자연계열 A 3번

(1안) $\displaystyle F(x)$가 연속함수이므로 $\displaystyle \lim_{x\to1}F(x)=F(1)=a$이다. 즉 $\displaystyle \lim_{x\to1}\frac{f(x)-xe^{x^2}}{x-1}=a$인데 $\displaystyle \lim_{x\to1}(x-1)=0$이므로 $\displaystyle \lim_{x\to1}(f(x)-xe^{x^2})=0$이다. $\displaystyle f(x)$가 연속함수이므로 $\displaystyle f(1)=e$이다.

$\displaystyle \begin{aligned}f\prime(1)&=\lim_{x\to1}\frac{f(x)-f(1)}{x-1}=\lim_{x\to1}\frac{(x-1)F(x)+xe^{x^2}-e}{x-1}\\&=\lim_{x\to1}\frac{(x-1)F(x)}{x-1}+\lim_{x\to1}\frac{xe^{x^2}-e}{x-1}\\&=\lim_{x\to1}F(x)+\lim_{x\to1}\frac{xe^{x^2}-e}{x-1}=a+3e\end{aligned}$

$\displaystyle \begin{aligned}g\prime(1)&=\lim_{x\to1}\frac{g(x)-g(1)}{x-1}=\lim_{x\to1}\frac{f\prime(x)-f\prime(1)}{x-1}\\&=\lim_{x\to1}\frac{F(x)+(x-1)F\prime(x)+e^{x^2}+2x^2e^{x^2}-(a+3e)}{x-1}\\&=\lim_{x\to1}\left(\frac{F(x)-F(1)}{x-1}+F\prime(x)+\frac{e^{x^2}+2x^2e^{x^2}-3e}{x-1}\right)\end{aligned}$

$\displaystyle =F\prime(1)+F\prime(1)+10e=6+10e$

$\displaystyle u=1-x$로 치환하면 $\displaystyle \int_0^1xF(1-x)\,dx=\int_1^0(1-u)F(u)(-du)=-\int_0^1(u-1)F(u)\,du$를 얻는다. 따라서 $\displaystyle \int_0^1(x-1)F(x)\,dx=0$이다.

그러므로$\displaystyle \int_0^1f(x)\,dx=\int_0^1\{(x-1)F(x)+xe^{x^2}\}\,dx=\int_0^1xe^{x^2}\,dx=\Biggl[\frac12e^{x^2}\Biggr]_0^1=\frac12(e-1)$이다.