Mathematics

# Let $f(x) = \displaystyle \int_0^2 |x - t| dt (t > 0)$ the minimum value of f(x) is

$1$

##### SOLUTION
$f(x) = \int_0^2 |x - t| dt = \int_0^x (x - t) dt + \int_x^2 (t - x) dt$
$\displaystyle = tx - \frac{t^2}{2} \left |_{0}^{x} + \frac{t^2}{2} - tx \right |_x^2$
$= x^2 - \displaystyle \frac{x^2}{2} + [2 - 2x] - \left [ \frac{x^2}{2} - x^2\right ]$
$= x - 2x + 2 = (x - 1)^2 + 1$
$\therefore$ Minimum value of f(x) = $1$.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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