intlimits_a^b {{mathop{rm sgn}} ,x} ,,dx = (w | vedclass

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Question

$I=\int_{a}^{b} \operatorname{sgn}(x) d x$

$a<0$     $b>0$

$I=\int_{a}^{b} \int \operatorname{sgn}(x) d x= \int_{a}^{0

Vedclass · Accepted Answer

Both $(A)$ and $(C)$

Vedclass · Answer

$I=\int_{a}^{b} \operatorname{sgn}(x) d x$

$a<0$     $b>0$

$I=\int_{a}^{b} \int \operatorname{sgn}(x) d x= \int_{a}^{0}(-1) d x+\int_{0}^{b}(1) d x$

$\geqslant-[x]_{a}^{0}+[x]_{0}^{b}$

$>-(0-a)+(b-0)$

$x a^{0}+b$

If $a \cdot \operatorname{sgn}(a)+b \operatorname{sgn}(b)$

$a>0, \quad b>0$

$I=\int_{a}^{b} 1 d x=[x]_{a}^{b}=b-a$

$\int_{a}^{b} \operatorname{sgn}(x) d x=|b|-|a|$

$\int\limits_a^b {{\mathop{\rm sgn}} \,x} \,\,dx$ $=$ (where $a, b \in R$)

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