삼차함수 극댓값과 삼각형 넓이의 극한

$f' \left( x \right)$$= 6 x ^{2} - 2 \left( t + 3 \right) x + 2 t$이므로 $f' \left( a \right)$$= 6 a ^{2} - 2 \left( t + 3 \right) a + 2 t = 0$ $\left( 6 a - 2 t \right) \left( a - 1 \right) = 0$에서 $\displaystyle a = \frac{t}{3} , a = 1$ $g \left( t \right)$$\displaystyle = \frac{a}{2} f \left( a \right)$ $\displaystyle = a ^{4} - \frac{3}{2} \left( a + 1 \right) a ^{3} + 3 a ^{3}$ $\displaystyle = - \frac{1}{2} a ^{4} + \frac{3}{2} a ^{3}$ 그런데, $t = 3 a$이므로 $t \rightarrow 0$일 때 $a \rightarrow 0$이므로 $\displaystyle \int _{0} ^{a} {} f \left( x \right) dx$$\displaystyle = \left[ \frac{1}{2} x ^{4} - \frac{1}{3} \left( t + 3 \right) x ^{3} + tx ^{2} \right] _{0} ^{a}$ $\displaystyle = \frac{1}{2} a ^{4} - \left( a + 1 \right) a ^{3} + 3 a ^{3}$ $\displaystyle = - \frac{1}{2} a ^{4} + 2 a ^{3}$ $\therefore$ $\displaystyle \lim\limits _{t \rightarrow 0} {} \frac{1}{g \left( t \right)} \int _{0} ^{a} {} f \left( x \right) dx$$\displaystyle = \lim\limits _{a \rightarrow 0} {} \frac{- a ^{4} + 4 a ^{3}}{- a ^{4} + 3 a ^{3}} = \frac{4}{3}$