Let-f-x-0-2-x-t-dt-x-gt-0-then-minimum-value-of-f-x-is-

Question Number 39342 by rahul 19 last updated on 05/Jul/18

Let f (x) = \int_{0}^{2} ∣ x - t ∣ dt (x > 0), then

minimum value of f (x) is ?

Answered by MrW3 last updated on 05/Jul/18

f o r 0 < x ⩽ 2 :

f (x) = \int_{0}^{2} ∣ x - t ∣ dt

= \int_{0}^{x} (x - t) dt + \int_{x}^{2} (x - t) d t

= {[x t - \frac{t^{2}}{2}]}_{0}^{x} - {[x t - \frac{t^{2}}{2}]}_{x}^{2}

= (x^{2} - \frac{x^{2}}{2}) - (2 x - 2 - x^{2} + \frac{x^{2}}{2})

= x^{2} - 2 x + 2

= {(x - 1)}^{2} + 1 ⩾ 1

f o r 2 ⩽ x :

f (x) = \int_{0}^{2} ∣ x - t ∣ dt

= \int_{0}^{2} (x - t) dt

= {[x t - \frac{t^{2}}{2}]}_{0}^{2}

= 2 (x - 1) ⩾ 2

m i n . f (x) = 1 a t x = 1 .

Commented by rahul 19 last updated on 05/Jul/18

Thank you sir ! ��