≈12.291≈12.3 (to 3 s.f.)\lim_{n \to \infty} V_1 \approx 12.291 \approx 12.3 \text{ (to 3 s.f.)}

答案

12.312.3

10(d)

1. Express the equation of the second curve with $y$ as the subject.

(x−1)2+(y−3)2=9(x-1)^2 + (y-3)^2 = 9\n(y−3)2=9−(x−1)2(y-3)^2 = 9 - (x-1)^2\ny−3=±9−(x−1)2y - 3 = \pm\sqrt{9 - (x-1)^2}\nSince it is given that y≤3y \le 3, we must take the negative square root to represent the lower semicircle:\ny=3−9−(x−1)2y = 3 - \sqrt{9 - (x-1)^2}

2. Find the $x$-coordinates of the points of intersection.

Using a graphing calculator, we find the points of intersection between y=f(x)y = \text{f}(x) and the lower semicircle y=3−9−(x−1)2y = 3 - \sqrt{9 - (x-1)^2} for x≥0x \ge 0.\nx≈0.33838orx≈3.8744x \approx 0.33838 \quad \text{or} \quad x \approx 3.8744

3. Formulate the definite integral for the area and evaluate it using a graphing calculator.

Observe from the graphing calculator that the curve y=f(x)y = \text{f}(x) lies above the semicircle between the intersection points.\nArea=∫0.338383.8744(x21+ex−(3−9−(x−1)2))dx\text{Area} = \int_{0.33838}^{3.8744} \left( \frac{x^2}{\sqrt{1 + \text{e}^x}} - \left( 3 - \sqrt{9 - (x-1)^2} \right) \right) \text{d}x\nEvaluating this via the graphing calculator gives:\nArea≈3.00 (to 3 s.f.)\text{Area} \approx 3.00 \text{ (to 3 s.f.)}

答案

3.003.00

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