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Question Number 93038 by i jagooll last updated on 10/May/20

∫_1 ^5 (√(x^3 +1)) dx = ?

$$\underset{\mathrm{1}} {\overset{\mathrm{5}} {\int}}\:\sqrt{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}\:\mathrm{dx}\:=\:? \\ $$

Commented by abdomathmax last updated on 14/May/20

I =∫_1 ^5 (√(x^3 +1))dx changemement (√(x^3 +1))=x^(3/2) +t give x^3 +1 =(x^(3/2) +t)^2 ⇒x^3 +1 =x^3 +2t x^(3/2) +t^2 ⇒ 2t x^(3/2) +t^2 −1 =0 ⇒2t x^(3/2) =1−t^2 ⇒ x^(3/2) =((1−t^2 )/(2t)) ⇒x^3 =(((1−t^2 )/(2t)))^2 ⇒x =(((1−t^2 )/(2t)))^(2/3) ⇒ dx =(2/3)((1/(2t))−(t/2))^′ (((1−t^2 )/(2t)))^(−(1/3)) =(2/3)(−(1/(2t^2 ))−(1/2))(((1−t^2 )/(2t)))^(−(1/3)) =−(1/3)((1/t^2 )+1)(((1−t^2 )^(−(1/3)) )/(2^(−(1/3)) t^(−(1/3)) ))=−(2^(1/3) /3)(((1+t^2 )(1−t^2 )^(−(1/3)) )/t^(5/3) ) I =−(2^(1/3) /3)∫_((√2)−1) ^((√(1+5^3 ))−5^(3/2) ) (((1−t^2 )/(2t))+t)×(((1+t^2 )(1−t^2 )^(−(1/3)) )/t^(5/3) )dt =−(2^(1/3) /6)∫_((√2)−1) ^((√(1+5^3 ))−5^(3/2) ) (((1+t^2 )^2 (1−t^2 )^(−(1/3)) )/t^(8/3) ) dt after we do the changement t =sinθ be continued ....

$${I}\:\:=\int_{\mathrm{1}} ^{\mathrm{5}} \:\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}{dx}\:\:\:{changemement}\:\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}={x}^{\frac{\mathrm{3}}{\mathrm{2}}} \:+{t} \\ $$$${give}\:{x}^{\mathrm{3}} \:+\mathrm{1}\:=\left({x}^{\frac{\mathrm{3}}{\mathrm{2}}} \:+{t}\right)^{\mathrm{2}} \:\Rightarrow{x}^{\mathrm{3}} \:+\mathrm{1}\:={x}^{\mathrm{3}} \:+\mathrm{2}{t}\:{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \:+{t}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{2}{t}\:{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \:+{t}^{\mathrm{2}} −\mathrm{1}\:=\mathrm{0}\:\:\Rightarrow\mathrm{2}{t}\:{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \:=\mathrm{1}−{t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}^{\frac{\mathrm{3}}{\mathrm{2}}} \:=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}\:\Rightarrow{x}^{\mathrm{3}} \:=\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}\right)^{\mathrm{2}} \:\Rightarrow{x}\:=\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \:\Rightarrow \\ $$$${dx}\:=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}{t}}−\frac{{t}}{\mathrm{2}}\right)^{'} \left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(−\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}\right)\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{3}}} \:{t}^{−\frac{\mathrm{1}}{\mathrm{3}}} }=−\frac{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{3}}\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{3}}} }{{t}^{\frac{\mathrm{5}}{\mathrm{3}}} } \\ $$$${I}\:=−\frac{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{3}}\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\sqrt{\mathrm{1}+\mathrm{5}^{\mathrm{3}} }−\mathrm{5}^{\frac{\mathrm{3}}{\mathrm{2}}} } \left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}+{t}\right)×\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{3}}} }{{t}^{\frac{\mathrm{5}}{\mathrm{3}}} }{dt} \\ $$$$=−\frac{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{6}}\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\sqrt{\mathrm{1}+\mathrm{5}^{\mathrm{3}} }−\mathrm{5}^{\frac{\mathrm{3}}{\mathrm{2}}} } \:\:\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{3}}} }{{t}^{\frac{\mathrm{8}}{\mathrm{3}}} }\:{dt} \\ $$$${after}\:{we}\:{do}\:{the}\:{changement}\:{t}\:={sin}\theta\:\:{be}\:{continued} \\ $$$$.... \\ $$

Answered by TANMAY PANACEA ... last updated on 10/May/20

f(x)=(√(x^3 +1)) f(1)=m=(√2) f(5)=M=(√(126)) b=5 a=1 (b−a)=4 M(b−a)>∫_1 ^5 (√(x^3 +1)) dx>m(b−a) (√(126)) ×4>I>(√2) ×4

$${f}\left({x}\right)=\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\: \\ $$$${f}\left(\mathrm{1}\right)={m}=\sqrt{\mathrm{2}}\: \\ $$$${f}\left(\mathrm{5}\right)={M}=\sqrt{\mathrm{126}}\: \\ $$$${b}=\mathrm{5}\:\:\:{a}=\mathrm{1}\:\:\:\left({b}−{a}\right)=\mathrm{4} \\ $$$${M}\left({b}−{a}\right)>\int_{\mathrm{1}} ^{\mathrm{5}} \sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:{dx}>{m}\left({b}−{a}\right) \\ $$$$\sqrt{\mathrm{126}}\:×\mathrm{4}>{I}>\sqrt{\mathrm{2}}\:×\mathrm{4} \\ $$

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