The value of ∫ limits04 |x-1|dx is

Question

The value of ∫ limits04 |x-1|dx is (A)  (5/2) (B) 5 (C) 4 (D) 1

Tardigrade Admin · Accepted Answer

∫ limits04|x-1| d x =∫ limits01-(x-1) d x+∫ limits14(x-1) d x =-[(x2/2)-x]01+[(x2/2)-x]14 =-[1 / 2-1]+[8-4-1 / 2+1] =- (-1 / 2)+(4+1 / 2) =1 / 2+9 / 2 =5

Q. The value of $0 \int 4 ∣ x - 1∣ d x$ is

$\frac{5}{2}$

$5$

$4$

$1$

$0 \int 4 ∣ x - 1∣ d x = 0 \int 1 - (x - 1) d x + 1 \int 4 (x - 1) d x$
$= - [\frac{x ^{2}}{2} - x]_{0}^{1} + [\frac{x ^{2}}{2} - x]_{1}^{4}$
$= - [1/2 - 1] + [8 - 4 - 1/2 + 1]$
$= - (- 1/2) + (4 + 1/2)$
$= 1/2 + 9/2$
$= 5$

Q. The value of 0∫4​∣x−1∣dx is

25​

5

4

1

0∫4​∣x−1∣dx=0∫1​−(x−1)dx+1∫4​(x−1)dx =−[2x2​−x]01​+[2x2​−x]14​ =−[1/2−1]+[8−4−1/2+1] =−(−1/2)+(4+1/2) =1/2+9/2 =5

Q. The value of $0 \int 4 ∣ x - 1∣ d x$ is

$\frac{5}{2}$

$5$

$4$

$1$

$0 \int 4 ∣ x - 1∣ d x = 0 \int 1 - (x - 1) d x + 1 \int 4 (x - 1) d x$
$= - [\frac{x ^{2}}{2} - x]_{0}^{1} + [\frac{x ^{2}}{2} - x]_{1}^{4}$
$= - [1/2 - 1] + [8 - 4 - 1/2 + 1]$
$= - (- 1/2) + (4 + 1/2)$
$= 1/2 + 9/2$
$= 5$