1. **Problem statement:** We are given a piecewise probability density function (pdf) for the time $T$ taken by runners to complete a marathon, defined on $[2.25,7.5]$ hours. We need to solve part (a): find the value of $\int_{2.25}^{4.5} f(t) \, dt$. 2. **Recall the pdf:** $$ f(t) = \begin{cases} \frac{4}{21}\left(1 - \cos\left(\frac{4\pi}{9}(t - 2.25)\right)\right), & 2.25 \leq t < 4.5 \\ \frac{4}{21}\left(1 + \cos\left(\frac{\pi}{3}(t - 4.5)\right)\right), & 4.5 \leq t < 7.5 \\ 0, & \text{otherwise} \end{cases} $$ 3. **Integral to find:** $$ I = \int_{2.25}^{4.5} \frac{4}{21} \left(1 - \cos\left(\frac{4\pi}{9}(t - 2.25)\right)\right) dt $$ 4. **Split the integral:** $$ I = \frac{4}{21} \int_{2.25}^{4.5} 1 \, dt - \frac{4}{21} \int_{2.25}^{4.5} \cos\left(\frac{4\pi}{9}(t - 2.25)\right) dt $$ 5. **Calculate the first integral:** $$ \int_{2.25}^{4.5} 1 \, dt = 4.5 - 2.25 = 2.25 $$ 6. **Calculate the second integral:** Let $u = \frac{4\pi}{9}(t - 2.25)$, then $du = \frac{4\pi}{9} dt$ or $dt = \frac{9}{4\pi} du$. When $t=2.25$, $u=0$; when $t=4.5$, $$ u = \frac{4\pi}{9}(4.5 - 2.25) = \frac{4\pi}{9} \times 2.25 = \frac{4\pi}{9} \times \frac{9}{4} = \pi $$ So, $$ \int_{2.25}^{4.5} \cos\left(\frac{4\pi}{9}(t - 2.25)\right) dt = \int_0^{\pi} \cos(u) \cdot \frac{9}{4\pi} du = \frac{9}{4\pi} \int_0^{\pi} \cos(u) du $$ 7. **Evaluate the cosine integral:** $$ \int_0^{\pi} \cos(u) du = \sin(u) \Big|_0^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0 $$ 8. **Therefore, the second integral is zero:** $$ \int_{2.25}^{4.5} \cos\left(\frac{4\pi}{9}(t - 2.25)\right) dt = 0 $$ 9. **Putting it all together:** $$ I = \frac{4}{21} \times 2.25 - \frac{4}{21} \times 0 = \frac{4}{21} \times 2.25 = \frac{4}{21} \times \frac{9}{4} = \frac{9}{21} = \frac{3}{7} \approx 0.4286 $$ **Final answer:** $$ \int_{2.25}^{4.5} f(t) dt = \frac{3}{7} $$