To find the area bounded by the x-axis and the graph of y=2-3x+x^2, we need to calculate the definite integral of the function with respect to x from the x-coordinate of the points where the function intersects the x-axis.
First, let's find the x-intercepts of the function:
0 = 2 - 3x + x^2
=> x^2 - 3x + 2 = 0
=> (x-1)(x-2) = 0
=> x = 1, x = 2
The area bounded by the x-axis and the graph of y=2-3x+x^2 can be calculated as:
Area = ∫[1,2] (2-3x+x^2) dx
= [2x - (3/2)x^2 + (1/3)x^3] | from 1 to 2
= 2(2) - (3/2)(2^2) + (1/3)(2^3) - [2(1) - (3/2)(1^2) + (1/3)(1^3)]
= 4 - 6 + 8/3 - (2 - 3/2 + 1/3)
= 8/3 + 1/2
= (16+3)/6
= 19/6
Therefore, the exact area bounded by the x-axis and the graph of y=2-3x+x^2 is 19/6 square units.
Calculate the exact area bounded by the x
-axis and the graph of y=2−3x+x2
.
area =
1 answer
1 answer