04. 리만적분The Riemann Integral

Solution. The subintervals of the partition $P = \{0, 1.2, 1.5, 2.3, 2.6, 3\}$ are $$[0, 1.2],\quad [1.2, 1.5],\quad [1.5, 2.3],\quad [2.3, 2.6],\quad [2.6, 3],$$ five in total.

The lengths of the subintervals are $$\Delta x_1 = 1.2,\quad \Delta x_2 = 0.3,\quad \Delta x_3 = 0.8,\quad \Delta x_4 = 0.3,\quad \Delta x_5 = 0.4.$$

The norm $\| P \|$ is the maximum of these lengths, so $\| P \| = 1.2$.

Proof. The partition obtained by dividing $[a,b]$ into $n$ equal subintervals is $$P_n = \left\{ a,\ a+\frac{b-a}{n},\ a+\frac{2(b-a)}{n},\ \dots,\ a+\frac{k(b-a)}{n},\ \dots,\ b \right\},$$ that is, if we set $x_k = a + \dfrac{k(b-a)}{n}$ for $k=0,1,\dots,n$, then $P_n = \{x_0, x_1, \dots, x_n\}$.

The length of each subinterval $[x_{k-1}, x_k]$ is $$\Delta x_k = x_k - x_{k-1} = \frac{k(b-a)}{n} - \frac{(k-1)(b-a)}{n} = \frac{b-a}{n},$$ which is the same for all $k=1,\dots,n$, so the norm is $$\| P_n \| = \max_{1\le k\le n} \Delta x_k = \frac{b-a}{n}.$$

Since $b-a$ is a fixed constant, $$\lim_{n\to\infty}\| P_n \| = \lim_{n\to\infty}\frac{b-a}{n} = 0.$$

Solution. (1) Since $P$ is a partition of $[0,2]$ and $c_k \in [x_{k-1}, x_k]$, the sum $\displaystyle\sum_{k=1}^n c_k^2\,\Delta x_k$ is a Riemann sum for $f(x)=x^2$ on $[0,2]$. As $f$ is continuous on $[0,2]$, it is Riemann integrable, and by definition $$\lim_{\|P\|\to 0}\sum_{k=1}^n c_k^2\,\Delta x_k = \int_0^2 x^2\,dx.$$

(2) Likewise, since $P$ is a partition of $[0,\pi/4]$, the sum $\displaystyle\sum_{k=1}^n(\tan c_k)\,\Delta x_k$ is a Riemann sum for $g(x)=\tan x$ on $[0,\pi/4]$. As $g$ is continuous on $[0,\pi/4]$, $$\lim_{\|P\|\to 0}\sum_{k=1}^n(\tan c_k)\,\Delta x_k = \int_0^{\pi/4} \tan x\,dx.$$

Solution. (1) Dividing $[a,b]$ into $n$ equal parts gives subintervals of constant length $\Delta x = \dfrac{b-a}{n}$, and the midpoint of the $k$th subinterval $\left[a+(k-1)\dfrac{b-a}{n},\ a+k\dfrac{b-a}{n}\right]$ is $$a+(k-1)\frac{b-a}{n}+\frac{\Delta x}{2} = a+\frac{(2k-1)(b-a)}{2n},$$ so the expression $a+(2k-1)\dfrac{b-a}{2n}$ in the given limit is exactly this midpoint.

Hence $$\sum_{k=1}^n f\left(a+(2k-1)\frac{b-a}{2n}\right)\frac{1}{n} = \frac{1}{b-a}\sum_{k=1}^n f\left(a+(2k-1)\frac{b-a}{2n}\right)\cdot\frac{b-a}{n} = \frac{1}{b-a}\sum_{k=1}^n f(c_k)\,\Delta x_k,$$ which is $\dfrac{1}{b-a}$ times the midpoint Riemann sum for $f$. Since $f$ is continuous on $[a,b]$, as $n\to\infty$ the norm $\Delta x = (b-a)/n \to 0$, and this Riemann sum converges to $\displaystyle\int_a^b f(x)\,dx$. Therefore $$\lim_{n\to\infty}\sum_{k=1}^n f\left(a+(2k-1)\frac{b-a}{2n}\right)\frac{1}{n} = \frac{1}{b-a}\int_a^b f(x)\,dx.$$

