1. **State the problem:** We want to calculate the coefficient $b_n$ for a signal defined by the piecewise integral $$b_n=\frac{2}{T}\left[\int_0^{\frac{T}{4}} 5 \sin(2\pi n f t)\,dt + \int_{\frac{T}{4}}^{\frac{T}{2}} 2.5 \sin(2\pi n f t)\,dt + \int_{\frac{T}{2}}^{\frac{3T}{4}} -2.5 \sin(2\pi n f t)\,dt + \int_{\frac{3T}{4}}^{T} -5 \sin(2\pi n f t)\,dt \right]$$ 2. **Simplify notation:** Let $\omega = 2\pi n f$. Then the integrals become $$b_n=\frac{2}{T}\left[5 \int_0^{\frac{T}{4}} \sin(\omega t) dt + 2.5 \int_{\frac{T}{4}}^{\frac{T}{2}} \sin(\omega t) dt - 2.5 \int_{\frac{T}{2}}^{\frac{3T}{4}} \sin(\omega t) dt - 5 \int_{\frac{3T}{4}}^{T} \sin(\omega t) dt \right]$$ 3. **Integrate each term:** Recall that $$\int \sin(\omega t) dt = -\frac{\cos(\omega t)}{\omega} + C$$ Calculate each integral: - $\int_0^{\frac{T}{4}} \sin(\omega t) dt = \left[-\frac{\cos(\omega t)}{\omega}\right]_0^{\frac{T}{4}} = -\frac{\cos\left(\omega \frac{T}{4}\right) - \cos(0)}{\omega} = \frac{1 - \cos\left(\frac{\omega T}{4}\right)}{\omega}$ - $\int_{\frac{T}{4}}^{\frac{T}{2}} \sin(\omega t) dt = \left[-\frac{\cos(\omega t)}{\omega}\right]_{\frac{T}{4}}^{\frac{T}{2}} = \frac{\cos\left(\frac{\omega T}{4}\right) - \cos\left(\frac{\omega T}{2}\right)}{\omega}$ - $\int_{\frac{T}{2}}^{\frac{3T}{4}} \sin(\omega t) dt = \left[-\frac{\cos(\omega t)}{\omega}\right]_{\frac{T}{2}}^{\frac{3T}{4}} = \frac{\cos\left(\frac{\omega T}{2}\right) - \cos\left(\frac{3 \omega T}{4}\right)}{\omega}$ - $\int_{\frac{3T}{4}}^{T} \sin(\omega t) dt = \left[-\frac{\cos(\omega t)}{\omega}\right]_{\frac{3T}{4}}^{T} = \frac{\cos\left(\frac{3 \omega T}{4}\right) - \cos(\omega T)}{\omega}$ 4. **Substitute back into $b_n$:** $$b_n = \frac{2}{T} \cdot \frac{1}{\omega} \left[5(1 - \cos(\frac{\omega T}{4})) + 2.5(\cos(\frac{\omega T}{4}) - \cos(\frac{\omega T}{2})) - 2.5(\cos(\frac{\omega T}{2}) - \cos(\frac{3 \omega T}{4})) - 5(\cos(\frac{3 \omega T}{4}) - \cos(\omega T)) \right]$$ 5. **Simplify the expression inside brackets:** Group terms: $$= 5 - 5 \cos\left(\frac{\omega T}{4}\right) + 2.5 \cos\left(\frac{\omega T}{4}\right) - 2.5 \cos\left(\frac{\omega T}{2}\right) - 2.5 \cos\left(\frac{\omega T}{2}\right) + 2.5 \cos\left(\frac{3 \omega T}{4}\right) - 5 \cos\left(\frac{3 \omega T}{4}\right) + 5 \cos(\omega T)$$ Combine like terms: $$= 5 - 2.5 \cos\left(\frac{\omega T}{4}\right) - 5 \cos\left(\frac{\omega T}{2}\right) - 2.5 \cos\left(\frac{3 \omega T}{4}\right) + 5 \cos(\omega T)$$ 6. **Recall that $\omega = 2 \pi n f$ and $f = \frac{1}{T}$, so** $$\omega T = 2 \pi n f T = 2 \pi n$$ Therefore, $$\cos(\omega T) = \cos(2 \pi n) = 1$$ Similarly, $$\cos\left(\frac{\omega T}{4}\right) = \cos\left(\frac{2 \pi n}{4}\right) = \cos\left(\frac{\pi n}{2}\right)$$ $$\cos\left(\frac{\omega T}{2}\right) = \cos(\pi n) = (-1)^n$$ $$\cos\left(\frac{3 \omega T}{4}\right) = \cos\left(\frac{3 \pi n}{2}\right)$$ 7. **Substitute these back:** $$b_n = \frac{2}{T} \cdot \frac{1}{2 \pi n f} \left[5 - 2.5 \cos\left(\frac{\pi n}{2}\right) - 5 (-1)^n - 2.5 \cos\left(\frac{3 \pi n}{2}\right) + 5 \cdot 1 \right]$$ Since $f = \frac{1}{T}$, then $T f = 1$, so $$b_n = \frac{2}{T} \cdot \frac{1}{2 \pi n \frac{1}{T}} \left[10 - 2.5 \cos\left(\frac{\pi n}{2}\right) - 5 (-1)^n - 2.5 \cos\left(\frac{3 \pi n}{2}\right) \right] = \frac{2}{T} \cdot \frac{T}{2 \pi n} \left[10 - 2.5 \cos\left(\frac{\pi n}{2}\right) - 5 (-1)^n - 2.5 \cos\left(\frac{3 \pi n}{2}\right) \right]$$ Simplify: $$b_n = \frac{1}{\pi n} \left[10 - 2.5 \cos\left(\frac{\pi n}{2}\right) - 5 (-1)^n - 2.5 \cos\left(\frac{3 \pi n}{2}\right) \right]$$ 8. **Use trigonometric identities:** Note that $$\cos\left(\frac{3 \pi n}{2}\right) = \cos\left(\frac{\pi n}{2} + \pi n\right) = \cos\left(\frac{\pi n}{2}\right) \cos(\pi n) - \sin\left(\frac{\pi n}{2}\right) \sin(\pi n)$$ Since $\sin(\pi n) = 0$ for integer $n$, and $\cos(\pi n) = (-1)^n$, we get $$\cos\left(\frac{3 \pi n}{2}\right) = \cos\left(\frac{\pi n}{2}\right) (-1)^n$$ 9. **Substitute this back:** $$b_n = \frac{1}{\pi n} \left[10 - 2.5 \cos\left(\frac{\pi n}{2}\right) - 5 (-1)^n - 2.5 \cos\left(\frac{\pi n}{2}\right) (-1)^n \right]$$ Group terms: $$= \frac{1}{\pi n} \left[10 - 5 (-1)^n - 2.5 \cos\left(\frac{\pi n}{2}\right) (1 + (-1)^n) \right]$$ 10. **Final expression:** $$\boxed{b_n = \frac{1}{\pi n} \left[10 - 5 (-1)^n - 2.5 \cos\left(\frac{\pi n}{2}\right) (1 + (-1)^n) \right]}$$ This formula gives the coefficient $b_n$ for the given piecewise signal. **Summary:** We integrated the piecewise function, simplified using trigonometric identities, and expressed $b_n$ in terms of $n$ only.