1. **Stating the problem:** We need 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 a given integer $n$. 2. **Understanding the problem:** This is a sum of two definite integrals involving sine functions multiplied by linear expressions in $x$. The factor outside is $5$. 3. **Integral formulas and rules:** We will use integration by parts for integrals of the form $\int x \sin(ax) \, dx$ and the linearity of integrals. Recall integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ 4. **First integral:** $$I_1 = \int_0^5 0.8x \sin(10 n \pi x) \, dx = 0.8 \int_0^5 x \sin(10 n \pi x) \, dx$$ Let $a = 10 n \pi$. Set $u = x$, $dv = \sin(ax) dx$. Then $du = dx$, $v = -\frac{\cos(ax)}{a}$. So, $$\int x \sin(ax) dx = -\frac{x \cos(ax)}{a} + \frac{1}{a} \int \cos(ax) dx = -\frac{x \cos(ax)}{a} + \frac{\sin(ax)}{a^2} + C$$ Evaluate from 0 to 5: $$\int_0^5 x \sin(ax) dx = \left[-\frac{x \cos(ax)}{a} + \frac{\sin(ax)}{a^2} \right]_0^5 = -\frac{5 \cos(5a)}{a} + \frac{\sin(5a)}{a^2} - 0$$ Since $\cos(0) = 1$ and $\sin(0) = 0$, the lower limit terms vanish. Therefore, $$I_1 = 0.8 \left(-\frac{5 \cos(5a)}{a} + \frac{\sin(5a)}{a^2} \right)$$ 5. **Second integral:** $$I_2 = \int_5^{10} (8 - 0.8x) \sin(ax) \, dx = 8 \int_5^{10} \sin(ax) dx - 0.8 \int_5^{10} x \sin(ax) dx$$ Calculate each separately: - First part: $$8 \int_5^{10} \sin(ax) dx = 8 \left[-\frac{\cos(ax)}{a} \right]_5^{10} = 8 \left(-\frac{\cos(10a)}{a} + \frac{\cos(5a)}{a} \right) = \frac{8}{a} (\cos(5a) - \cos(10a))$$ - Second part: $$\int_5^{10} x \sin(ax) dx = \left[-\frac{x \cos(ax)}{a} + \frac{\sin(ax)}{a^2} \right]_5^{10} = \left(-\frac{10 \cos(10a)}{a} + \frac{\sin(10a)}{a^2} \right) - \left(-\frac{5 \cos(5a)}{a} + \frac{\sin(5a)}{a^2} \right)$$ Simplify: $$= -\frac{10 \cos(10a)}{a} + \frac{\sin(10a)}{a^2} + \frac{5 \cos(5a)}{a} - \frac{\sin(5a)}{a^2}$$ Multiply by $-0.8$: $$-0.8 \int_5^{10} x \sin(ax) dx = 0.8 \frac{10 \cos(10a)}{a} - 0.8 \frac{\sin(10a)}{a^2} - 0.8 \frac{5 \cos(5a)}{a} + 0.8 \frac{\sin(5a)}{a^2}$$ 6. **Sum of second integral parts:** $$I_2 = \frac{8}{a} (\cos(5a) - \cos(10a)) + 0.8 \frac{10 \cos(10a)}{a} - 0.8 \frac{\sin(10a)}{a^2} - 0.8 \frac{5 \cos(5a)}{a} + 0.8 \frac{\sin(5a)}{a^2}$$ Group terms: Cosine terms: $$\frac{8}{a} \cos(5a) - \frac{8}{a} \cos(10a) + \frac{8}{a} \cos(10a) - \frac{4}{a} \cos(5a) = \left(\frac{8}{a} - \frac{4}{a} \right) \cos(5a) + \left(-\frac{8}{a} + \frac{8}{a} \right) \cos(10a) = \frac{4}{a} \cos(5a)$$ Sine terms: $$-0.8 \frac{\sin(10a)}{a^2} + 0.8 \frac{\sin(5a)}{a^2} = 0.8 \frac{\sin(5a) - \sin(10a)}{a^2}$$ So, $$I_2 = \frac{4}{a} \cos(5a) + 0.8 \frac{\sin(5a) - \sin(10a)}{a^2}$$ 7. **Total integral:** $$I = I_1 + I_2 = 0.8 \left(-\frac{5 \cos(5a)}{a} + \frac{\sin(5a)}{a^2} \right) + \frac{4}{a} \cos(5a) + 0.8 \frac{\sin(5a) - \sin(10a)}{a^2}$$ Simplify cosine terms: $$0.8 \left(-\frac{5 \cos(5a)}{a} \right) + \frac{4}{a} \cos(5a) = -\frac{4}{a} \cos(5a) + \frac{4}{a} \cos(5a) = 0$$ Simplify sine terms: $$0.8 \frac{\sin(5a)}{a^2} + 0.8 \frac{\sin(5a) - \sin(10a)}{a^2} = 0.8 \frac{2 \sin(5a) - \sin(10a)}{a^2}$$ 8. **Recall $a = 10 n \pi$ and multiply by 5:** $$b_n = 5 \times I = 5 \times 0.8 \frac{2 \sin(5a) - \sin(10a)}{a^2} = 4 \frac{2 \sin(5a) - \sin(10a)}{a^2}$$ Substitute back $a = 10 n \pi$: $$b_n = \frac{4}{(10 n \pi)^2} \left(2 \sin(50 n \pi) - \sin(100 n \pi) \right)$$ 9. **Evaluate sine terms:** Since $\sin(k \pi) = 0$ for any integer $k$, both $\sin(50 n \pi)$ and $\sin(100 n \pi)$ are zero. Therefore, $$b_n = 0$$ **Final answer:** $$\boxed{b_n = 0}$$ This means the given expression evaluates to zero for all integer $n$.