Question Number 66801 by mathmax by abdo last updated on 19/Aug/19
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{{x}+{t}^{\mathrm{2}} }{dt}\:\:\:{with}\:{x}\geqslant\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{dt}}{\:\sqrt{{x}+{t}^{\mathrm{2}} }} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\left[{of}\:\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{\mathrm{4}+{t}^{\mathrm{2}} }{dt}\:{and}\:\int_{\mathrm{0}} ^{\mathrm{2}} \frac{{dt}}{\:\sqrt{\mathrm{3}+{t}^{\mathrm{2}} }}\right. \\ $$$$\left.\mathrm{4}\right)\:{give}\:{g}^{'} \left({x}\right)\:{at}\:{form}\:{of}\:{integral}. \\ $$
Commented by mathmax by abdo last updated on 21/Aug/19
![1) f(x)=∫_0 ^2 (√(x+t^2 ))dt changement t =(√x)u give f(x)=∫_0 ^(2/( (√x))) (√x)(√(1+u^2 ))(√x)du =x ∫_0 ^(2/( (√x))) (√(1+u^2 ))du changement u=shz give f(x) =x ∫_0 ^(argsh((2/( (√x))))) ch(z)ch(z)ez =(x/2) ∫_0 ^(argsh((2/( (√)x)))) (ch(2z)+1)dz =(x/4)[sh(2z)]_0 ^(ln((2/( (√x)))+(√(1+(4/x))))) +(x/2)ln((2/( (√x)))+(√(1+(4/x)))) =(x/8)[e^(2z) −e^(−2z) ]_0 ^(ln(((2+(√(x+4)))/( (√x) )))) +(x/2)ln(((2+(√(x+4)))/( (√x)))) ⇒ f(x)=(x/8){(((2+(√(x+4)))/( (√x))))^2 −(1/((((2+(√(x+4)))/( (√x))))^2 ))}+(x/2)ln(((2+(√(x+4)))/( (√x)))) 2)we have f^′ (x)= ∫_0 ^2 (1/(2(√(x+t^2 )))) dt =(1/2)g(x) ⇒g(x) =2f^′ (x) rest to calculate f^′ (x) ..be continued...](https://www.tinkutara.com/question/Q66940.png)
$$\left.\mathrm{1}\right)\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{{x}+{t}^{\mathrm{2}} }{dt}\:\:\:{changement}\:{t}\:=\sqrt{{x}}{u}\:{give}\: \\ $$$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\mathrm{2}}{\:\sqrt{{x}}}} \:\:\sqrt{{x}}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\sqrt{{x}}{du}\:={x}\:\int_{\mathrm{0}} ^{\frac{\mathrm{2}}{\:\sqrt{{x}}}} \sqrt{\mathrm{1}+{u}^{\mathrm{2}} }{du}\:\:{changement}\:{u}={shz} \\ $$$${give}\:{f}\left({x}\right)\:={x}\:\int_{\mathrm{0}} ^{{argsh}\left(\frac{\mathrm{2}}{\:\sqrt{{x}}}\right)} {ch}\left({z}\right){ch}\left({z}\right){ez}\:=\frac{{x}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{argsh}\left(\frac{\mathrm{2}}{\:\sqrt{}{x}}\right)} \left({ch}\left(\mathrm{2}{z}\right)+\mathrm{1}\right){dz} \\ $$$$=\frac{{x}}{\mathrm{4}}\left[{sh}\left(\mathrm{2}{z}\right)\right]_{\mathrm{0}} ^{{ln}\left(\frac{\mathrm{2}}{\:\sqrt{{x}}}+\sqrt{\mathrm{1}+\frac{\mathrm{4}}{{x}}}\right)} \:+\frac{{x}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}}{\:\sqrt{{x}}}+\sqrt{\mathrm{1}+\frac{\mathrm{4}}{{x}}}\right) \\ $$$$=\frac{{x}}{\mathrm{8}}\left[{e}^{\mathrm{2}{z}} −{e}^{−\mathrm{2}{z}} \right]_{\mathrm{0}} ^{{ln}\left(\frac{\mathrm{2}+\sqrt{{x}+\mathrm{4}}}{\:\sqrt{{x}}\:}\right)} \:+\frac{{x}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}+\sqrt{{x}+\mathrm{4}}}{\:\sqrt{{x}}}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{{x}}{\mathrm{8}}\left\{\left(\frac{\mathrm{2}+\sqrt{{x}+\mathrm{4}}}{\:\sqrt{{x}}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\left(\frac{\mathrm{2}+\sqrt{{x}+\mathrm{4}}}{\:\sqrt{{x}}}\right)^{\mathrm{2}} }\right\}+\frac{{x}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}+\sqrt{{x}+\mathrm{4}}}{\:\sqrt{{x}}}\right) \\ $$$$\left.\mathrm{2}\right){we}\:{have}\:{f}^{'} \left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+{t}^{\mathrm{2}} }}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}{g}\left({x}\right)\:\Rightarrow{g}\left({x}\right)\:=\mathrm{2}{f}^{'} \left({x}\right) \\ $$$${rest}\:{to}\:{calculate}\:{f}^{'} \left({x}\right)\:..{be}\:{continued}… \\ $$
Commented by mathmax by abdo last updated on 21/Aug/19
![3) we have ∫_0 ^2 (√(4+t^2 ))dt =f(4) =(1/2){(((2+2(√2))/2))^2 −(1/((((2+2(√2))/2))^2 )) +2ln(((2+2(√2))/2)) =(1/2){ (1+(√2))^2 −(1/((1+(√2))^2 ))} +2ln(1+(√2)) ∫_0 ^2 (dt/( (√(3+t^2 )))) =_(t=(√3)u) ∫_0 ^(2/( (√3))) (((√3)du)/( (√3)(√(1+u^2 )))) =∫_0 ^(2/( (√3))) (du/( (√(1+u^2 )))) =[ln(u+(√(1+u^2 )))]_0 ^(2/( (√3))) =ln((2/( (√3))) +(√(1+(4/3)))) =ln(((2+(√7))/( (√3))))](https://www.tinkutara.com/question/Q66941.png)
$$\left.\mathrm{3}\right)\:{we}\:{have}\:\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{\mathrm{4}+{t}^{\mathrm{2}} }{dt}\:={f}\left(\mathrm{4}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left(\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\left(\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} }\right. \\ $$$$+\mathrm{2}{ln}\left(\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\right\}\:+\mathrm{2}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\frac{{dt}}{\:\sqrt{\mathrm{3}+{t}^{\mathrm{2}} }}\:=_{{t}=\sqrt{\mathrm{3}}{u}} \:\:\:\:\int_{\mathrm{0}} ^{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}} \:\:\:\:\frac{\sqrt{\mathrm{3}}{du}}{\:\sqrt{\mathrm{3}}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:=\int_{\mathrm{0}} ^{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}} \:\:\:\frac{{du}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }} \\ $$$$=\left[{ln}\left({u}+\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\right)\right]_{\mathrm{0}} ^{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}} \:\:={ln}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:+\sqrt{\mathrm{1}+\frac{\mathrm{4}}{\mathrm{3}}}\right)\:={ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{7}}}{\:\sqrt{\mathrm{3}}}\right) \\ $$
Commented by mathmax by abdo last updated on 21/Aug/19
$$\left.\mathrm{4}\right)\:{we}\:{have}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}} \left({x}+{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{dt}\:\Rightarrow{g}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} {dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{{dt}}{\left({x}+{t}^{\mathrm{2}} \right)\sqrt{{x}+{t}^{\mathrm{2}} }}\:. \\ $$