f-x-2-0-x-2t-f-t-2-dt-then-1-2-f-x-dx-

Question Number 193768 by cortano12 last updated on 19/Jun/23

f (x) = 2 + \underset{0}{\int^{x}} {(2 t + f (t))}^{2} dt

then \underset{- 1}{\int^{2}} f (x) dx =

Answered by gatocomcirrose last updated on 20/Jun/23

f^{'} (x) = {(2 x + f (x))}^{2}

v (x) = 2 x + y (x) \Rightarrow v^{'} (x) = 2 + f^{'} (x) = 2 + v {(x)}^{2}

\frac{dv}{2 + v^{2}} = dx \Rightarrow \frac{1}{\sqrt{2}} \arctan (\frac{v (x)}{\sqrt{2}}) = x + c_{1}

\Rightarrow v (x) = \sqrt{2} \tan (\sqrt{2} x + C)

\Rightarrow f (x) = \sqrt{2} \tan (\sqrt{2} x + C) - 2 x

\int_{- 1}^{2} (\sqrt{2} \tan (\sqrt{2} x + C) - 2 x) dx

= \sqrt{2} \int_{- 1}^{2} \tan (\sqrt{2} x + C) dx - 3

= {[- \ln ∣ \cos (\sqrt{2} x + C) ∣]}_{- 1}^{2} - 3

= \ln ∣ \cos (C - \sqrt{2}) ∣ - \ln ∣ \cos (2 \sqrt{2} + C) ∣ - 3

= \ln ∣ \frac{\cos (C - \sqrt{2})}{\cos (C + 2 \sqrt{2})} ∣ - 3, C \in R

Commented by mr W last updated on 20/Jun/23

f r o m f (x) ∣_{x = 0} = 2 y o u c a n g e t C .

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