(2) Let $N=n^2$; as $n\to\infty$ we also have $N\to\infty$. Dividing $[0,1]$ into $N$ equal parts, the $k$th subinterval is $\left[\dfrac{k-1}{N}, \dfrac{k}{N}\right]$ with length $\dfrac1N$. Taking the sample point $$c_k = \frac{k-1}{N}+\frac{1}{N+1},$$ since $0 \lt \dfrac{1}{N+1} \lt \dfrac1N$ we have $$\frac{k-1}{N} \le c_k \lt \frac{k-1}{N}+\frac1N = \frac{k}{N},$$ so $c_k$ lies in the $k$th subinterval. Thus the given sum is $$\sum_{k=1}^{n^2}\tan\left(\frac{k-1}{n^2}+\frac{1}{n^2+1}\right)\frac{2}{n^2} = 2\sum_{k=1}^{N}\tan(c_k)\cdot\frac1N,$$ which is twice a Riemann sum for $\tan x$ on $[0,1]$ with sample points $c_k$.

Since $\tan x$ is continuous on $[0,1]$ and the norm $1/N = 1/n^2 \to 0$ as $n\to\infty$, this Riemann sum converges to $\displaystyle\int_0^1 \tan x\,dx$. Therefore $$\lim_{n\to\infty}\sum_{k=1}^{n^2}\tan\left(\frac{k-1}{n^2}+\frac{1}{n^2+1}\right)\frac{2}{n^2} = 2\int_0^1 \tan x\,dx.$$

Proof. Consider an arbitrary partition $P=\{x_0, x_1, \dots, x_n\}$ of $[0,1]$. Each subinterval $[x_{k-1}, x_k]$ has positive length, so by the density of the rationals it contains a rational number, and by the density of the irrationals it also contains an irrational number.

Hence, on each subinterval, $$M_k = \sup_{x\in[x_{k-1},x_k]} f(x) = 1, \qquad m_k = \inf_{x\in[x_{k-1},x_k]} f(x) = 0,$$ since $f=1$ at the rational points and $f=0$ at the irrational points of the subinterval.

Therefore the upper sum and lower sum of $P$ are $$U(P,f) = \sum_{k=1}^n M_k\,\Delta x_k = \sum_{k=1}^n 1\cdot\Delta x_k = 1, \qquad L(P,f) = \sum_{k=1}^n m_k\,\Delta x_k = \sum_{k=1}^n 0\cdot\Delta x_k = 0.$$ This holds regardless of the choice of $P$, so the upper and lower integrals are $$\overline{\int_0^1} f = \inf_P U(P,f) = 1, \qquad \underline{\int_0^1} f = \sup_P L(P,f) = 0.$$

Since the upper and lower integrals differ ($1 \neq 0$), $f$ is not Riemann integrable on $[0,1]$.

Proof. Let $h=\dfrac{\pi}{2n}$; then $\displaystyle\lim_{n\to\infty}\sum_{k=1}^n \sin\left(\frac{\pi k}{2n}\right)\frac{\pi}{2n} = \lim_{n\to\infty} h\sum_{k=1}^n \sin(kh)$.

Substituting $m=n$ into the identity from the hint, $$\sum_{k=1}^n \sin(kh) = \frac{\cos(h/2)-\cos\left((n+\frac12)h\right)}{2\sin(h/2)}.$$ Here $h/2=\dfrac{\pi}{4n}$ and $$\left(n+\frac12\right)h = \left(n+\frac12\right)\frac{\pi}{2n} = \frac{\pi}{2}+\frac{\pi}{4n},$$ so $$\cos\left(\left(n+\frac12\right)h\right) = \cos\left(\frac{\pi}{2}+\frac{\pi}{4n}\right) = -\sin\left(\frac{\pi}{4n}\right).$$ Hence $$\sum_{k=1}^n \sin(kh) = \frac{\cos(\pi/4n)+\sin(\pi/4n)}{2\sin(\pi/4n)}.$$

