The area enclosed between the curves is 3 square units.
To find the area enclosed between the curves y = x^2 + 1, y = 2x − 2, and the ordinates x = −1 and x = 2, you can set up an integral and evaluate it.
The area (A) can be found by taking the definite integral of the difference between the upper and lower curves with respect to x within the given bounds:
A = ∫_{-1}^{2} [(2x - 2) - (x^2 + 1)] dx
Now, let's integrate this expression:
A = ∫_{-1}^{2} (2x - 2 - x^2 - 1) dx
Combine like terms:
A = ∫_{-1}^{2} (-x^2 + 2x - 3) dx
Now, integrate each term separately:
A = [-x^3/3 + x^2 - 3x]_{-1}^{2}
Now, substitute the upper and lower bounds and subtract:
A = [-2^3/3 + 2^2 - 3(2)] - [(-1)^3/3 + (-1)^2 - 3(-1)]
A = [-8/3 + 4 - 6] - [1/3 + 1 + 3]
A = [-2/3] - [7/3]
Now, combine the terms:
A = -2/3 - 7/3
A = -9/3
A = -3
The negative sign indicates that the area is below the x-axis. However, the area is always positive, so take the absolute value of the result: |A| =3
Therefore, the area enclosed between the curves is 3 square units.
Question
Find the area enclosed between the curves
y=x2+1,
y=2x−2 and the ordinates x=−1 and x = 2 .