The value of displaystyle∫-12 (|x|/x)dx is

Question

The value of displaystyle∫-12 (|x|/x)dx is (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

Let l= displaystyle∫-12 (|x|/x) d x= displaystyle∫-10(-(x/x)) d x+ displaystyle∫02((x/x)) d x (∵|x|= begincasesx, text if x ≥ 0 -x1 text if x<0 endcases) =- displaystyle∫-10 d x+∫02 d x =-[x]-10+[x]02 =-[0+1]+[2-0] =-1+2=1

Q. The value of $\int_{- 1}^{2} \frac{∣ x ∣}{x} d x$ is

0

1

2

3

Let $l = \int_{- 1}^{2} \frac{∣ x ∣}{x} d x = \int_{- 1}^{0} (- \frac{x}{x}) d x + \int_{0}^{2} (\frac{x}{x}) d x$
$(∵ ∣ x ∣ = {x, - x_{1} if if x \geq 0 x < 0)$
$= - \int_{- 1}^{0} d x + \int_{0}^{2} d x$
$= - [x]_{- 1}^{0} + [x]_{0}^{2}$
$= - [0 + 1] + [2 - 0]$
$= - 1 + 2 = 1$

Q. The value of ∫−12​x∣x∣​dx is

0

1

2

3

Let l=∫−12​x∣x∣​dx=∫−10​(−xx​)dx+∫02​(xx​)dx (∵∣x∣={x,−x1​​ if if ​x≥0x<0​) =−∫−10​dx+∫02​dx =−[x]−10​+[x]02​ =−[0+1]+[2−0] =−1+2=1

Q. The value of $\int_{- 1}^{2} \frac{∣ x ∣}{x} d x$ is

Let $l = \int_{- 1}^{2} \frac{∣ x ∣}{x} d x = \int_{- 1}^{0} (- \frac{x}{x}) d x + \int_{0}^{2} (\frac{x}{x}) d x$
$(∵ ∣ x ∣ = {x, - x_{1} if if x \geq 0 x < 0)$
$= - \int_{- 1}^{0} d x + \int_{0}^{2} d x$
$= - [x]_{- 1}^{0} + [x]_{0}^{2}$
$= - [0 + 1] + [2 - 0]$
$= - 1 + 2 = 1$