Therefore the sum in question is $$h\sum_{k=1}^n \sin(kh) = \frac{\pi}{2n}\cdot\frac{\cos(\pi/4n)+\sin(\pi/4n)}{2\sin(\pi/4n)} = \frac{\pi/4n}{\sin(\pi/4n)}\left[\cos\left(\frac{\pi}{4n}\right)+\sin\left(\frac{\pi}{4n}\right)\right].$$

As $n\to\infty$, $\dfrac{\pi}{4n}\to 0$, and since $\displaystyle\lim_{t\to 0}\frac{t}{\sin t}=1$, $$\lim_{n\to\infty}\frac{\pi/4n}{\sin(\pi/4n)} = 1,$$ while continuity of $\cos$ and $\sin$ gives $$\lim_{n\to\infty}\left[\cos\left(\frac{\pi}{4n}\right)+\sin\left(\frac{\pi}{4n}\right)\right] = \cos 0+\sin 0 = 1.$$ Therefore $$\lim_{n\to\infty}\sum_{k=1}^n \sin\left(\frac{\pi k}{2n}\right)\frac{\pi}{2n} = 1\cdot 1 = 1.$$

On the other hand, this sum is a Riemann sum for $\sin x$ on $[0,\pi/2]$ with right endpoints as sample points, so it converges to $\displaystyle\int_0^{\pi/2}\sin x\,dx$. Combining the two computations, $$\int_0^{\pi/2}\sin x\,dx = 1.$$

Solution. (1) The average value of $f(x)=x^2-1$ on $[0,\sqrt3]$ is $$\operatorname{av}(f) = \frac{1}{\sqrt3-0}\int_0^{\sqrt3}(x^2-1)\,dx = \frac{1}{\sqrt3}\left[\frac{x^3}{3}-x\right]_0^{\sqrt3}.$$ Since $(\sqrt3)^3 = 3\sqrt3$, $$\left[\frac{x^3}{3}-x\right]_0^{\sqrt3} = \frac{3\sqrt3}{3}-\sqrt3 = \sqrt3-\sqrt3 = 0,$$ so $\operatorname{av}(f) = \dfrac{1}{\sqrt3}\cdot 0 = 0$.

(2) The average value of $g(x)=3x^2-3$ on $[0,1]$ is $$\operatorname{av}(g) = \frac{1}{1-0}\int_0^1(3x^2-3)\,dx = \left[x^3-3x\right]_0^1 = (1-3)-0 = -2.$$

Proof. The average value of a function is defined by $\operatorname{av}(f) = \dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx$, and $\operatorname{av}(f)$ is a constant not depending on $x$.

(1) Since $\operatorname{av}(f)$ is constant, $$\int_a^b \operatorname{av}(f)\,dx = \operatorname{av}(f)\cdot(b-a) = \frac{1}{b-a}\int_a^b f(x)\,dx\cdot(b-a) = \int_a^b f(x)\,dx.$$

(2) By linearity of the integral, $$\operatorname{av}(f+g) = \frac{1}{b-a}\int_a^b (f+g)(x)\,dx = \frac{1}{b-a}\left[\int_a^b f(x)\,dx+\int_a^b g(x)\,dx\right] = \operatorname{av}(f)+\operatorname{av}(g).$$

(3) Since $k$ is a constant, linearity of the integral gives $$\operatorname{av}(kf) = \frac{1}{b-a}\int_a^b kf(x)\,dx = \frac{k}{b-a}\int_a^b f(x)\,dx = k\operatorname{av}(f).$$

(4) If $f(x)\le g(x)$ on $[a,b]$, then monotonicity of the integral gives $\displaystyle\int_a^b f(x)\,dx \le \int_a^b g(x)\,dx$. Since $b-a \gt 0$, dividing both sides by $b-a$ gives $$\operatorname{av}(f) = \frac{1}{b-a}\int_a^b f(x)\,dx \le \frac{1}{b-a}\int_a^b g(x)\,dx = \operatorname{av}(g).$$