To help restore a beach, sand is being added to the beach at a rate of $s(t)=65+24\sin(0.3t)$

Key concept: This is a rate-over-time problem: total sand added equals the integral of the rate function over the time interval.

Step by step

The rate is

$s(t)=65+24\sin(0.3t)$

tons per hour, where $t$ is measured in hours since 5:00 A.M.

We need the total amount added from 7:00 A.M. to 10:00 A.M.

• 7:00 A.M. is $t=2$
• 10:00 A.M. is $t=5$

So the total sand added is

$\int_2^5 \left(65+24\sin(0.3t)\right)\,dt.$

Compute the integral:

=65t-\frac{24}{0.3}\cos(0.3t).$$ Since $\frac{24}{0.3}=80$, $$=65t-80\cos(0.3t).$$ Now evaluate from 2 to 5: $$\left[65t-80\cos(0.3t)\right]_2^5 =\left(325-80\cos(1.5)\right)-\left(130-80\cos(0.6)\right).$$ $$=195-80\cos(1.5)+80\cos(0.6).$$ Numerically, $$\approx 255.368.$$ So the answer is $$\boxed{255.368}$$ which is choice **A**. ### Pitfall alert Do not confuse the time origin. Since $t$ is measured from 5:00 A.M., 7:00 A.M. corresponds to $t=2$, not $t=7$. Another common mistake is treating the rate function as the total amount instead of integrating it. ### Try different conditions If the interval changed, the process would be the same: convert the clock times into hours since 5:00 A.M., then integrate $s(t)$ over the new bounds. A longer interval would simply add more area under the rate curve. ### Further reading rate function, definite integral, accumulation