find the area of the region bounded by the function y=-x^2+5x-4 and the x-axis
y = -(x-1)(x-4)
so you want to integrate between the roots, in the interval [1,4]
a = ∫[1,4] -x^2+5x-4 dx
= -1/3 x^3 + 5/2 x^2 - 4x [1,4]
Now just evaluate at the ends and subtract.
To find the area of the region bounded by the function y = -x^2 + 5x - 4 and the x-axis, we need to integrate the function between the x-values where the function intersects the x-axis.
First, let's find the x-values where y = 0:
0 = -x^2 + 5x - 4
We can solve this quadratic equation by factoring or using the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -1, b = 5, and c = -4. Plugging these values into the formula, we get:
x = (-5 ± √(5^2 - 4(-1)(-4))) / (2(-1))
x = (-5 ± √(25 - 16)) / (-2)
x = (-5 ± √9) / (-2)
x = (-5 ± 3) / (-2)
This gives us two possible solutions for x:
x = (-5 + 3) / (-2) = -1
x = (-5 - 3) / (-2) = -4
So the function intersects the x-axis at x = -1 and x = -4.
To find the area, we integrate the function between these two x-values. Since the function is below the x-axis in this interval, we'll integrate the negative of the function to find the area:
A = ∫[-4,-1] (-(-x^2 + 5x - 4)) dx
A = ∫[-4,-1] (x^2 - 5x + 4) dx
Now we integrate the function:
A = [(1/3)x^3 - (5/2)x^2 + 4x] from -4 to -1
Evaluating this integral at the limits:
A = [(1/3)(-1)^3 - (5/2)(-1)^2 + 4(-1)] - [(1/3)(-4)^3 - (5/2)(-4)^2 + 4(-4)]
A = [(1/3)(-1) - (5/2) + (-4)] - [(1/3)(-64) - (5/2)(16) - 16]
A = [-1/3 - 5/2 - 4] - [-64/3 + 40 - 16]
A = [-1/3 - 5/2 - 12/3] - [-64/3 + 40 - 16]
A = [-1/3 - 15/6 - 12/3] - [-64/3 + 24 - 16]
A = [-1/3 - 30/6 - 36/6] - [-64/3 + 8]
A = [-1/3 - 5 - 6] - [-64/3 + 8]
A = [-1/3 - 91/3] - [-64/3 + 8]
A = [-92/3] - [-64/3 + 8]
A = [-92/3] - [-64/3 + 24/3]
A = -92/3 + 40/3
A = -52/3
Therefore, the area of the region bounded by the function y = -x^2 + 5x - 4 and the x-axis is -52/3 square units.