Y = cosx, X belongs to [0,3 π / 2], find the area enclosed by the curve and the coordinate axis

Y = cosx, X belongs to [0,3 π / 2], find the area enclosed by the curve and the coordinate axis

The function y = cosx, X ∈ [0,3 π / 2],
The area bounded by the curve and the coordinate axis is s = ∫ (0 → π / 2) cosx DX + ∫ (π / 2,3 π / 2) - cosx DX
∴S=sinx(0→π/2)-sinx(π/2→3π/2)
=1+2
=3.
Therefore, the area enclosed by the curve and the coordinate axis is 3

Calculate the area s of the figure surrounded by the curve y = x ^ 2 + 1, the line x + y = 3 and the two coordinate axes

x^2+y^2=x+y
(x^2-x+1/4)+(y^2-y+1/4)=1/2
(x-1/2)^2+(y-1/2)^2=1/2
So the curve represents a circle, and the radius is the root sign (1 / 2)
So the area is: П R ^ 2 = П * (√ (1 / 2)) ^ 2 = П / 2

The area enclosed by the curve y = cosx (0 ≤ x ≤ 1.5) and the coordinate axis is

The area enclosed by the curve y = cosx (0 ≤ x ≤ 1.5 pie) and the coordinate axis is equal to three times of the area enclosed by the curve y = cosx (0 ≤ x ≤ 0.5 pie) and the coordinate axis
S = 3 ∫ cosxdx = 3sinx {integral time limit 0 to π / 2} = 3

Observe the sine curve and cosine curve, write the interval of X satisfying the following conditions: (1) SiNx < 1 / 2 (2) SiNx > cosx

(1)(-7π/6+2kπ,π/6+2kπ),k∈Z
(2)(π/4+2kπ,5π/4+2kπ),k∈Z