Definite Integration and Area
DAILY_CHALLENGE
Grade None
Question:
Let $f:[0,1]\to[0,1]$ be the function defined by $f(x)=\dfrac{x^3}{3}-x^2+\dfrac{5}{9}x+\dfrac{17}{36}$. Consider the square region $S=[0,1]\times[0,1]$. Let $G=\{(x,y)\in S:y>f(x)\}$ be called the green region and $R=\{(x,y)\in S:y<f(x)\}$ be called the red region. Let $L_h=\{(x,h)\in S:x\in[0,1]\}$ be the horizontal line drawn at a height $h\in[0,1]$. Then which of the following statements is(are) true?
There exists an $h\in\left[\dfrac{1}{4},\dfrac{2}{3}\right]$ such that the area of the green region above the line $L_h$ equals the area of the green region below the line $L_h$
There exists an $h\in\left[\dfrac{1}{4},\dfrac{2}{3}\right]$ such that the area of the red region above the line $L_h$ equals the area of the red region below the line $L_h$
There exists an $h\in\left[\dfrac{1}{4},\dfrac{2}{3}\right]$ such that the area of the green region above the line $L_h$ equals the area of the red region below the line $L_h$
There exists an $h\in\left[\dfrac{1}{4},\dfrac{2}{3}\right]$ such that the area of the red region above the line $L_h$ equals the area of the green region below the line $L_h$
Step-by-Step Solution
Key Concept: Green above minus Red below = 1/2 − h; use this identity to find h values directly
$\int_0^1 f(x)dx = \left[\dfrac{x^4}{12}-\dfrac{x^3}{3}+\dfrac{5x^2}{18}+\dfrac{17x}{36}\right]_0^1 = \dfrac{1}{12}-\dfrac{1}{3}+\dfrac{5}{18}+\dfrac{17}{36}=\dfrac{3-12+10+17}{36}=\dfrac{18}{36}=\dfrac{1}{2}$.
So green area = red area = 1/2.
Note: $f'(x)=x^2-2x+5/9=(x-1/3)(x-5/3)$. On $[0,1]$, only critical point $x=1/3$ is a local max. $f(0)=17/36\approx0.47$, $f(1/3)\approx0.56$, $f(1)=13/36\approx0.36$. So $f(x)\in[13/36,~181/324]\subset(1/4,2/3)$ for all $x\in[0,1]$.
**Key identity**: (Green above $L_h$) $-$ (Red below $L_h$) $= \int_0^1(1-h-f(x))dx = \dfrac{1}{2}-h$.
Similarly (Red above) $-$ (Green below) $= \int_0^1(f(x)-h)dx = \dfrac{1}{2}-h$.
(C) Green above $=$ Red below $\Leftrightarrow h=1/2\in[1/4,2/3]$. TRUE.
(D) Red above $=$ Green below $\Leftrightarrow h=1/2\in[1/4,2/3]$. TRUE.
(B) Red above $=$ Red below: total red $=1/2$, each $=1/4$. Since $f(x)>1/4$ for all $x$ (as $f\geq13/36>1/4$), at $h=1/4$: Red above $=\int_0^1(f-1/4)dx=1/4$ and Red below $=\int_0^1(1/4)dx=1/4$. So $h=1/4\in[1/4,2/3]$ works. TRUE.
(A) Green above $=$ Green below: Since $f(x)>1/4$ for all $x$, Green below $L_{1/4}=0$ and Green above $=1/2>0$. At $h=2/3$: $f<2/3$ everywhere, so Green above $=\int_0^1(1-2/3)dx=1/3$ and Green below $=\int_0^1(2/3-f)dx=1/6$. Both unequal. The equality point lies outside $[1/4,2/3]$. FALSE.
Correct Answer: B, C, D