Question Number 54826 by rahul 19 last updated on 13/Feb/19
$$\left.\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\left(\left(\mathrm{1}−{x}^{\mathrm{7}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{7}}} \right){dx}\:=\:? \\ $$$$\left.\mathrm{2}\right)\:{If}\:{f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{4}\:{then}\:{the}\:{value}\:{of} \\ $$$$\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} {f}\left({x}\right){dx}\:+\:\int_{\mathrm{0}} ^{\:\mathrm{4}} {f}^{\:−\mathrm{1}} \left({x}\right){dx}\:=\:? \\ $$$$\left.\mathrm{3}\right)\:\int_{−\pi} ^{\:\pi} \left(\mathrm{1}+{cosx}+{cos}\mathrm{2}{x}+….+{cos}\mathrm{13}{x}\right)\left(\mathrm{1}+{sinx}+…+{sin}\mathrm{13}{x}\right){dx}=? \\ $$$$\left.\mathrm{4}\right)\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\left(\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:+\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\right){dx}\:=? \\ $$$$\left({This}\:{time}\:{func}.\:{are}\:{not}\:{inverse}\:{of}\:{each}\right. \\ $$$$\left.{other},{right}\:?\right) \\ $$
Commented by maxmathsup by imad last updated on 12/Feb/19
![2) we have ∫_(−1) ^1 f(x)dx =∫_(−1) ^1 (x^3 +3x+4)dx =[(x^4 /4) +(3/2)x^2 +4x]_(−1) ^1 =(1/4) +(3/2) +4 −(1/4) −(3/2) +4 =8 let find f^(−1) (x) f(x)=y ⇔x^3 +3x +4 =y ⇔x^3 +3x+4−y =0 changement x =u+v give u^3 +3uv(u+v) +v^3 +3(u+v) +4−y =0 ⇒ u^3 +v^3 +4−y + (u+v)(3uv+3) =0 ⇒u^3 +v^3 =y−4 and uv =−1 ⇒ so u^3 and v^3 are solution of the equation X^2 −(y−4)X −1 =0 ⇒ Δ =(y−4)^2 +4>0 ⇒X_1 =((y−4 +(√((y−4)^2 +4)))/2) and X_2 =((y−4 −(√((y−4)^2 )4)))/2) ⇒x =^3 (√X_1 ) +(√X_2 ) ⇒f^(−1) (x)=(((x−4+(√((x−4)^2 +4)))/2))^(1/3) +(((x−4 −(√((x−4)^2 +4)))/2))^(1/3) ⇒ ∫_0 ^4 f^(−1) (x)dx =(1/((^3 (√2))))∫_0 ^4 (x−4+(√((x−4)^2 +4)))^(1/3) dx +(1/((^3 (√2))))∫_0 ^4 (x−4−(√((x−4)^2 +4)))^(1/3) dx let I =∫_0 ^4 (x−4 +(√((x−4)^2 +4)))^(1/3) dx chang .x−4 =2sh(t) give I = ∫_(−argsh(2)) ^0 (2sh(t) +2 ch(t))^(1/3) 2ch(t)dt =2(^3 (√2)) ∫_(−ln(2+(√5))) ^0 (sh(t)+ch(t)^(1/3) ch(t)dt ....be continued... let A =∫_0 ^4 f^(−1) (x)dx changement f^(−1) (x)= t give x=f(t) ⇒ A =∫_(f^(−1) (0)) ^(f^(−1) (4)) t f^′ (t)dt = [t f(t)]_(f^(−1) (0)) ^(f^(−1) (4)) −∫_(f^(−1) (0)) ^(f^(−1) (4)) f(t)dt but f^(−1) (4) =1 −1 =0 and f^(−1) (0) =(((−4 +(√(20)))/2))^(1/3) +(((−4−(√(20)))/2))^(1/3) =(−2+(√5))^(1/3) −(2+(√5))^(1/(3 )) =α_0 ⇒ A =[tf(t)]_α_0 ^0 −∫_α_0 ^0 f(t)dt =−α_0 f(α_0 ) +∫_0 ^α_0 ( t^3 +3t+4)dt =−α_0 f(α_0 ) +(α_0 ^3 /3) +(3/2) α_0 ^2 +4α_0 so the value of ∫_0 ^4 f^(−1) (x) dx is known .](https://www.tinkutara.com/question/Q54833.png)
$$\left.\mathrm{2}\right)\:{we}\:{have}\:\int_{−\mathrm{1}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \left({x}^{\mathrm{3}} \:+\mathrm{3}{x}+\mathrm{4}\right){dx}\:=\left[\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\:+\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \:+\mathrm{4}{x}\right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{3}}{\mathrm{2}}\:+\mathrm{4}\:−\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{\mathrm{3}}{\mathrm{2}}\:+\mathrm{4}\:=\mathrm{8}\:\:\:{let}\:{find}\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$${f}\left({x}\right)={y}\:\Leftrightarrow{x}^{\mathrm{3}} \:+\mathrm{3}{x}\:+\mathrm{4}\:={y}\:\Leftrightarrow{x}^{\mathrm{3}} \:+\mathrm{3}{x}+\mathrm{4}−{y}\:=\mathrm{0}\:\:\:{changement}\:{x}\:={u}+{v}\:{give} \\ $$$${u}^{\mathrm{3}} \:+\mathrm{3}{uv}\left({u}+{v}\right)\:+{v}^{\mathrm{3}} \:\:+\mathrm{3}\left({u}+{v}\right)\:+\mathrm{4}−{y}\:=\mathrm{0}\:\Rightarrow \\ $$$${u}^{\mathrm{3}} \:+{v}^{\mathrm{3}} \:+\mathrm{4}−{y}\:+\:\left({u}+{v}\right)\left(\mathrm{3}{uv}+\mathrm{3}\right)\:=\mathrm{0}\:\Rightarrow{u}^{\mathrm{3}} \:+{v}^{\mathrm{3}} \:={y}−\mathrm{4}\:{and}\:{uv}\:=−\mathrm{1}\:\Rightarrow \\ $$$${so}\:{u}^{\mathrm{3}} \:{and}\:{v}^{\mathrm{3}} \:{are}\:{solution}\:{of}\:{the}\:{equation}\:{X}^{\mathrm{2}} −\left({y}−\mathrm{4}\right){X}\:−\mathrm{1}\:=\mathrm{0}\:\Rightarrow \\ $$$$\Delta\:=\left({y}−\mathrm{4}\right)^{\mathrm{2}} \:+\mathrm{4}>\mathrm{0}\:\Rightarrow{X}_{\mathrm{1}} =\frac{{y}−\mathrm{4}\:+\sqrt{\left({y}−\mathrm{4}\right)^{\mathrm{2}} \:+\mathrm{4}}}{\mathrm{2}}\:\:{and} \\ $$$${X}_{\mathrm{2}} =\frac{{y}−\mathrm{4}\:−\sqrt{\left.\left({y}−\mathrm{4}\right)^{\mathrm{2}} \:\right)\mathrm{4}}}{\mathrm{2}}\:\:\Rightarrow{x}\:=^{\mathrm{3}} \sqrt{{X}_{\mathrm{1}} }\:+\sqrt{{X}_{\mathrm{2}} } \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\left(\frac{{x}−\mathrm{4}+\sqrt{\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:+\left(\frac{{x}−\mathrm{4}\:−\sqrt{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \:+\mathrm{4}}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{4}} \:{f}^{−\mathrm{1}} \left({x}\right){dx}\:=\frac{\mathrm{1}}{\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)}\int_{\mathrm{0}} ^{\mathrm{4}} \:\left({x}−\mathrm{4}+\sqrt{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \:+\mathrm{4}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx}\:+\frac{\mathrm{1}}{\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)}\int_{\mathrm{0}} ^{\mathrm{4}} \left({x}−\mathrm{4}−\sqrt{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \:+\mathrm{4}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx} \\ $$$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{4}} \left({x}−\mathrm{4}\:+\sqrt{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \:+\mathrm{4}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx}\:\:\:\:{chang}\:.{x}−\mathrm{4}\:=\mathrm{2}{sh}\left({t}\right)\:{give} \\ $$$${I}\:=\:\int_{−{argsh}\left(\mathrm{2}\right)} ^{\mathrm{0}} \:\left(\mathrm{2}{sh}\left({t}\right)\:+\mathrm{2}\:{ch}\left({t}\right)\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{2}{ch}\left({t}\right){dt} \\ $$$$=\mathrm{2}\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)\:\:\int_{−{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} ^{\mathrm{0}} \:\left({sh}\left({t}\right)+{ch}\left({t}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:{ch}\left({t}\right){dt}\:\:….