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Question Number 133973 by liberty last updated on 26/Feb/21

Given {\begin{cases} f (x) = \sqrt[3]{x + \sqrt{x^{2} + \frac{1}{27}}} + \sqrt[3]{x - \sqrt{x^{2} + \frac{1}{27}}} \\ g (x) = x^{3} + x + 1 \end{cases}

Find \underset{0}{\int^{4}} (g \circ f \circ g) (x) dx .

Answered by EDWIN88 last updated on 26/Feb/21

recall {(a + b)}^{3} = a^{3} + b^{3} + 3 a b (a + b)

from f (x) = \sqrt[3]{x + \sqrt{x^{2} + \frac{1}{27}}} + \sqrt[3]{x - \sqrt{x^{2} + \frac{1}{27}}}

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

this yields (g \circ f) (x) = {(f (x))}^{3} + f (x) + 1 = 2 x + 1

\Rightarrow (g \circ f \circ g) (x) = 2 g (x) + 1 = {2 x}^{3} + 2 x + 3

then λ = \underset{0}{\int^{4}} ({2 x}^{3} + 2 x + 3) dx = {[\frac{x^{4}}{2} + x^{2} + 3 x]}_{0}^{4}

\underset{―}{= 128 + 16 + 12 = 156}

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