The value of the definite integral int_{19}^{ | vedclass

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(B) Let $f(x) = \{x\}^2 + 3 \sin(2\pi x)$.Since $\{x\}$ is periodic with period $T = 1$ and $\sin(2\pi x)$ is periodic with period $T = 1$,the functio

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$6$

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(B) Let $f(x) = \{x\}^2 + 3 \sin(2\pi x)$.Since $\{x\}$ is periodic with period $T = 1$ and $\sin(2\pi x)$ is periodic with period $T = 1$,the function $f(x)$ is periodic with period $1$.We use the property $\int_{a}^{b} f(x) \, dx = (b-a) \int_{0}^{1} f(x) \, dx$ for a periodic function with period $1$.Here,$a = 19$ and $b = 37$,so $b - a = 37 - 19 = 18$.Thus,the integral becomes $18 \int_{0}^{1} (x^2 + 3 \sin(2\pi x)) \, dx$.Evaluating the integral:$18 \left[ \frac{x^3}{3} - \frac{3}{2\pi} \cos(2\pi x) ight]_{0}^{1}$$= 18 \left( (\frac{1}{3} - \frac{3}{2\pi} \cos(2\pi)) - (0 - \frac{3}{2\pi} \cos(0)) ight)$$= 18 \left( \frac{1}{3} - \frac{3}{2\pi} - (0 - \frac{3}{2\pi}) ight)$$= 18 \left( \frac{1}{3} - \frac{3}{2\pi} + \frac{3}{2\pi} ight)$$= 18 	imes \frac{1}{3} = 6$.

The value of the definite integral $\int_{19}^{37} (\{x\}^2 + 3 \sin(2\pi x)) \, dx$,where $\{x\}$ denotes the fractional part function.

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