Question

STATEMENT-1 : $\int\limits_0^{2 \pi} \tan ^2 x dx =4 \int\limits_0^{\pi / 2} \tan ^2 x dx$
STATEMENT-2 : $\int\limits_0^{n T} f(x) d x=n \int\limits_0^T f(x) d x$, where $n$ is an integer and $T$ is a period of $f(x)$

• A. Both the statements are true.
• B. Statement-I is true, but Statement-II is false.
• C. Statement-I is false, but Statement-II is true.
• D. Both the statements are false.

Answer 1

Answer: (A)

$\int\limits_0^{2 \pi} \tan ^2 x d x=2 \int\limits_0^\pi \tan ^2 x d x$
$=2\left[\int\limits_0^{\pi / 2} \tan ^2 x d x+\int\limits_0^{\pi / 2} \tan ^2(\pi-x) d x\right]=4 \int\limits_0^{\pi / 2} \tan ^2 x d x$
$\therefore $ Statement 1 is true
statement-2 $\int\limits_0^{n T} f(x) d x=\int\limits_0^T f(x) d x+\int\limits_T^{2 T} f(x) d x+\ldots . .+\int\limits_{(n-1) T}^{n T} f(x) d x$
$=\int\limits_0^T f(x) d x+\int\limits_0^T f(x+T) d x+\ldots \ldots+\int\limits_0^T f(x+(n-1) T) d x$
$=\int\limits_0^T f(x) d x+\int\limits_0^T f(x) d x+\ldots . .+\int\limits_0^T f(x) d x$
$(\because f$ has a period $T)$
$=n \int\limits_0^T f(x) d x$