the region bounded by the x - axis and the part of the graph of $y = \\cos x$ between $x=-\frac{\\pi}{2}$…

Explanation:

Step1: Recall the area - under - curve formula

The area under the curve (y = f(x)) from (x=a) to (x = b) is given by (A=\int_{a}^{b}|f(x)|dx). Here, (y = \cos x) and for (x\in[-\frac{\pi}{2},\frac{\pi}{2}]), (\cos x\geq0). The area under (y = \cos x) from (x=-\frac{\pi}{2}) to (x=\frac{\pi}{2}) is (A=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos xdx). Since (\int\cos xdx=\sin x + C), then (A=\sin x\big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\sin\frac{\pi}{2}-\sin(-\frac{\pi}{2})=1 - (- 1)=2).

Step2: Set up the relationship between the two sub - areas

Let (A_1=\int_{-\frac{\pi}{2}}^{k}\cos xdx) and (A_2=\int_{k}^{\frac{\pi}{2}}\cos xdx). We know that (A_1 = 3A_2), and also (A_1+A_2=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos xdx = 2). Substituting (A_1 = 3A_2) into (A_1+A_2 = 2), we get (3A_2+A_2=2), so (4A_2 = 2), then (A_2=\frac{1}{2}) and (A_1=\frac{3}{2}).

Step3: Calculate (k) using the definite - integral

Since (A_1=\int_{-\frac{\pi}{2}}^{k}\cos xdx=\sin x\big|_{-\frac{\pi}{2}}^{k}=\sin k-\sin(-\frac{\pi}{2})=\sin k + 1). And (A_1=\frac{3}{2}), so (\sin k+1=\frac{3}{2}), then (\sin k=\frac{1}{2}). Since (x\in[-\frac{\pi}{2},\frac{\pi}{2}]), then (k=\frac{\pi}{6}).

Answer:

(\frac{\pi}{6})