정적분과 급수(구분구적법)

Using the definition of a right Riemann sum,

$$\begin{aligned} \displaystyle\sum_{i=1}^{10} 3(7+2i)^2\sin(4i) &= \displaystyle\sum_{i=1}^{10} \Delta x f(a+i\Delta x) \\ \text{Since $\Delta x = 10$ and $a=-5$,} \\ \displaystyle\sum_{i=1}^{10} 3(7+2i)^2\sin(4i) &= \displaystyle\sum_{i=1}^{10} 10 f(-5+10i) \\ \text{Dividing both expressions by 10,} \\ \displaystyle\sum_{i=1}^{10} \frac{3}{10}(7+2i)^2\sin(4i) &= \displaystyle\sum_{i=1}^{10} f(-5+10i) \\ \text{So, we have an expression for $f(-5+10i)$:} \\ f(-5+10i) &= \frac{3}{10}(7+2i)^2\sin(4i) \\ \text{In order to find $f(x)$, let $x=-5+10i$. Then $i=\frac{x}{10}+\frac{1}{2}$.} \\ f(x) &= \frac{3}{10}\left(7+2\left(\frac{x}{10}+\frac{1}{2}\right)\right)^2\sin\left(4\left(\frac{x}{10}+\frac{1}{2}\right)\right)\\ &=\frac{3}{10}\left(\frac{x}{5}+8\right)^2\sin\left(\frac{2x}{5}+2\right)\ . \end{aligned}$$

First, we note $y=\sqrt{1-x^2}$ is the upper half of a circle of radius 1, centred at the origin. We're taking the area under the curve from 0 to $a$, so the area in question is as shown in the picture below.

In order to use geometry to find this area, we break it up into two pieces: a sector of a circle, and a triangle, shown below.

• Area of sector: The sector is a portion of a circle with radius 1, with inner angle $\theta$. So, its area is $\frac{\theta}{2\pi}\left(\text{area of circle}\right) = \frac{\theta}{2\pi}\left(\pi\right) = \frac{\theta}{2}$.

Our job now is to find $\theta$ in terms of $a$. Note $\frac{\pi}{2}-\theta$ is the inner angle of the red triangle, which lies in the unit circle. So, $\cos\left(\frac{\pi}{2}-\theta\right)=a$. Then $\frac{\pi}{2}-\theta= \arccos(a)$, and so $\theta = \frac{\pi}{2} - \arccos(a)$.

Then the area of the sector is $\frac{\pi}{4} - \frac{1}{2}\arccos(a)$ square units.

• Area of triangle: The triangle has base $a$. Its height is the $y$-value of the function when $x=a$, so its height is $\sqrt{1-a^2}$. Then the area of the triangle is $\frac{1}{2}a\sqrt{1-a^2}$.

We conclude $\displaystyle\int_0^a \sqrt{1-x^2}\ \mathrm{d}x = \frac{\pi}{4} -\frac{1}{2} \arccos(a) + \frac{1}{2}a\sqrt{1-a^2}$.

(a) The function is being evaluated at $x_1 = 1.4$, $x_2 = 1.8$, $x_3 = 2.2$, $x_4 = 2.6$, and $x_5 = 3.0$. Thus, since we know this is a right Riemann sum, we are using $5$ subintervals with $\triangle x = 0.4$, and $x_0 = 1.0$. From the pattern, we observe that the function must be $f(x) = x^2 + 1$ on the interval $[1,3]$. (b) If $S$ is a left Riemann sum instead, then the $x_i$'s shift left: $x_0 = 1.4$, $x_1 = 1.8$, $\ldots$, $x_5 = 3.4$. In this setting, the function is identical, but we are considering it on the interval $[1.4, 3.4]$. If $S$ is a middle Riemann sum, then (for example) $1.4$ is the midpoint of an interval $\triangle x = 0.4$ units long, so $x_0 = 1.2$ and $x_1 = 1.6$. Continuing that trend, we'd be considering the function $f(x) = x^2 + 1$ on the interval $[1.2, 3.2]$. (c) $S$ is an approximation of the area that lies under $f(x) = x^2 + 1$ on $[1,3]$. (d) To use twice as many subintervals, we move to $\triangle x = 0.2$ and thus $x_1 = 1.2$, $x_2 = 1.4$, and so on, with the general right endpoint being given by $x_i = 1 + 0.2 \cdot i$. We'll also have $n = 10$, so $$R_{10} = \sum_{i=1}^{10} \left( (1+0.2i)^2 + 1 \right) \cdot 0.2\,.$$