The area S of the region enclosed by the curve y = f(x), the x-axis, x = a, and x = b is
S = ∫_a^b |f(x)| dx
If f(x) ≧ 0, use it as it is without changing its sign; if f(x) ≦ 0, reverse its sign; if the sign changes, divide the interval and calculate.
Example ① The area enclosed by y = x², the x-axis, x = 0, and x = 2.
Solution:Since f ≧ 0 on the interval, S = ∫_0^2 x² dx = [x³/3]_0^2 = 8/3
Solution:Since f ≧ 0 on the interval, S = ∫_0^2 x² dx = [x³/3]_0^2 = 8/3
Example ② The area enclosed by y = sin x, the x-axis, x=0, and x=2π.
Solution:On [0, π], sin x ≧ 0; on [π, 2π], sin x ≦ 0.
S = ∫_0^π sin x dx + ∫_π^{2π} (−sin x) dx = 2 + 2 = 4
Solution:On [0, π], sin x ≧ 0; on [π, 2π], sin x ≦ 0.
S = ∫_0^π sin x dx + ∫_π^{2π} (−sin x) dx = 2 + 2 = 4
Key Point
- Area = ∫ |f| dx: pay attention to the sign
- Always draw the graph and divide the interval at points where the sign changes
Practice Problems
- The area enclosed by y = x² − 1, the x-axis, x=0, and x=2
Answer and Explanation
- Answer:2
Negative on [0,1] and positive on [1,2]. ∫_0^1(1−x²)dx + ∫_1^2(x²−1)dx = 2/3 + 4/3 = 2.