int_{a}^{b} operatorname{sgn}(x) , dx = dots | vedclass

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(D) We know that the signum function is defined as $\operatorname{sgn}(x) = \begin{cases} 1, & x > 0 \ 0, & x = 0 \ -1, & x < 0 \end{cases}$.Also,no

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) We know that the signum function is defined as $\operatorname{sgn}(x) = \begin{cases} 1, & x > 0 \ 0, & x = 0 \ -1, & x Also,note that $x \operatorname{sgn}(x) = |x|$.Let $I = \int_{a}^{b} \operatorname{sgn}(x) \, dx$.Consider the antiderivative of $\operatorname{sgn}(x)$,which is $f(x) = |x|$.By the Fundamental Theorem of Calculus,$\int_{a}^{b} f'(x) \, dx = f(b) - f(a)$.Since $\frac{d}{dx}(|x|) = \operatorname{sgn}(x)$ for $x 
eq 0$,the integral is $\int_{a}^{b} \operatorname{sgn}(x) \, dx = [|x|]_{a}^{b} = |b| - |a|$.Alternatively,since $|x| = x \operatorname{sgn}(x)$,we have $|b| - |a| = b \operatorname{sgn}(b) - a \operatorname{sgn}(a)$.Therefore,both expressions $(A)$ and $(C)$ are equivalent.Thus,the correct option is $(D)$.

$\int_{a}^{b} \operatorname{sgn}(x) \, dx = \dots$ (where $a, b \in \mathbb{R}$)

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