Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 60050 by maxmathsup by imad last updated on 17/May/19

calculate ∫_0 ^1 (x^3 −2)(√(x^2 +3))dx

$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{3}} −\mathrm{2}\right)\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx} \\ $$

Commented by maxmathsup by imad last updated on 18/May/19

let A =∫_0 ^1 (x^3 −2)(√(x^2 +3))dx ⇒A =∫_0 ^1 x^3 (√(x^2 +3))dx −2 ∫_0 ^1 (√(x^2 +3))dx=H−K changement (√(x^2 +3))=t give x^2 +3 =t^2 ⇒2xdx =2tdt ⇒ A =∫_(√3) ^2 (t^2 −3)t tdt =∫_(√3) ^2 t^2 (t^2 −3)dt =∫_(√3) ^2 (t^4 −3t^2 )dt =[(1/5)t^5 −t^3 ]_(√3) ^2 =((32)/5) −8−(1/5)((√3))^5 +((√3))^3 ∫_0 ^1 (√(x^2 +3))dx =_(x =(√3)sh(t)) ∫_0 ^(argsh((1/(√3)))) (√3)ch(t)(√3)ch(t)dt =3 ∫_0 ^(ln((1/(√3)) +(√(1+(1/3))))) ch^2 (t)dt =(3/2) ∫_0 ^(ln((1/(√3))+(2/(√3)))) (1+ch(2t))dt =(3/2)ln((√3)) +(3/4)[sh(2t)]_0 ^(ln((√3))) =(3/2)ln((√3))+(3/8)[e^(2t) −e^(−2t) ]_0 ^(ln((√3))) =(3/4)ln(3) +(3/8){ ((√3))^2 −(1/(((√3))^2 ))} =(3/4)ln(3)+(3/8){3−(1/3)} =(3/4)ln(3) +1 so the value of A is determined .

$${let}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{3}} −\mathrm{2}\right)\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\:\Rightarrow{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\:−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}={H}−{K} \\ $$$${changement}\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}={t}\:{give}\:\:{x}^{\mathrm{2}} \:+\mathrm{3}\:={t}^{\mathrm{2}} \:\Rightarrow\mathrm{2}{xdx}\:=\mathrm{2}{tdt}\:\Rightarrow \\ $$$${A}\:=\int_{\sqrt{\mathrm{3}}} ^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{3}\right){t}\:{tdt}\:=\int_{\sqrt{\mathrm{3}}} ^{\mathrm{2}} {t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{3}\right){dt}\:=\int_{\sqrt{\mathrm{3}}} ^{\mathrm{2}} \left({t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} \right){dt} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{5}}{t}^{\mathrm{5}} −{t}^{\mathrm{3}} \right]_{\sqrt{\mathrm{3}}} ^{\mathrm{2}} \:=\frac{\mathrm{32}}{\mathrm{5}}\:−\mathrm{8}−\frac{\mathrm{1}}{\mathrm{5}}\left(\sqrt{\mathrm{3}}\right)^{\mathrm{5}} \:+\left(\sqrt{\mathrm{3}}\right)^{\mathrm{3}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\:=_{{x}\:=\sqrt{\mathrm{3}}{sh}\left({t}\right)} \:\:\:\int_{\mathrm{0}} ^{{argsh}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)} \sqrt{\mathrm{3}}{ch}\left({t}\right)\sqrt{\mathrm{3}}{ch}\left({t}\right){dt} \\ $$$$=\mathrm{3}\:\int_{\mathrm{0}} ^{{ln}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}}\right)} {ch}^{\mathrm{2}} \left({t}\right){dt}\:=\frac{\mathrm{3}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{ln}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}+\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\right)} \left(\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\sqrt{\mathrm{3}}\right)\:+\frac{\mathrm{3}}{\mathrm{4}}\left[{sh}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{{ln}\left(\sqrt{\mathrm{3}}\right)} \:=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\sqrt{\mathrm{3}}\right)+\frac{\mathrm{3}}{\mathrm{8}}\left[{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} \right]_{\mathrm{0}} ^{{ln}\left(\sqrt{\mathrm{3}}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{ln}\left(\mathrm{3}\right)\:+\frac{\mathrm{3}}{\mathrm{8}}\left\{\:\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\right\}\:=\frac{\mathrm{3}}{\mathrm{4}}{ln}\left(\mathrm{3}\right)+\frac{\mathrm{3}}{\mathrm{8}}\left\{\mathrm{3}−\frac{\mathrm{1}}{\mathrm{3}}\right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{ln}\left(\mathrm{3}\right)\:+\mathrm{1}\:\:{so}\:{the}\:{value}\:{of}\:{A}\:{is}\:{determined}\:. \\ $$$$ \\ $$

