1-x-3-5x-f-t-dt-2x-then-f-18-

Question Number 124853 by bemath last updated on 06/Dec/20

\int_{1}^{x^{3} + 5 x} f (t) d t = 2 x

t h e n f (18) = ?

Answered by liberty last updated on 06/Dec/20

\frac{d}{d x} [\int_{1}^{x^{3} + 5 x} f (t) d t] = \frac{d}{d x} (2 x)

(3 x^{2} + 5) f (x^{3} + 5 x) = 2 \Rightarrow f (x^{3} + 5 x) = \frac{2}{3 x^{2} + 5}

p u t x^{3} + 5 x = 18; (x - 2) (x^{2} + 2 x + 9) = 0

\Rightarrow x_{1} = 2 \land f (18) = \frac{2}{3 \times 4 + 5} = \frac{2}{17}

\Rightarrow x_{2, 3} = \frac{- 2 \pm 4 i \sqrt{2}}{2} = - 1 \pm 2 i \sqrt{2}

w e g e t f (18) = \frac{2}{3 (- 7 - 4 i \sqrt{2}) + 5} = \frac{2}{- 16 - 12 i \sqrt{2}} = \frac{1}{- 8 - 6 i \sqrt{2}}

a n d f (18) = \frac{2}{3 (- 7 + 4 i \sqrt{2}) + 5} = \frac{2}{- 16 + 12 i \sqrt{2}} = \frac{1}{- 8 + 6 i \sqrt{2}}

Answered by mathmax by abdo last updated on 06/Dec/20

by derivation we get ({3 x}^{2} + 5) f (x^{3} + 5 x) = 2

x = 2 \Rightarrow (12 + 5) f (8 + 10) = 2 \Rightarrow 17 f (18) = 2 \Rightarrow f (18) = \frac{2}{17}