Write an equation of a line that
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Solve for the variable y (move the x to the right of equal with changed sign)
M is the coefficient before the x and b is the known number
-3y = -12 - 6x
Since the y can't be negative, multiply every term for -1, therefore change their sign
3y = 6x + 12
To find y, divide both terms for 3
y = 2x+4
Therefore your answer is y = 2x+4
M is the coefficient before the x and b is the known number
-3y = -12 - 6x
Since the y can't be negative, multiply every term for -1, therefore change their sign
3y = 6x + 12
To find y, divide both terms for 3
y = 2x+4
Therefore your answer is y = 2x+4
Check the picture below.
thus, if one side on the square is x - x², then the area for the square is (x - x²)(x - x²), or (x - x²)²
now... what is the "x" value for the region, let's check their intersection.
![\bf \begin{cases} y=x\\ y=x^2 \end{cases}\implies x=x^2\implies 0=x^2-x\implies 0=x(x-1) \\\\\\ x= \begin{cases} 0\\ 1 \end{cases}\\\\ -------------------------------\\\\ \displaystyle \int\limits_{0}^{1}\ (x-x^2)^2\cdot dx\implies \int\limits_{0}^{1}\ (x^2-2x^3+x^4)\cdot dx \\\\\\ \cfrac{x^2}{2}-\cfrac{2x^4}{4}+\cfrac{x^5}{5}\implies \left. \cfrac{x^2}{2}-\cfrac{x^4}{2}+\cfrac{x^5}{5} \right]_{0}^{1}\implies \left[ \cfrac{1}{2}-\cfrac{1}{2}+\cfrac{1}{5} \right]-[0]\implies \cfrac{1}{5}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ay%3Dx%5C%5C%0Ay%3Dx%5E2%0A%5Cend%7Bcases%7D%5Cimplies%20x%3Dx%5E2%5Cimplies%200%3Dx%5E2-x%5Cimplies%200%3Dx%28x-1%29%0A%5C%5C%5C%5C%5C%5C%0Ax%3D%0A%5Cbegin%7Bcases%7D%0A0%5C%5C%0A1%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Cdisplaystyle%20%5Cint%5Climits_%7B0%7D%5E%7B1%7D%5C%20%28x-x%5E2%29%5E2%5Ccdot%20dx%5Cimplies%20%5Cint%5Climits_%7B0%7D%5E%7B1%7D%5C%20%28x%5E2-2x%5E3%2Bx%5E4%29%5Ccdot%20dx%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bx%5E2%7D%7B2%7D-%5Ccfrac%7B2x%5E4%7D%7B4%7D%2B%5Ccfrac%7Bx%5E5%7D%7B5%7D%5Cimplies%20%5Cleft.%20%5Ccfrac%7Bx%5E2%7D%7B2%7D-%5Ccfrac%7Bx%5E4%7D%7B2%7D%2B%5Ccfrac%7Bx%5E5%7D%7B5%7D%20%5Cright%5D_%7B0%7D%5E%7B1%7D%5Cimplies%20%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D-%5Ccfrac%7B1%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B5%7D%20%5Cright%5D-%5B0%5D%5Cimplies%20%5Ccfrac%7B1%7D%7B5%7D)
thus, if one side on the square is x - x², then the area for the square is (x - x²)(x - x²), or (x - x²)²
now... what is the "x" value for the region, let's check their intersection.