Question

fine the area of the region bounded by the curves y=x²-4x+3 and the X-axis​

Answer 1

$\huge\boxed{Answer}$

y=-x^2+4x-3

y=0 → 0=-x^2+4x-3

→delta = 4^2–4(-3)(-1)= 16–4*3=4

x=(-4+2)/(-2)= +1

x=(-4-2)/(-2)= +3

Area= ∫-x^2+4x-3 dx from x=1 to x=3

Area= [-(1/3)x^3+2x^2-3x] from x=1 to x=3

Area= ([-(1/3)x^3+2x^2-3x] for x=3)

-([-(1/3)x^3+2x^2-3x] for x=1)

Area= [-(1/3)(3)^3+2(3)^2-3(3)] -[-(1/3)(1)^3+2(1)^2-3(1)]

Area= [-(1/3)*27+2*9-3(3)] -[-(1/3)(1)+2(1)-3(1)]

Area= [-9+18-9] -[-(1/3)+2-3]

Area= [0] -[-(4/3)]

Area= 4/3