Let f : [0,1] → [0,1] be the function defined by f(x) = x³/3 - x² + 5/9x + 17/36 . Consider the square region S = [0,1]×[0,1] . Let G = {(x,y) ∈ S : y > f(x)} be called the green region and R = {(x,y) ∈ S : y < f(x)} be called the red region. Let Lₕ = {(x,h) ∈ S : x ∈ [0,1]} be the horizontal line drawn at a height h ∈ [0,1] . Then which of the following statements is(are) true?

Let f : [0,1] → [0,1] be the function defined by f(x) = x³/3 - x² + 5/9x + 17/36 . Consider the square region S = [0,1]×[0,1] . Let G = {(x,y) ∈ S : y > f(x)} be called the green region and R = {(x,y) ∈ S : y f(x) (green); R: y + A_R^> = 1 - h A_G^ = 1/2, A_R^ = 1/2 A_G^> - A_R^ - A_G^ = A_R^ = A_G^ = A_R^ 1/4 ⇒ A_R^ = A_G^< iff A_G^< = 1/4 f(x) < 2/3 ⇒ A_G^<(2/3) = 2/3 - 1/2 = 1/6 < 1/4 ⇒ (1) FALSE TRUE: (2), (3), (4) boxed(2), (3), (4)

Let $f : [0,1] \to [0,1]$ be the function defined by $f(x) = \frac{x^3}{3} - x^2 + \frac{5}{9}x + \frac{17}{36}$ . 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?

Answers: B. There exists an $h \in \left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $L_h$ equals the area of the red region below the line $L_h$ · C. There exists an $h \in \left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $L_h$ equals the area of the red region below the line $L_h$ · D. There exists an $h \in \left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $L_h$ equals the area of the green region below the line $L_h$

Applications of Integrals Let

f : [0,1] \to [0,1]

be the function defined by

f(x) = \frac{x^3}{3} - x^2 + \frac{5}{9}x + \frac{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?

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Written solution

$f(x) = \dfrac{x^3}{3} - x^2 + \dfrac{5}{9}x + \dfrac{17}{36}, \quad x \in [0,1]$

$S = [0,1]^2;\quad G: y > f(x)\ \text{(green)};\quad R: y < f(x)\ \text{(red)};\quad L_h: y = h$

$\text{To find: which statements (1)--(4) are true}$

$f'(x) = x^2 - 2x + \dfrac{5}{9} = (x-\tfrac13)(x-\tfrac53)$

$\int_0^1 f(x)\,dx = \dfrac{1}{12} - \dfrac13 + \dfrac{5}{18} + \dfrac{17}{36} = \dfrac12$

$\Rightarrow \operatorname{area}(R) = \operatorname{area}(G) = \dfrac12$

$A_G^< + A_R^< = h, \qquad A_G^> + A_R^> = 1 - h$

$A_G^< + A_G^> = \dfrac12, \qquad A_R^< + A_R^> = \dfrac12$

$A_G^> - A_R^< = \dfrac12 - h, \qquad A_R^> - A_G^< = \dfrac12 - h$

$h = \dfrac12 \Rightarrow A_G^> = A_R^< \text{ and } A_R^> = A_G^< \Rightarrow (3), (4)\ \text{TRUE}$

$(2):\ A_R^> = A_R^< \iff A_R^< = \dfrac14$

$f(x) \ge \dfrac{13}{36} > \dfrac14 \Rightarrow A_R^<\!\left(\dfrac14\right) = \dfrac14 \Rightarrow (2)\ \text{TRUE}$

$(1):\ A_G^> = A_G^< \iff A_G^< = \dfrac14$

$f(x) < \dfrac23 \Rightarrow A_G^<\!\left(\dfrac23\right) = \dfrac23 - \dfrac12 = \dfrac16 < \dfrac14 \Rightarrow (1)\ \text{FALSE}$

$\text{TRUE: } (2),\ (3),\ (4)$

$\boxed{(2),\ (3),\ (4)}$

Written solution

More from Applications of Integrals