{be}\:{continued}…\right. \\ $$$${let}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{4}} \:{f}^{−\mathrm{1}} \left({x}\right){dx}\:\:{changement}\:{f}^{−\mathrm{1}} \left({x}\right)=\:{t}\:{give}\:{x}={f}\left({t}\right)\:\Rightarrow \\ $$$${A}\:=\int_{{f}^{−\mathrm{1}} \left(\mathrm{0}\right)} ^{{f}^{−\mathrm{1}} \left(\mathrm{4}\right)} \:\:{t}\:{f}^{'} \left({t}\right){dt}\:=\:\left[{t}\:{f}\left({t}\right)\right]_{{f}^{−\mathrm{1}} \left(\mathrm{0}\right)} ^{{f}^{−\mathrm{1}} \left(\mathrm{4}\right)} \:−\int_{{f}^{−\mathrm{1}} \left(\mathrm{0}\right)} ^{{f}^{−\mathrm{1}} \left(\mathrm{4}\right)} \:{f}\left({t}\right){dt} \\ $$$${but}\:{f}^{−\mathrm{1}} \left(\mathrm{4}\right)\:=\mathrm{1}\:−\mathrm{1}\:=\mathrm{0}\:\:{and}\:{f}^{−\mathrm{1}} \left(\mathrm{0}\right)\:=\left(\frac{−\mathrm{4}\:+\sqrt{\mathrm{20}}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:+\left(\frac{−\mathrm{4}−\sqrt{\mathrm{20}}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\left(−\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:−\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\frac{\mathrm{1}}{\mathrm{3}\:}} \:\:=\alpha_{\mathrm{0}} \:\Rightarrow \\ $$$${A}\:=\left[{tf}\left({t}\right)\right]_{\alpha_{\mathrm{0}} } ^{\mathrm{0}} \:−\int_{\alpha_{\mathrm{0}} } ^{\mathrm{0}} \:{f}\left({t}\right){dt}\:=−\alpha_{\mathrm{0}} {f}\left(\alpha_{\mathrm{0}} \right)\:+\int_{\mathrm{0}} ^{\alpha_{\mathrm{0}} } \left(\:{t}^{\mathrm{3}} \:+\mathrm{3}{t}+\mathrm{4}\right){dt} \\ $$$$=−\alpha_{\mathrm{0}} {f}\left(\alpha_{\mathrm{0}} \right)\:+\frac{\alpha_{\mathrm{0}} ^{\mathrm{3}} }{\mathrm{3}}\:+\frac{\mathrm{3}}{\mathrm{2}}\:\alpha_{\mathrm{0}} ^{\mathrm{2}} \:+\mathrm{4}\alpha_{\mathrm{0}} \\ $$$${so}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{4}} \:{f}^{−\mathrm{1}} \left({x}\right)\:{dx}\:\:{is}\:{known}\:. \\ $$
Commented by maxmathsup by imad last updated on 12/Feb/19
![in general let find ∫_a ^b f^(−1) (x)dx chang.f^(−1) (x) =t give x =f(t) ⇒ ∫_a ^b f^(−1) (x)dx =∫_(f^(−1) (a)) ^(f^(−1) (b)) t f^′ (t)dt =[t f(t)]_(f^(−1) (a)) ^(f^(−1) (b)) −∫_(f^(−1) (a)) ^(f^(−1) (b)) f(t)dt =bf^(−1) (b)−af^(−1) (a) −∫_(f^(−1) (a)) ^(f^(−1) (b)) f(t)dt .](https://www.tinkutara.com/question/Q54844.png)
$${in}\:{general}\:{let}\:{find}\:\int_{{a}} ^{{b}} \:{f}^{−\mathrm{1}} \left({x}\right){dx}\:\:{chang}.{f}^{−\mathrm{1}} \left({x}\right)\:={t}\:{give}\:{x}\:={f}\left({t}\right)\:\Rightarrow \\ $$$$\int_{{a}} ^{{b}} \:{f}^{−\mathrm{1}} \left({x}\right){dx}\:=\int_{{f}^{−\mathrm{1}} \left({a}\right)} ^{{f}^{−\mathrm{1}} \left({b}\right)} \:{t}\:{f}^{'} \left({t}\right){dt}\:=\left[{t}\:{f}\left({t}\right)\right]_{{f}^{−\mathrm{1}} \left({a}\right)} ^{{f}^{−\mathrm{1}} \left({b}\right)} \:−\int_{{f}^{−\mathrm{1}} \left({a}\right)} ^{{f}^{−\mathrm{1}} \left({b}\right)} \:{f}\left({t}\right){dt} \\ $$$$={bf}^{−\mathrm{1}} \left({b}\right)−{af}^{−\mathrm{1}} \left({a}\right)\:−\int_{{f}^{−\mathrm{1}} \left({a}\right)} ^{{f}^{−\mathrm{1}} \left({b}\right)} {f}\left({t}\right){dt}\:. \\ $$
Commented by kaivan.ahmadi last updated on 12/Feb/19
