Area of Definite integration example of solved equation
Example:
Find the area enclosed by the curve y = x^2 + 1, the x-axis, and the vertical lines x = 0 and x = 2.
To find the area using definite integration, we need to integrate the function from x = 0 to x = 2.
∫[0,2] (x^2 + 1) dx
Integrating this function, we get:
∫[0,2] (x^2 + 1) dx = (1/3)x^3 + x ∣[0,2]
Plugging in the limits of integration, we get:
(1/3)(2)^3 + 2 - (1/3)(0)^3 - 0 = (8/3) + 2 - 0 = 8/3 + 2 = 8/3 + 6/3 = 14/3
Therefore, the area enclosed by the curve y = x^2 + 1, the x-axis, and the vertical lines x = 0 and x = 2 is 14/3 square units.