The value of sum_{r=1}^{20} left( sqrt{pi lef | vedclass

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(B) Let $I_r = \int_0^r x |\sin \pi x| dx$ ...$(1)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we have:$I_r = \int_0^r (r-x) |\sin \pi

Vedclass · Accepted Answer

$210$

Vedclass · Answer

(B) Let $I_r = \int_0^r x |\sin \pi x| dx$ ...$(1)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we have:$I_r = \int_0^r (r-x) |\sin \pi (r-x)| dx = \int_0^r (r-x) |\sin \pi x| dx$ ...$(2)$Adding $(1)$ and $(2)$:$2I_r = \int_0^r r |\sin \pi x| dx \Rightarrow I_r = \frac{r}{2} \int_0^r |\sin \pi x| dx$Since $\int_0^n |\sin \pi x| dx = \frac{2n}{\pi}$, we have $I_r = \frac{r}{2} \cdot \frac{2r}{\pi} = \frac{r^2}{\pi}$.The expression becomes $\sum_{r=1}^{20} \sqrt{\pi \cdot \frac{r^2}{\pi}} = \sum_{r=1}^{20} r$.Sum $= \frac{20(21)}{2} = 210$.

The value of $\sum_{r=1}^{20} \left( \sqrt{\pi \left( \int_0^r x |\sin \pi x| dx \right)} \right)$ is . . . . . . .

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