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Question Number 127481 by Mathgreat last updated on 30/Dec/20

Commented by hknkrc46 last updated on 30/Dec/20

$g (x) = x^{3} + 1 \Rightarrow g^{- 1} (x) = \sqrt[3]{x - 1}$ $f (x^{3} + 1) ∣_{\sqrt[3]{x - 1}} = (x^{5} + 4 x + 2) ∣_{\sqrt[3]{x - 1}}$ $f (x) = {(x - 1)}^{\frac{5}{3}} + 4 {(x - 1)}^{\frac{1}{3}} + 2$ $\underset{0}{\int^{1}} f (x) d x = \underset{0}{\int^{1}} [{(x - 1)}^{\frac{5}{3}} + 4 {(x - 1)}^{\frac{1}{3}} + 2]$ $= {[\frac{3}{8} {(x - 1)}^{\frac{8}{3}} + 3 {(x - 1)}^{\frac{4}{3}} + 2 x]}_{0}^{1}$ $= 2 - (\frac{3}{8} + 3) = - 1 - \frac{3}{8} = - \frac{11}{8}$
	Answered by Olaf last updated on 30/Dec/20

	$Ω = \int_{0}^{1} f (x) d x$ $Let x = u^{3} + 1$ $Ω = \int_{- 1}^{0} f (u^{3} + 1) (3 u^{2} d u)$ $Ω = 3 \int_{- 1}^{0} (u^{5} + 4 u + 2) u^{2} d u$ $Ω = 3 {[\frac{x^{8}}{8} + x^{4} + \frac{2}{3} x^{3}]}_{- 1}^{0}$ $Ω = - 3 [\frac{1}{8} + 1 - \frac{2}{3}]$ $Ω = - 3 \times \frac{11}{24} = - \frac{11}{8}$
	Answered by mathmax by abdo last updated on 31/Dec/20
	$x^3 +1 =t ⇒x^3 =t−1 ⇒x=(t−1)^(1/3) ⇒f(t)=x^5 +4x+2 =(t−1)^(5/3) +4(t−1)^(1/3) +2 ⇒∫_0 ^1 f(t)dt =∫_0 ^1 (t−1)^(5/3) +4∫_0 ^1 (t−1)^(1/3) dt +2∫_0 ^1 dt =(1/(1+(5/3)))[(t−1)^(1+(5/3)) ]_0 ^1 +4 [(1/(1+(1/3)))(t−1)^((1/3)+1) ]_0 ^1 +2[t]_0 ^1 =(3/8)[(t−1)^(8/3) ]_0 ^1 +4×(3/4)[(t−1)^(4/3) ]_0 ^1 +2 =(3/8){−(−1)^(8/3) }+3{−(−1)^(4/3) } +2$
	$x^{3} + 1 = t \Rightarrow x^{3} = t - 1 \Rightarrow x = {(t - 1)}^{\frac{1}{3}} \Rightarrow f (t) = x^{5} + 4 x + 2$ $= {(t - 1)}^{\frac{5}{3}} + 4 {(t - 1)}^{\frac{1}{3}} + 2 \Rightarrow \int_{0}^{1} f (t) dt$ $= \int_{0}^{1} {(t - 1)}^{\frac{5}{3}} + 4 \int_{0}^{1} {(t - 1)}^{\frac{1}{3}} dt + 2 \int_{0}^{1} dt$ $= \frac{1}{1 + \frac{5}{3}} {[{(t - 1)}^{1 + \frac{5}{3}}]}_{0}^{1} + 4 {[\frac{1}{1 + \frac{1}{3}} {(t - 1)}^{\frac{1}{3} + 1}]}_{0}^{1} + 2 {[t]}_{0}^{1}$ $= \frac{3}{8} {[{(t - 1)}^{\frac{8}{3}}]}_{0}^{1} + 4 \times \frac{3}{4} {[{(t - 1)}^{\frac{4}{3}}]}_{0}^{1} + 2$ $= \frac{3}{8} {- {(- 1)}^{\frac{8}{3}}} + 3 {- {(- 1)}^{\frac{4}{3}}} + 2$
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