If f(x)= min (|x-1|,|x|,|x+1|), then the value of 6(∫ limits-11 f(x) d x) is

Question

If f(x)= min (|x-1|,|x|,|x+1|), then the value of 6(∫ limits-11 f(x) d x) is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-wcqotqlwuzjedork.jpg" /> ∫ limits - 11 f (x) d x= shaded area =2((1/2) baseâ height) =2((1/2) ⋅ 1 ⋅ (1/2))=(1/2)

Q. If $f (x) = min (∣ x - 1∣, ∣ x ∣, ∣ x + 1∣)$ , then the value of $6 (- 1 \int 1 f (x) d x)$ is

$- 1 \int 1 f (x) d x =$ shaded area
$= 2 (\frac{1}{2}$ base⋅ height)
$= 2 (\frac{1}{2} \cdot 1 \cdot \frac{1}{2}) = \frac{1}{2}$

Q. If f(x)=min(∣x−1∣,∣x∣,∣x+1∣), then the value of 6(−1∫1​f(x)dx) is

−1∫1​f(x)dx= shaded area =2(21​ base⋅ height) =2(21​⋅1⋅21​)=21​

Q. If $f (x) = min (∣ x - 1∣, ∣ x ∣, ∣ x + 1∣)$ , then the value of $6 (- 1 \int 1 f (x) d x)$ is

$- 1 \int 1 f (x) d x =$ shaded area
$= 2 (\frac{1}{2}$ base⋅ height)
$= 2 (\frac{1}{2} \cdot 1 \cdot \frac{1}{2}) = \frac{1}{2}$