Find the area enclosed by the curve y=x^2-2x-3 and the line y=X+1
To find the area enclosed by the curve y = x^2 - 2x - 3 and the line y = x + 1, we need to find the points of intersection of the two functions and then calculate the area between them.
First, we set the two equations equal to each other:
x^2 - 2x - 3 = x + 1
x^2 - 3x - 4 = 0
(x - 4)(x + 1) = 0
Therefore, the points of intersection are x = 4 and x = -1.
Next, we need to find the y-coordinates for these x-values in each function:
For y = x^2 - 2x - 3:
When x = 4,
y = 4^2 - 2(4) - 3
y = 16 - 8 - 3
y = 5
When x = -1,
y = (-1)^2 - 2(-1) - 3
y = 1 + 2 - 3
y = 0
For y = x + 1:
When x = 4,
y = 4 + 1
y = 5
When x = -1,
y = -1 + 1
y = 0
Now, we can integrate the two functions and find the area between them using the points of intersection:
∫[0, 4] (x^2 - 2x - 3) dx - ∫[-1, 4] (x + 1) dx
= [x^3/3 - x^2 - 3x] from 0 to 4 - [x^2/2 + x] from -1 to 4
= [(4^3/3 - 4^2 - 3(4)) - (0) - 0] - [(4^2/2 + 4) - (-1^2/2 - 1)]
= [(64/3 - 16 - 12) - 0] - [(8 - 4) - (1/2 - 1)]
= [64/3 - 28) - 0] - [4 - 3/2]
= 36/3 - 25/2
= 12 - 12.5
= -0.5
Therefore, the area enclosed by the curve y = x^2 - 2x - 3 and the line y = x + 1 is 0.5 square units.