자연계열 1번

(문제 1-1) 배점 10점

(정답) $\displaystyle a=\frac1e$, $\displaystyle b=\frac12$

(풀이)

구간 $\displaystyle (0,4)$에서 미분가능 하려면 $\displaystyle x=2$에서 연속이어야 한다.

$\displaystyle 2\times2-2=a\times2\times e^{2b}$, $\displaystyle 1=ae^{2b}$

$\displaystyle x=2$에서 좌미분계수는

$\displaystyle \lim_{h\to0^-}\frac{f(2+h)-f(2)}h=\lim_{h\to0^-}\frac{(2\times(2+h)-2)-(2\times2-2)}h=\lim_{h\to0^-}\frac{2h}h=2$

우미분계수는

$\displaystyle \begin{aligned}\lim_{h\to0^+}\frac{f(2+h)-f(2)}h&=\lim_{h\to0^+}\frac{a(2+h)e^{b(2+h)}-a(2)e^{b(2)}}h\\&=\lim_{h\to0^+}\frac{2ae^{b(2+h)}+ahe^{b(2+h)}-2ae^{2b}}h\\&=\lim_{h\to0^+}\frac{2ae^{2b}(e^{bh}-1)+ahe^{2b}e^{bh}}h\\&=\lim_{h\to0^+}\frac{2ae^{2b}(e^{bh}-1)+ahe^{2b}e^{bh}}h\\&=2ae^{2b}\lim_{h\to0^+}\frac{e^{bh}-1}h+\lim_{h\to0^+}ae^{2b}e^{bh}=2abe^{2b}+ae^{2b}\quad\left(\because\lim_{h\to0^+}\frac{e^{bh}-1}{bh}b=b\right)\end{aligned}$

이다.

구간 $\displaystyle (0,4)$에서 미분가능 하려면 좌미분계수와 우미분계수가 같아야 하므로 $\displaystyle 2=ae^{2b}+2abe^{2b}$

$\displaystyle a,b$값을 구하면

$\displaystyle 2=1(1+2b)$이므로 $\displaystyle b=\frac12$

$\displaystyle 1=ae^{2\times\frac12}$이므로 $\displaystyle a=\frac1e$

(문제 1-2) 배점 20점

(정답) $\displaystyle 4e+16$

(풀이)

$\displaystyle x<1$, $\displaystyle x>3$인 경우, $\displaystyle g(x)=(x-1)(x-3)-5$이므로

$\displaystyle g(x)=x^2-4x+3-5=x^2-4x-2$

$\displaystyle 1<x<3$인 경우, $\displaystyle g(x)=-(x-1)(x-3)-5$이므로

$\displaystyle g(x)=-x^2+4x-3-5=-x^2+4x-8$

적분구간을 4개로 나누어서 표현하면

$\displaystyle \begin{aligned}\int_0^4(f(x)-g(x))\,dx&=\int_0^1((2x-2)-(x^2-4x-2))\,dx+\int_1^2((2x-2)-(-x^2+4x-8))\,dx\\&\quad+\int_2^3\left(xe^{\frac12(x-2)}-(-x^2+4x-8)\right)\,dx+\int_3^4\left(xe^{\frac12(x-2)}-(x^2-4x-2)\right)\,dx\\&=\int_0^1(-x^2+6x)\,dx+\int_1^2(x^2-2x+6)\,dx\\&\quad+\int_2^3\left(xe^{\frac12(x-2)}+x^2-4x+8\right)\,dx+\int_3^4\left(xe^{\frac12(x-2)}-x^2+4x+2\right)\,dx\end{aligned}$

$\displaystyle \begin{aligned}&=\Biggl[-\frac{x^3}3+3x^2\Biggr]_0^1+\Biggl[\frac{x^3}3-x^2+6x\Biggr]_1^2\\&\quad+\Biggl[2xe^{\frac12(x-2)}\Biggr]_2^3-\int_2^32e^{\frac12(x-2)}\,dx+\Biggl[\frac{x^3}3-2x^2+8x\Biggr]_2^3\\&\quad+\Biggl[2xe^{\frac12(x-2)}\Biggr]_3^4-\int_3^42e^{\frac12(x-2)}\,dx+\Biggl[-\frac{x^3}3+2x^2+2x\Biggr]_3^4\\&=\Biggl[-\frac{x^3}3+3x^2\Biggr]_0^1+\Biggl[\frac{x^3}3-x^2+6x\Biggr]_1^2\\&\quad+\Biggl[2xe^{\frac12(x-2)}\Biggr]_2^3-\Biggl[4e^{\frac12(x-2)}\Biggr]_2^3+\Biggl[\frac{x^3}3-2x^2+8x\Biggr]_2^3\\&\quad+\Biggl[2xe^{\frac12(x-2)}\Biggr]_3^4-\Biggl[4e^{\frac12(x-2)}\Biggr]_3^4+\Biggl[-\frac{x^3}3+2x^2+2x\Biggr]_3^4\\&=\left(-\frac13+3\right)+\left(\frac83-4+12-\left(\frac13-1+6\right)\right)\\&\quad+\left(6e^{0.5}-4-(4e^{0.5}-4)+\frac{27}3-18+24-\left(\frac83-8+16\right)\right)\\&\quad+\left(8e-6e^{0.5}-(4e-4e^{0.5})-\frac{64}3+32+8-\left(-\frac{27}3+18+6\right)\right)\\&=4e+16\end{aligned}$