자연계열 2번

2-1. $\displaystyle f(x)=\begin{cases}3x&(0\le x<1)\\3&(1\le x<5)\\-\frac35x+6&(5\le x<10)\end{cases}$$\displaystyle \begin{aligned}f(\sqrt{x+1}-1)&=\begin{cases}3(\sqrt{x+1}-1)&(0\le\sqrt{x+1}-1<1)\\3&(1\le\sqrt{x+1}-1<5)\\-\frac35(\sqrt{x+1}-1)+6&(5\le\sqrt{x+1}-1<10)\end{cases}\\&=\begin{cases}3(\sqrt{x+1}-1)&(0\le x<3)\\3&(3\le x<35)\\-\frac35(\sqrt{x+1}-1)+6&(35\le x<120)\end{cases}\end{aligned}$

$\displaystyle \begin{aligned}\int_0^{10}f(\sqrt{x+1}-1)dx&=\int_0^3f(\sqrt{x+1}-1)dx+\int_3^{10}f(\sqrt{x+1}-1)dx\\&=\int_0^33(\sqrt{x+1}-1)dx+\int_3^{10}3dx\\&=\Biggl[2(x+1)^{\frac32}-3x\Biggr]_0^3+\Biggl[3x\Biggr]_3^{10}\\&=16-9-2+30-9\\&=26\end{aligned}$

2-2. 1-2 와 같은 방법으로$\displaystyle f\left(\frac{x}{2}\right)=\begin{cases}\frac32x&(0\le x<2)\\3&(2\le x<10)\\-\frac3{10}x+6&(10\le x<20)\end{cases}$$\displaystyle \begin{aligned}\int_0^{20}e^xf\left(\frac{x}{2}\right)dx&=\int_0^2e^xf\left(\frac{x}{2}\right)dx+\int_2^{10}e^xf\left(\frac{x}{2}\right)dx+\int_{10}^{20}e^xf\left(\frac{x}{2}\right)dx\\&=\int_0^2e^x\left(\frac{3x}{2}\right)dx+\int_2^{10}e^x(3)dx+\int_{10}^{20}e^x\left(-\frac{3x}{10}+6\right)dx\end{aligned}$제시문 <나>를 이용하면,$\displaystyle \begin{aligned}&=\Biggl[\frac32xe^x\Biggr]_0^2-\int_0^2\frac32e^xdx+3\int_2^{10}e^xdx+6\int_{10}^{20}e^xdx-\Biggl[\frac{3x}{10}e^x\Biggr]_{10}^{20}+\int_{10}^{20}\frac3{10}e^xdx\\&=3e^2-\frac32e^2+\frac32+3e^{10}-3e^2+6e^{20}-6e^{10}-6e^{20}+3e^{10}+\frac3{10}e^{20}-\frac3{10}e^{10}\\&=\frac3{10}e^{20}-\frac3{10}e^{10}-\frac32e^2+\frac32\end{aligned}$