1. **State the problem:** We want to evaluate the expression $$b_n = \frac{5}{1} \left[ \int_0^5 0.8x \sin(10 n \pi x) \, dx + \int_5^{10} (8 - 0.8x) \sin(10 n \pi x) \, dx \right]$$ for $n=5$. 2. **Rewrite the problem:** Since $b_n = 5 \left[ \int_0^5 0.8x \sin(10 n \pi x) \, dx + \int_5^{10} (8 - 0.8x) \sin(10 n \pi x) \, dx \right]$, substitute $n=5$: $$b_5 = 5 \left[ \int_0^5 0.8x \sin(50 \pi x) \, dx + \int_5^{10} (8 - 0.8x) \sin(50 \pi x) \, dx \right]$$ 3. **Evaluate the first integral:** $$I_1 = \int_0^5 0.8x \sin(50 \pi x) \, dx$$ Use integration by parts: Let $u = 0.8x$, $dv = \sin(50 \pi x) dx$. Then $du = 0.8 dx$, and $$v = -\frac{\cos(50 \pi x)}{50 \pi}$$ So, $$I_1 = uv \bigg|_0^5 - \int_0^5 v du = \left(-\frac{0.8x \cos(50 \pi x)}{50 \pi}\right)_0^5 + \int_0^5 \frac{0.8}{50 \pi} \cos(50 \pi x) dx$$ Evaluate the boundary term: At $x=5$, $\cos(50 \pi \times 5) = \cos(250 \pi) = 1$ (since $\cos(k \pi)$ is 1 if $k$ is even, and 250 is even). At $x=0$, $\cos(0) = 1$. So, $$-\frac{0.8 \times 5 \times 1}{50 \pi} + \frac{0.8 \times 0 \times 1}{50 \pi} = -\frac{4}{50 \pi} = -\frac{2}{25 \pi}$$ Now the integral: $$\int_0^5 \frac{0.8}{50 \pi} \cos(50 \pi x) dx = \frac{0.8}{50 \pi} \left( \frac{\sin(50 \pi x)}{50 \pi} \right)_0^5 = \frac{0.8}{50 \pi} \times \frac{\sin(250 \pi) - \sin(0)}{50 \pi} = 0$$ because $\sin(k \pi) = 0$ for integer $k$. Therefore, $$I_1 = -\frac{2}{25 \pi}$$ 4. **Evaluate the second integral:** $$I_2 = \int_5^{10} (8 - 0.8x) \sin(50 \pi x) \, dx = \int_5^{10} 8 \sin(50 \pi x) \, dx - \int_5^{10} 0.8x \sin(50 \pi x) \, dx$$ First part: $$\int_5^{10} 8 \sin(50 \pi x) \, dx = 8 \left( -\frac{\cos(50 \pi x)}{50 \pi} \right)_5^{10} = -\frac{8}{50 \pi} [\cos(500 \pi) - \cos(250 \pi)]$$ Since $\cos(500 \pi) = 1$ and $\cos(250 \pi) = 1$, this is $$-\frac{8}{50 \pi} (1 - 1) = 0$$ Second part: Use integration by parts again for $$\int_5^{10} 0.8x \sin(50 \pi x) \, dx$$ Let $u = 0.8x$, $dv = \sin(50 \pi x) dx$, so $du = 0.8 dx$, $v = -\frac{\cos(50 \pi x)}{50 \pi}$. Then $$\int_5^{10} 0.8x \sin(50 \pi x) dx = \left(-\frac{0.8x \cos(50 \pi x)}{50 \pi}\right)_5^{10} + \int_5^{10} \frac{0.8}{50 \pi} \cos(50 \pi x) dx$$ Evaluate boundary term: At $x=10$, $\cos(500 \pi) = 1$, so $$-\frac{0.8 \times 10 \times 1}{50 \pi} = -\frac{8}{50 \pi} = -\frac{4}{25 \pi}$$ At $x=5$, $\cos(250 \pi) = 1$, so $$-\frac{0.8 \times 5 \times 1}{50 \pi} = -\frac{4}{50 \pi} = -\frac{2}{25 \pi}$$ Difference: $$-\frac{4}{25 \pi} + \frac{2}{25 \pi} = -\frac{2}{25 \pi}$$ Integral term: $$\int_5^{10} \frac{0.8}{50 \pi} \cos(50 \pi x) dx = \frac{0.8}{50 \pi} \left( \frac{\sin(50 \pi x)}{50 \pi} \right)_5^{10} = \frac{0.8}{50 \pi} \times \frac{\sin(500 \pi) - \sin(250 \pi)}{50 \pi} = 0$$ So, $$\int_5^{10} 0.8x \sin(50 \pi x) dx = -\frac{2}{25 \pi}$$ Therefore, $$I_2 = 0 - \left(-\frac{2}{25 \pi}\right) = \frac{2}{25 \pi}$$ 5. **Sum the integrals and multiply by 5:** $$b_5 = 5 (I_1 + I_2) = 5 \left(-\frac{2}{25 \pi} + \frac{2}{25 \pi}\right) = 5 \times 0 = 0$$ **Final answer:** $$\boxed{b_5 = 0}$$