Total oil pumped via net change theorem

By the Net Change Theorem, the total oil pumped equals $\int_0^{24} R'(t)\, dt$. Split the integral: $\int_0^{24} 100\, dt + \int_0^{24} 20\cos\left(\frac{\pi t}{12}\right) dt$. For the second integral, let $u = \frac{\pi t}{12}$, so $du = \frac{\pi}{12} dt$, meaning $dt = \frac{12}{\pi} du$.

This gives $\int 20\cos(u) \cdot \frac{12}{\pi} du = \frac{240}{\pi}\sin(u) = \frac{240}{\pi}\sin\left(\frac{\pi t}{12}\right)$. Evaluating from $0$ to $24$: the sine term is $\sin(2\pi) - \sin(0) = 0$, since $t=24$ corresponds to $u = 2\pi$. So only the constant term contributes: $100(24) - 100(0) = 2400$ barrels.

Therefore the total oil pumped over the 24 hours is $2400$ barrels — the cosine term integrates to zero over a full period, leaving only the constant average rate of $100$ barrels/hour times $24$ hours.