Answered by tanmay last updated on 17/May/19

∫x^3 (√(x^2 +3)) dx−2∫(√(x^2 +3)) dx I_1 =∫x^3 (√(x^2 +3)) dx t^2 =x^2 +3 2tdt=2xdx ∫x^2 ×x(√(x^2 +3)) dx ∫(t^2 −3)×t×tdt ∫(t^4 −3t^2 )dt =(t^5 /5)−t^3 +c =(((x^2 +3)^(5/2) )/5)−(x^2 +3)^(3/2) +c I_2 =∫(√(x^2 +3)) dx→use formula =((x(√(x^2 +3)))/2)+(3/2)ln(x+(√(x^2 +3)) )+c_1 I=I_1 −2I_2 I=∣(((x^2 +3)^(5/2) )/5)−(x^2 +3)^(3/2) −x(√(x^2 +3)) −3ln(x+(√(x^2 +3)) )∣_0 ^1 =(1/5){(4)^(5/2) −(3)^(5/2) }−{(4)^(3/2) −3^(3/2) }−{(√4) −0}−3ln((3/(√3))) =((32)/5)−(9/5)(√3) −8+3(√3) −2−(3/2)ln3 =((32)/5)−10+3(√3) (1−(3/5))−(3/2)ln3 =((−18)/5)+(6/5)(√3) −(3/2)ln3

$$\int{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\:{dx}−\mathrm{2}\int\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\:{dx} \\ $$$${I}_{\mathrm{1}} =\int{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\:{dx} \\ $$$${t}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{3} \\ $$$$\mathrm{2}{tdt}=\mathrm{2}{xdx} \\ $$$$\int{x}^{\mathrm{2}} ×{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\:{dx} \\ $$$$\int\left({t}^{\mathrm{2}} −\mathrm{3}\right)×{t}×{tdt} \\ $$$$\int\left({t}^{\mathrm{4}} −\mathrm{3}{t}^{\mathrm{2}} \right){dt} \\ $$$$=\frac{{t}^{\mathrm{5}} }{\mathrm{5}}−{t}^{\mathrm{3}} +{c} \\ $$$$=\frac{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{5}}−\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +{c} \\ $$$${I}_{\mathrm{2}} =\int\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\:{dx}\rightarrow{use}\:{formula} \\ $$$$=\frac{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\:\right)+{c}_{\mathrm{1}} \\ $$$${I}={I}_{\mathrm{1}} −\mathrm{2}{I}_{\mathrm{2}} \\ $$$${I}=\mid\frac{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{5}}−\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:−{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\:−\mathrm{3}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\:\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left\{\left(\mathrm{4}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} −\left(\mathrm{3}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \right\}−\left\{\left(\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{3}^{\frac{\mathrm{3}}{\mathrm{2}}} \right\}−\left\{\sqrt{\mathrm{4}}\:−\mathrm{0}\right\}−\mathrm{3}{ln}\left(\frac{\mathrm{3}}{\sqrt{\mathrm{3}}}\right) \\ $$$$=\frac{\mathrm{32}}{\mathrm{5}}−\frac{\mathrm{9}}{\mathrm{5}}\sqrt{\mathrm{3}}\:−\mathrm{8}+\mathrm{3}\sqrt{\mathrm{3}}\:−\mathrm{2}−\frac{\mathrm{3}}{\mathrm{2}}{ln}\mathrm{3} \\ $$$$=\frac{\mathrm{32}}{\mathrm{5}}−\mathrm{10}+\mathrm{3}\sqrt{\mathrm{3}}\:\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{5}}\right)−\frac{\mathrm{3}}{\mathrm{2}}{ln}\mathrm{3} \\ $$$$=\frac{−\mathrm{18}}{\mathrm{5}}+\frac{\mathrm{6}}{\mathrm{5}}\sqrt{\mathrm{3}}\:−\frac{\mathrm{3}}{\mathrm{2}}{ln}\mathrm{3} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com