Given-f-x-0-x-t-4-t-2-t-2-1-dt-Find-minimum-value-of-f-x-

Question Number 134391 by bramlexs22 last updated on 03/Mar/21

Given f (x) = \int_{0}^{x} (\frac{t^{4} - t^{2}}{t^{2} + 1}) d t .

Find minimum value of f (x) .

Answered by liberty last updated on 03/Mar/21

Given f (x) = \int_{0}^{x} [\frac{t^{4} - t^{2}}{t^{2} + 1}] d t .

Find minimum value of f (x) .

(∙) \frac{df (x)}{dx} = \frac{x^{4} - x^{2}}{x^{2} + 1} = 0

x^{2} (x - 1) = 0 \to {\begin{cases} x = 0 \\ x = 1 \end{cases}

(∙ ∙) \frac{d^{2} f (x)}{{dx}^{2}} ∣_{x = 1} = \frac{({4 x}^{3} - 2 x) (x^{2} + 1) - 2 x (x^{4} - x^{2})}{{(x^{2} + 1)}^{2}} > 0 for x = 1

so minimum value is f (1)

(∙ ∙ ∙) f (1) = \int_{0}^{1} \frac{t^{4} - t^{2}}{t^{2} + 1} dt

f (1) = \int_{0}^{1} (t^{2} - 2 + \frac{2}{t^{2} + 1}) dt

f (1) = {[\frac{t^{3}}{3} - 2 t + 2 \arctan t]}_{0}^{1}

f (1) = \frac{1}{3} - 2 + 2 (\frac{π}{4}) = \frac{π}{2} - \frac{5}{6}

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