The equation of a line that passes thruogh the point (x1,y1) and has a slope of m is
y-y1=m(x-x1)
so
passing through (a,b) and having slope of b is
y-b=b(x-a)
Answer: 6x^7+ 12x^6+ 3x^5
Hope that helped
Answer:
I am pretty sure it is dependent
Let me know if it is wrong.
Adjacent angles are angles that are next to each other and share a common vertex and a common side.
The two curves meet in the first quadrant when
![x^2 = 2-x \implies x^2 + x - 2 = (x + 2) (x - 1) = 0 \implies x=1](https://tex.z-dn.net/?f=x%5E2%20%3D%202-x%20%5Cimplies%20x%5E2%20%2B%20x%20-%202%20%3D%20%28x%20%2B%202%29%20%28x%20-%201%29%20%3D%200%20%5Cimplies%20x%3D1)
Then the integral in question is
![\displaystyle \iint_R x \, dA = \int_0^1 \int_{x^2}^{2-x} x \, dy \, dx \\\\ ~~~~~~~~ = \int_0^1 x (2-x-x^2) \, dx \\\\ ~~~~~~~~ = \int_0^1 (2x - x^2 - x^3) \, dx = 1 - \frac13 - \frac14 = \boxed{\frac5{12}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_R%20x%20%5C%2C%20dA%20%3D%20%5Cint_0%5E1%20%5Cint_%7Bx%5E2%7D%5E%7B2-x%7D%20x%20%5C%2C%20dy%20%5C%2C%20dx%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%20%5Cint_0%5E1%20x%20%282-x-x%5E2%29%20%5C%2C%20dx%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%20%5Cint_0%5E1%20%282x%20-%20x%5E2%20-%20x%5E3%29%20%5C%2C%20dx%20%3D%201%20-%20%5Cfrac13%20-%20%5Cfrac14%20%3D%20%5Cboxed%7B%5Cfrac5%7B12%7D%7D)