Question Number 163891 by HongKing last updated on 11/Jan/22
$$\mathrm{if}\:\:\:\mathrm{f}\left(\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{1}\right)\:=\:\mathrm{x}^{\mathrm{5}} \:+\:\mathrm{4x}\:+\:\mathrm{2} \\ $$$$\mathrm{find}\:\:\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\ $$
Answered by Ar Brandon last updated on 11/Jan/22
![x^3 +1=y⇒x=((y−1))^(1/3) ⇒f(y)=(y−1)^(5/3) +4(y−1)^(1/3) +2 ⇒f(x)=(x−1)^(5/3) +4(x−1)^(1/3) +2 ∫_0 ^1 f(x)dx=[(3/8)(x−1)^(8/3) +3(x−1)^(4/3) +2x]_0 ^1 =2−((3/8)+3)=−((11)/8)](https://www.tinkutara.com/question/Q163898.png)
$${x}^{\mathrm{3}} +\mathrm{1}={y}\Rightarrow{x}=\sqrt[{\mathrm{3}}]{{y}−\mathrm{1}} \\ $$$$\Rightarrow{f}\left({y}\right)=\left({y}−\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{3}}} +\mathrm{4}\left({y}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{2} \\ $$$$\Rightarrow{f}\left({x}\right)=\left({x}−\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{3}}} +\mathrm{4}\left({x}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{2} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\left[\frac{\mathrm{3}}{\mathrm{8}}\left({x}−\mathrm{1}\right)^{\frac{\mathrm{8}}{\mathrm{3}}} +\mathrm{3}\left({x}−\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} +\mathrm{2}{x}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{2}−\left(\frac{\mathrm{3}}{\mathrm{8}}+\mathrm{3}\right)=−\frac{\mathrm{11}}{\mathrm{8}} \\ $$
Commented by HongKing last updated on 11/Jan/22
$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Answered by mr W last updated on 11/Jan/22
![∫_0 ^1 f(x)dx =∫_(−1) ^0 f(t^3 +1)d(t^3 +1) =∫_(−1) ^0 3t^2 (t^5 +4t+2)dt =3[(t^8 /8)+t^4 +((2t^3 )/3)]_(−1) ^0 =3(−(1/8)−1+(2/3)) =−((11)/8)](https://www.tinkutara.com/question/Q163910.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} \mathrm{f}\left(\mathrm{t}^{\mathrm{3}} +\mathrm{1}\right)\mathrm{d}\left(\mathrm{t}^{\mathrm{3}} +\mathrm{1}\right) \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} \mathrm{3t}^{\mathrm{2}} \left(\mathrm{t}^{\mathrm{5}} +\mathrm{4t}+\mathrm{2}\right)\mathrm{dt} \\ $$$$=\mathrm{3}\left[\frac{\mathrm{t}^{\mathrm{8}} }{\mathrm{8}}+\mathrm{t}^{\mathrm{4}} +\frac{\mathrm{2t}^{\mathrm{3}} }{\mathrm{3}}\right]_{−\mathrm{1}} ^{\mathrm{0}} \\ $$$$=\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$$=−\frac{\mathrm{11}}{\mathrm{8}} \\ $$