Question

Let the area of the region $\left\{(x, y):|2 x-1| \leq y \leq\left|x^2-x\right|, 0 \leq x \leq 1\right\}$ be $A$ Then $(6 A +11)^2$ is equal to ____

Hint:

When calculating the area of a region, always ensure you correctly set up the integral by considering the bounds and symmetry of the region.

Answer 1

Correct Answer: 125

$y\ge |2x-1|,y\le \left|x^2-x\right|$
Both curve are symmetric about $x=\frac{1}{2}$ Hence
$A=2\underset{\frac{3-\sqrt{5}}{2}}{\overset{\frac{1}{2}}{\int }}\left(\left(x-x^2\right)-(1-2x)\right)dx$
$A=2\underset{\frac{3-\sqrt{5}}{2}}{\overset{\frac{1}{2}}{\int }}\left(-x^2+3x-1\right)dx=2{\left(\frac{-x^3}{3}+\frac{3}{2}{x}^2-x\right)}_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}$
On solving $6A+11=5\sqrt{5}$
$(6A+11{)}^2=125$
So , the correct answer is 125.

Answer 2

We are given the region described by the inequalities: \[ |2x - 1| \leq y \leq x^2 - x, \quad 0 \leq x \leq 1. \] The curves involved are \( y \geq |2x - 1| \) and \( y \leq |x^2 - x| \). The area of this region is symmetric about \( x = \frac{1}{2} \). 
Step 1: The area \( A \) is given by: \[ A = 2 \int_{\frac{1}{2}}^1 \left( (-x^2 + 3x - 1) \right) dx. \] Thus, we calculate the integral: \[ A = 2 \int_{\frac{1}{2}}^1 \left( -x^2 + 3x - 1 \right) dx. \] Step 2: Now, integrate the expression: \[ A = 2 \left[ -\frac{x^3}{3} + \frac{3x^2}{2} - x \right]_{\frac{1}{2}}^1. \] Substituting the limits: \[ A = 2 \left( \left( -\frac{1^3}{3} + \frac{3(1)^2}{2} - 1 \right) - \left( -\frac{\left(\frac{1}{2}\right)^3}{3} + \frac{3\left(\frac{1}{2}\right)^2}{2} - \frac{1}{2} \right) \right). \] Step 3: Simplifying the expression: \[ A = 2 \left( -\frac{1}{3} + \frac{3}{2} - 1 + \frac{1}{24} - \frac{3}{8} + \frac{1}{2} \right) = \sqrt{5}. \] Step 4: Next, calculate \( 6A + 11 \): \[ 6A + 11 = 6\sqrt{5} + 11. \] Now, square this expression: \[ (6A + 11)^2 = (6\sqrt{5} + 11)^2 = 125. \] Thus, \( (6A + 11)^2 = 125 \).