Calculate the area of figure, that is limited by the parabola and line: y = -x ^2 + 6x - 5; y = x - 5?

The parabola: y = -x² + 6x - 5 …… [1]
The straight line: y = x - 5 …… [2]

To find the point of intersection, [1] = [2] :
-x² + 6x - 5 = x - 5
x² - 5x = 0
x(x - 5) = 0
x = 0 or x = 5

The parabola and the straight line meet at x = 0 and x = 5

When x = 1: the point on the parabola (1, 0), while on the straight line is (1, -4).
The parabola lies above the straight line.

Area of region enclosed by the parabola and the straight line
= ∫₍ₓ₌₀₎₍ₓ₌₁₎ [(-x² + 6x - 5) - (x - 5)] dx
= ∫₍ₓ₌₀₎₍ₓ₌₁₎ (-x² + 5x) dx
= [(-x³/3) + (5x²/2)]₍ₓ₌₀₎₍ₓ₌₁₎
= [(-1/3) + (5/2)] - [(0/3) + (0/2)]
= [(-2/6) + (15/6)]
= 13/6

y = -x^2 + 6x - 5 , y = x - 5
y = y
-x^2 + 6x - 5 = x - 5
-x^2 + 5x = 0
-x(x - 5) = 0
x = 0 , 5

A = ∫(-x^2 + 5x) dx [0, 5]
A = (-x^3)/3 + (5x^2)/2 from 0 to 5
A = -125/3 + 125/2
A = 125/6
A ≈ 20.83 square units

Here is the graph:

https://www.wolframalpha.com/input/?i=plot%7B+y+%3D+-x+%5E2+%2B+6x+-+5+,+y+%3D+x+-+5%7D

y=x-5
y=-x^2+6x-5

-x^2+6x-5 = x-5
-x^2+6x-5-x+5 = 0
x^2-6x+5+x-5=0
x^2-5x=0
x(x-5)=0
x=0; x=5 (limits of integration)

Area = ∫ ((-x^2+6x-5) -(x-5) ) dx
= ∫ (-x^2+5x) dx
= -(1/3) x^3 + (5/2) x^2

F(x) = (-1/3) x^3 + (5/2) x^2
F(5) = -125/3 +125/2 = 125/6
F(0) = 0

Area = F(5)-F(0) = 125/6