during a rainfall, the depth of water in a rain gauge increases at a rate modeled by r(t)=0.5 + t…

Explanation:

Step1: Recall the definite - integral formula

The increase in the depth of water from (t = a) to (t = b) is given by (\int_{a}^{b}R(t)dt). Here, (a = 0), (b = 3), and (R(t)=0.5 + t\cos(\frac{\pi t^{3}}{80})), so we need to calculate (\int_{0}^{3}(0.5 + t\cos(\frac{\pi t^{3}}{80}))dt=\int_{0}^{3}0.5dt+\int_{0}^{3}t\cos(\frac{\pi t^{3}}{80})dt).

Step2: Calculate (\int_{0}^{3}0.5dt)

Using the power - rule for integration (\int kdt=kt + C) (where (k) is a constant), we have (\int_{0}^{3}0.5dt=0.5t\big|_{0}^{3}=0.5\times(3 - 0)=1.5).

Step3: Calculate (\int_{0}^{3}t\cos(\frac{\pi t^{3}}{80})dt)

Let (u=\frac{\pi t^{3}}{80}), then (du=\frac{3\pi t^{2}}{80}dt), and (t dt=\frac{80}{3\pi}du). When (t = 0), (u = 0); when (t = 3), (u=\frac{\pi\times3^{3}}{80}=\frac{27\pi}{80}). So (\int_{0}^{3}t\cos(\frac{\pi t^{3}}{80})dt=\frac{80}{3\pi}\int_{0}^{\frac{27\pi}{80}}\cos(u)du). Since (\int\cos(u)du=\sin(u)+C), (\frac{80}{3\pi}\int_{0}^{\frac{27\pi}{80}}\cos(u)du=\frac{80}{3\pi}[\sin(u)]_{0}^{\frac{27\pi}{80}}=\frac{80}{3\pi}\sin(\frac{27\pi}{80})\approx0.466).

Step4: Find the total increase

Add the results of the two integrals: (1.5+\ 0.466 = 1.966).

Answer:

C. (1.966\ cm)