적분의 활용 전체 문제

$70\pi \sqrt{2}$

By definition, the average is

$$\begin{aligned} \frac{1}{e-1}\int_1^e x^2\log x\ \mathrm{d}x & \\ \text{To antidifferentiate, we use integration by parts with $u=\log x$ and $\mathrm{d}v=x^2\,\mathrm{d}x$, hence $\mathrm{d}u=\frac{1}{x}\,\mathrm{d}x$ and $v=\frac{1}{3}x^3$.} \\ \frac{1}{e-1}\int_1^e x^2\log x\ \mathrm{d}x&= \frac{1}{e-1}\left(\left[\frac{1}{3}x^3\log x\right]_1^e - \int_1^e \frac{1}{3}x^2\,\mathrm{d}x \right) \\&= \frac{1}{e-1}\bigg[\frac{x^3}{3}\log x-\frac{x^3}{9}\bigg]_{x=1}^{x=e}\\ &=\frac{1}{e-1}\bigg[\frac{e^3}{3}-\frac{e^3}{9}+\frac{1}{9}\bigg]\\ &=\frac{1}{e-1}\bigg[\frac{2}{9}e^3+\frac{1}{9}\bigg] \end{aligned}$$

By definition, the average value in question equals

$$\frac1{\pi/2-0} \int_0^{\pi/2} ( 3\cos^3x + 2\cos^2x ) \,\mathrm{d}x = \frac2\pi \bigg( \int_0^{\pi/2} 3\cos^3x \,\mathrm{d}x + \int_0^{\pi/2} 2\cos^2x \,\mathrm{d}x \bigg)$$

For the first integral we use the substitution $u = \sin x$, $\mathrm{d}u=\cos x\,\mathrm{d}x$, $\cos^2 x=1-\sin^2 x = 1-u^2$. Note that the endpoints $x=0$ and $x=\frac\pi2$ become $u=0$ and $u=1$, respectively.

$$\begin{aligned} \int_0^{\pi/2} 3 \cos^3x \,\mathrm{d}x &= \int_0^{\pi/2} 3 \cos^2 x \cos x \,\mathrm{d}x \\ &= \int_0^{1} 3(1-u^2) \,\mathrm{d}u \\ &= (3u-u^3)\Big|_0^1 = 2. \end{aligned}$$

For the second integral we use the trigonometric identity $\cos^2x\,\mathrm{d}x = \frac{1 + \cos(2x)}{2}$.

$$\begin{aligned} 2 \int_0^{\pi/2} \cos^2x\,\mathrm{d}x &= \int_0^{\pi/2} \big(1 + \cos(2x) \big) \,\mathrm{d}x \\ &= \bigg[ x + \frac{1}{2}\sin(2x) \bigg]_0^{\pi/2} = \frac\pi2 \end{aligned}$$

Therefore, the average value in question is

$$\frac2\pi \bigg( \int_0^{\pi/2} 3\cos^3x \,\mathrm{d}x + \int_0^{\pi/2} 2\cos^2x \,\mathrm{d}x \bigg) = \frac2\pi \bigg( 2 + \frac\pi2 \bigg) = \frac{4}{\pi}+1.$$

By definition, the average value in question equals

$$\text{Ave}=\frac1{\pi/k-0} \int_0^{\pi/k} \sin(kx) \,\mathrm{d}x$$

To evaluate the integral, we use the substitution $u = kx$, $\mathrm{d}u=k\,\mathrm{d}x$. Note that the endpoints $x=0$ and $x=\pi/k$ become $u=0$ and $u=\pi$, respectively. So

$$\begin{aligned} \text{Ave}=\frac{k}{\pi} \int_0^\pi \sin(u) \,\frac{\mathrm{d}u}{k} =\frac{1}{\pi}\Big[-\cos(u)\Big]_0^\pi =\frac{2}{\pi} \end{aligned}$$

Remark: the average does not depend on $k$. To see why this is, note that $\sin(kx)$ runs between $-1$ and $1$ as $x$ changes. When $x=0$, $kx=0$, and when $x=\pi/k$, $kx=\pi$. So, our function $\sin (kx)$ runs exactly from $\sin 0 =0$ to $\sin(\pi/2)=1$, then back down to $\sin\pi=0$.

By definition, the average value is

$$\begin{aligned} \frac{1}{e-1}\int_1^e\frac{\log x}{x}\ \mathrm{d}x& \\ \text{To integrate, we use the substitution $u=\log x$, $\mathrm{d}u=\frac{1}{x}\,\mathrm{d}x$. Then the limits of integration become 0 and 1, respectively.} \\ \frac{1}{e-1}\int_1^e\frac{\log x}{x}\ \mathrm{d}x& =\frac{1}{e-1}\int_0^1u\ \mathrm{d}u =\frac{1}{e-1}\bigg[\frac{u^2}{2}\bigg]_0^1 =\frac{1}{2(e-1)} \end{aligned}$$

By definition, the average value is:

$$\begin{aligned} \frac{1}{2\pi}\int_0^{2\pi}\cos^2x\,\mathrm{d}x =\frac{1}{2\pi}\cdot\frac{1}{2}\int_0^{2\pi}\big(\cos(2x)+1\big)\,\mathrm{d}x =\frac{1}{4\pi}\Big[\frac{\sin(2x)}{2}+x\Big]_0^{2\pi} =\frac{1}{4\pi}\cdot 2\pi =\frac{1}{2} \end{aligned}$$