Let f x - x- -x if x - 0 and f0 - 0 and a- b- R be such that a - b- Then value of I - - ab fx dx is

The correct option is A | b| | a| Note that f (x) ={ 1 amp;if nbsp; amp;x lt;0 0 nbsp; amp;if nbsp; amp;x=0 1 nbsp; ...

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Byju's Answer

Standard XII

Mathematics

Definite Integral as Limit of Sum

Let f x =| ...

Question

Let $f (x) = \frac{| x |}{x}$ if $x \neq 0$ and $f (0) = 0$ and $a, b \in R$ be such that $a < b .$ Then value of $I = \int_{a}^{b} f (x) d x$ is

| b | - | a |

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\frac{1}{2} (b^{2} - a^{2})

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M a x {| a |, | b |}

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M i n {| a |, | b |}

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Solution

The correct option is A $| b | - | a |$
Note that $f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} - 1 & i f & x < 0 0 & i f & x = 0 1 & i f & x > 0 \end{matrix}$
If $0 \leq a < b,$ then
$I = \int_{a}^{b} d x = b - a = | b | - | a |$
If $a < 0 \leq b$ then
$I = \int_{a}^{0} (- 1) d x + \int_{0}^{b} 1 d x = a + b = b - (- a)$
$= | b | - | a |$
If $a < b < 0,$ then
$I = \int_{a}^{b} (- 1) d x = - b + a = | b | - | a |$

Suggest Corrections

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Let f(x)=|x|x if x≠0 and f(0)=0 and a,b∈R be such that a<b. Then value of I=∫baf(x)dx is

The correct option is A |b|−|a|Note that f(x)=⎧⎪⎨⎪⎩−1ifx<00ifx=01ifx>0If 0≤a<b, thenI=∫badx=b−a=|b|−|a| If a<0≤b thenI=∫0a(−1)dx+∫b01dx=a+b=b−(−a)=|b|−|a| If a<b<0, thenI=∫ba(−1)dx=−b+a=|b|−|a|

Let $f (x) = \frac{| x |}{x}$ if $x \neq 0$ and $f (0) = 0$ and $a, b \in R$ be such that $a < b .$ Then value of $I = \int_{a}^{b} f (x) d x$ is