Mathematics

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


ANSWER

$$1$$


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
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