Help with the calculus problem please
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The area of the region is equal to the area under the curve y = cos(x) over the interval 0 ≤ x ≤ π/6, given by the integral
[tex]\displaystyle\int_0^{\frac\pi6}\cos(x)\,\mathrm dx = \sin(x)\bigg|_0^{\frac\pi6}=\sin\left(\frac\pi6\right)-\sin(0)=\boxed{\frac12}[/tex]
which follows from
• the fact that d/dx [sin(x)] = cos(x)
• the fundamental theorem of calculus