Definite Integrals Part 4

Concept:The sum involves an integral with $|\sin \pi x|$ which is periodic with period 1.By substitution and periodicity, the integral simplifies to a sum of odd numbers, yielding $r^2/\pi$.Explanation:Define $J_r = \int_0^r x |\sin \pi x| \, dx$.We need $I = \sum_{r=1}^{20} \sqrt{\pi J_r}$.Substitute $t = \pi x$, so $dx = dt/\pi$.Then $J_r = \int_0^{\pi r} \frac{t}{\pi} |\sin t| \frac{dt}{\pi} = \frac1{\pi^2} \int_0^{\pi r} t |\sin t| \, dt$.Since $|\sin t|$ has period $\pi$, break the integral into intervals of length $\pi$: $\int_0^{\pi r} t |\sin t|\, dt = \sum_{k=0}^{r-1} \int_{k\pi}^{(k+1)\pi} t |\sin t|\, dt$.In $(0,\pi)$, $|\sin t| = \sin t$; in $(\pi,2\pi)$, $|\sin t| = -\sin t$; pattern alternates with sign $(-1)^k$.Thus $\int_0^{\pi r} t|\sin t|\, dt = \sum_{k=0}^{r-1} (-1)^k \int_{k\pi}^{(k+1)\pi} t \sin t \, dt$.Integrate $\int t \sin t \, dt = -t\cos t + \sin t$ (by parts).Evaluate at limits: for interval $k\pi$ to $(k+1)\pi$, let $F(t)=-t\cos t+\sin t$.$F((k+1)\pi) - F(k\pi) = [-(k+1)\pi(-1)^{k+1} + 0] - [-k\pi(-1)^k + 0] = (k+1)\pi(-1)^k - k\pi(-1)^{k+1}$? Wait carefully: $\cos(k\pi)=(-1)^k$, $\sin(k\pi)=0$. So:$F((k+1)\pi) = -(k+1)\pi (-1)^{k+1} = (k+1)\pi (-1)^k$ (since $-(-1)^{k+1}=(-1)^k$).$F(k\pi) = -k\pi (-1)^k = -k\pi (-1)^k$.Difference = $(k+1)\pi (-1)^k + k\pi (-1)^k = (2k+1)\pi (-1)^k$.But in the alternating sum, we have $(-1)^k$ times this difference, so term becomes $(2k+1)\pi$.Thus $\int_0^{\pi r} t |\sin t|\, dt = \sum_{k=0}^{r-1} (2k+1)\pi = \pi \cdot (\text{sum of first } r \text{ odd numbers}) = \pi r^2$.Therefore $J_r = \frac1{\pi^2} \cdot \pi r^2 = \frac{r^2}{\pi}$.Now $I = \sum_{r=1}^{20} \sqrt{\pi \cdot \frac{r^2}{\pi}} = \sum_{r=1}^{20} r = \frac{20 \times 21}{2} = 210$.Answer:$210$ (option A).