정적분과 급수(구분구적법)

• Solution #1: Set $x_i^*=-2+\frac{2i}{n}$. Then $a=x_0=-2$ and $b=x_n=0$ and $\Delta x=\frac{2}{n}$. So

$$\begin{aligned} \lim_{n\rightarrow\infty} \sum_{i=1}^n\frac{2}{n}\sqrt{4-\left(-2+\frac{2i}{n}\right)^2} &= \lim_{n\rightarrow\infty} \sum_{i=1}^n f(x_i^*)\Delta x\qquad \text{ with $f(x) = \sqrt{4-x^2}$ and $\Delta x = \frac{2}{n}$ } \\ &=\int_{-2}^0 \sqrt{4-x^2}\,\,\mathrm{d}x \end{aligned}$$

For the integral $\int_{-2}^0 \sqrt{4-x^2}\,\,\mathrm{d}x$, $y=\sqrt{4-x^2}$ is equivalent to $x^2+y^2=4$, $y\ge 0$. So the integral represents the area between the upper half of the circle $x^2+y^2=4$ (which has radius $2$) and the $x$-axis with $-2\le x\le 0$, which is a quarter circle with area $\frac{1}{4}\cdot \pi\, 2^2 = \pi$.

• Solution #2: Set $x_i^*=\frac{2i}{n}$. Then $a=x_0=0$ and $b=x_n=2$ and $\Delta x=\frac{2}{n}$. So

$$\begin{aligned} \lim_{n\rightarrow\infty} \sum_{i=1}^n\frac{2}{n}\sqrt{4-\left(-2+\frac{2i}{n}\right)^2} &= \lim_{n\rightarrow\infty} \sum_{i=1}^n f(x_i^*)\Delta x\quad \text{ with $f(x) = \sqrt{4-(-2+x)^2}$, $\Delta x = \frac{2}{n}$ } \\ &=\int_0^2 \sqrt{4-(-2+x)^2}\,\,\mathrm{d}x \end{aligned}$$

For the integral $\int_{0}^2 \sqrt{4-(-2+x)^2}\,\,\mathrm{d}x$ , $y=\sqrt{4-(x-2)^2}$ is equivalent to $(x-2)^2+y^2=4$, $y\ge 0$. So the integral represents the area between the upper half of the circle $(x-2)^2+y^2=4$ (which is centered at $(2,0)$ and has radius $2$) and the $x$-axis with $0\le x\le 2$, which is a quarter circle with area $\frac{1}{4}\cdot \pi\, 2^2 = \pi$.