方程为 $y = 2x - 8x^{\frac{1}{2}}$ 的曲线有一个最低点 $A$，并与正 $x$ 轴相交于点 $B$。

(a) 求 $A$ 和 $B$ 的坐标。

(b)

图中显示方程为 $y = 2x - 8x^{\frac{1}{2}}$ 的曲线和直线 $AB$。已知 $AB$ 的方程为 $y = \frac{2x - 32}{3}$。

求曲线与直线之间阴影区域的面积。

在 $x>0$ 上用导数找最低点。另把原式因式分解，找正的横截距。

For $x>0$, $y'=2-4/\sqrt x$. The derivative is zero at $\sqrt x=2$, so $x=4$, and it changes from negative to positive there. This gives the minimum $A=(4,-8)$.

The intercept condition is $2\sqrt x(\sqrt x-4)=0$. Its positive root is $x=16$, so $B=(16,0)$.

阴影在 $x=4$ 与 $x=16$ 之间，上方是直线。把高度差化简后积分。

The line lies above the curve between $A$ and $B$. The area is

两条图像都在 $x$ 轴下方。先算曲线到 $x$ 轴的面积，再减去直线到 $x$ 轴的三角形。

The area between the curve and the axis on $[4,16]$ is

The triangle between the line and the axis has width $12$ and height $8$, so its area is $48$. Thus the shaded area is $\frac{176}3-48=\frac{32}3$.