넓이 조건 적분

$( 0 , 0 ) , ( t , f ( t ) ) , ( t + 1 , f ( t + 1 ) )$을 꼭짓점으로 하는 삼각형의 넓이는 $\displaystyle \frac{1}{2} \left| t f ( t + 1 ) - ( t + 1 ) f ( t ) \right| = \frac{t + 1}{t}$ 양변을 $t ( t + 1 )$로 나누면 $\displaystyle \frac{1}{2} \left| \frac{f ( t + 1 )}{t + 1} - \frac{f ( t )}{t} \right| = \frac{1}{t ^{2}}$ $f ( t )$는 감소함수이고 $f ( t ) > 0$이므로 $\displaystyle \frac{f ( t )}{t} > \frac{f ( t + 1 )}{t + 1}$ $\displaystyle \therefore \frac{f ( t )}{t} - \frac{f ( t + 1 )}{t + 1} = \frac{2}{t ^{2}}$ $\displaystyle \frac{f ( t + 1 )}{t + 1} - \frac{f ( t )}{t} = \left[ \frac{f ( x )}{x} \right] _{t} ^{t + 1}$$\displaystyle = \frac{d}{dt} \left( \int _{t} ^{t + 1} {\frac{f ( x )}{x} dx} \right) = - \frac{2}{t ^{2}}$ 양변을 적분하면 $\displaystyle \int _{t} ^{t + 1} {\frac{f ( x )}{x} = \frac{2}{t} + C}$ $t = 1$일 때, $\displaystyle \int _{1} ^{2} \frac{f ( x )}{x} dx = 2 + c = 2$에서 $C = 0$ $\displaystyle \therefore \int _{t} ^{t+1} \frac{f ( x )}{x} dx = \frac{2}{t}$ $\displaystyle \int _{\frac{7}{2}} ^{\frac{11}{2}} {\frac{f ( x )}{x}} dx = \int _{\frac{7}{2}} ^{\frac{9}{2}} {\frac{f ( x )}{x}} dx + \int _{\frac{9}{2}} ^{\frac{11}{2}} {\frac{f ( x )}{x}} dx$ $\displaystyle = \frac{4}{7} + \frac{4}{9}$$\displaystyle = \frac{64}{63}$ $\therefore p + q = 127$