7.2 definite integral — Maths STD 12 Science — Question

$a<0$     $b>0$

$I=\int_{a}^{b} \int \operatorname{sgn}(x) d x= \int_{a}^{0}(-1) d x+\int_{0}^{b}(1) d x$

$\geqslant-[x]_{a}^{0}+[x]_{0}^{b}$

$>-(0-a)+(b-0)$

$x a^{0}+b$

If $a \cdot \operatorname{sgn}(a)+b \operatorname{sgn}(b)$

$a>0, \quad b>0$

$I=\int_{a}^{b} 1 d x=[x]_{a}^{b}=b-a$

$\int_{a}^{b} \operatorname{sgn}(x) d x=|b|-|a|$