Question

Statement-1 : The value of the definite integral
${\int }_{19}^{37}({\left\{x\right\}}^2+3(sin2\pi x\left)\right)dx$ where {.} denotes the fractional part funtion, is $6$.
Statement-2 :
${\int }_{aT}^{bT}f\left(x\right)dx=(b-a){\int }_0^{T}f\left(x\right)dx$, where, $a$, $b$ $\in$ N such that $b>a$ and $T$ is period of $f\left(x\right)$.

• A. Statement -1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1
• B. Statement -1 is True, Statement-2 is True ; Statement-2 is not a correct explanation for Statement-1
• C. Statement-1 is True, Statement-2 is False
• D. Statement-1 is False, Statement-2 is True

Answer 1

The correct option is A Statement -1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1

Statement -2 :
${\int }_{aT}^{bT}f\left(x\right)dx={\int }_{aT}^{0}f\left(x\right)dx+{\int }_0^{bT}f\left(x\right)dx$
$=-{\int }_0^{aT}f\left(x\right)dx+{\int }_0^{bT}f\left(x\right)dx=(b-a){\int }_0^{T}f\left(x\right)dx$
$\therefore$ Statement -2 is True
Statement-1 :
${\int }_{19}^{37}\left\{\right(x-\left[x\right]{)}^2+3\left(sin2\pi x\right)\}dx=18{\int }_0^1(x^2+3sin2\pi x)dx=6$
( $\because$ both $\left\{x\right\}$ and $\sin 2\pi x$ are periodic with period $1$)
$=18{\left|[\frac{x^3}{3}-\frac{3}{2\pi }\cos 2\pi x]\right|}_0^1=6$
$\therefore$ Statement -1 is True, and is explained by statement-2