![y=f^(−1) (x)⇒f(y)=x⇒y^3 +3y+4=x,dx=(3y^2 +3)dy if x=0⇒y=−1 if x=4⇒y=0 or −3 ∫_0 ^4 f^(−1) (x)dx=∫_(−1) ^0 y(3y^2 +3)dy=[((3y^4 )/4)+((3y^2 )/2)]_(−1) ^0 =−((3/4)+(3/2))=−(9/4) is it true?](https://www.tinkutara.com/question/Q54845.png)
$$\mathrm{y}=\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)\Rightarrow\mathrm{f}\left(\mathrm{y}\right)=\mathrm{x}\Rightarrow\mathrm{y}^{\mathrm{3}} +\mathrm{3y}+\mathrm{4}=\mathrm{x},\mathrm{dx}=\left(\mathrm{3y}^{\mathrm{2}} +\mathrm{3}\right)\mathrm{dy} \\ $$$$\mathrm{if}\:\mathrm{x}=\mathrm{0}\Rightarrow\mathrm{y}=−\mathrm{1} \\ $$$$\mathrm{if}\:\mathrm{x}=\mathrm{4}\Rightarrow\mathrm{y}=\mathrm{0}\:\mathrm{or}\:−\mathrm{3}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{4}} \mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)\mathrm{dx}=\int_{−\mathrm{1}} ^{\mathrm{0}} \mathrm{y}\left(\mathrm{3y}^{\mathrm{2}} +\mathrm{3}\right)\mathrm{dy}=\left[\frac{\mathrm{3y}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{3y}^{\mathrm{2}} }{\mathrm{2}}\right]_{−\mathrm{1}} ^{\mathrm{0}} =−\left(\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}\right)=−\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{true}? \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 12/Feb/19
$$\left.\mathrm{3}\right)\:{let}\:{simlify}\:{A}\left({x}\right)=\left(\mathrm{1}+{cosx}\:+{cos}\left(\mathrm{2}{x}\right)+…+{cos}\left(\mathrm{3}{x}\right)\right)\left(\mathrm{1}+{sinx}\:+{sin}\left(\mathrm{2}{x}\right)+…+{sin}\left(\mathrm{13}{x}\right)\right) \\ $$$${let}\:{S}_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} {cos}\left({kx}\right)\:{and}\:{W}_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{sin}\left({kx}\right)\:\Rightarrow\:{S}_{{n}} \:+{iW}_{{n}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\left({e}^{{ix}} \right)^{{k}} \:=\frac{\mathrm{1}−\left({e}^{{ix}} \right)^{{n}+\mathrm{1}} }{\mathrm{1}−{e}^{{ix}} }\:=\:\frac{\mathrm{1}−{e}^{{i}\left({n}+\mathrm{1}\right){x}} }{\mathrm{1}−{cosx}\:−{isinx}} \\ $$$$=\frac{\mathrm{1}−{cos}\left({n}+\mathrm{1}\right){x}\:−{isin}\left({n}+\mathrm{1}\right){x}}{\mathrm{1}−{cosx}\:−{isinx}}\:=\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}−\mathrm{2}{i}\:{sin}\left(\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}\right){cos}\left(\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}\right)}{\mathrm{2}\:{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}{i}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$=\frac{−{i}\:{sin}\left(\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}\right)\:{e}^{{i}\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}} }{−{isin}\left(\frac{{x}}{\mathrm{2}}\right)\:{e}^{\frac{{ix}}{\mathrm{2}}} }\:=\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}\right){e}^{{i}\frac{{nx}}{\mathrm{2}}} }{{sin}\left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$=\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\left\{\:{cos}\left(\frac{{nx}}{\mathrm{2}}\right)\:+{isin}\left(\frac{{nx}}{\mathrm{2}}\right)\right\}\:\Rightarrow \\ $$$$\:{S}_{{n}} \left({x}\right)\:=\:\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}\right)\:{cos}\left(\frac{{nx}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\:{and}\:{W}_{{n}} \left({x}\right)=\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{2}}\right){sin}\left(\frac{{nx}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\:\Rightarrow \\ $$$$\mathrm{1}+{cosx}\:+{cos}\left(\mathrm{2}{x}\right)+…{cos}\left(\mathrm{13}{x}\right)=\frac{{sin}\left(\mathrm{7}{x}\right){cos}\left(\frac{\mathrm{13}{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\:{and}\: \\ $$$$\mathrm{1}+{sinx}\:+{sin}\left(\mathrm{2}{x}\right)+…+{sin}\left(\mathrm{13}{x}\right)\:=\mathrm{1}+\frac{{sin}\left(\mathrm{7}{x}\right){sin}\left(\frac{\mathrm{13}{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\:\Rightarrow \\ $$$$\left(…\right)×\left(…\right)\:=\frac{{sin}\left(\mathrm{7}{x}\right)\:{cos}\left(\frac{\mathrm{13}{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\:+\frac{{sin}^{\mathrm{2}} \left(\mathrm{7}{x}\right){sin}\left(\mathrm{13}{x}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}\:\:\Rightarrow \\ $$$$\int_{−\pi} ^{\pi} \:{A}\left({x}\right){dx}\:=\:\int_{−\pi} ^{\pi} \:\frac{{sin}\left(\mathrm{7}{x}\right){cos}\left(\frac{\mathrm{13}{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{dx}\:\:+\int_{−\pi} ^{\pi} \:\:\frac{{sin}^{\mathrm{2}} \left(\mathrm{7}{x}\right){sin}\left(\mathrm{13}{x}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{dx} \\ $$$$\int_{−\pi} ^{\pi} \:\:\frac{{sin}\left(\mathrm{7}{x}\right){cos}\left(\frac{\mathrm{13}{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\:{dx}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{sin}\left(\mathrm{7}{x}\right){cos}\left(\frac{\mathrm{13}{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{dx}\:\left({even}\:{function}\right) \\ $$$$\int_{−\pi} ^{\pi} \:\:\frac{{sin}^{\mathrm{2}} \left(\mathrm{7}{x}\right){sin}\left(\mathrm{13}{x}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{dx}\:=\mathrm{0}\:\left(\:{odd}\:{function}\right)\:…{be}\:{continued}…. \\ $$$$ \\ $$
Commented by Meritguide1234 last updated on 13/Feb/19
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Feb/19
![1)∫_0 ^1 (1−x^7 )^(1/4) dx−∫_0 ^1 (1−x^4 )^(1/7) dx I_1 −I_2 ∫_0 ^1 (1−x^a )^(1/b) dx x^a =sin^2 α x=(sinα)^(2/a) dx=(2/a)×(sinα)^((2/a)−1) (cosα)dα ∫_0 ^(π/2) (1−sin^2 α)^(1/b) ×(2/a)×(sinα)^((2/a)−1) (cosα)dα ∫_0 ^(π/2) (cosα)^(2/b) ×(2/a)×(sinα)^((2/a)−1) (cosα)dα (2/a)∫_0 ^(π/2) (sinα)^((2/a)−1) ×(cosα)^((2/b)+1) dα (1/a)×2∫_0 ^(π/2) (sinα)^(2×(1/a)−1) (cosα)^(2((1/b)+1)−1) dα formula 2∫_0 ^(π/2) (sinα)^(2p−1) (cosα)^(2q−1) dα ((⌈(p)⌈q))/(⌈(p+q))) so (1/a)×2∫_0 ^(π/2) (sinα)^(2×(1/a)−1) (cosα)^(2((1/b)+1)−1) dα =(1/a)×((⌈((1/a))⌈((1/b)+1))/(⌈((1/a)+(1/b)+1))) so I_1 =(1/7)×((⌈((1/7))⌈((1/4)+1))/(⌈((1/4)+(1/7)+1))) [a=7 b=4] I_2 =(1/4)×((⌈((1/4))⌈((1/7)+1))/(⌈((1/4)+(1/7)+1))) [a=4 b=7] required ans=I_1 −I_2 formula ⌈(n+1)=n! pls check upto this step... I_1 −I_2 =(1/(⌈(((11)/(28))+1)))[(1/7)×⌈((1/7))×⌈((1/4)+1)−(1/4)×⌈((1/4))×⌈((1/7)+1)] let a=(1/7) b=(1/4) ans=(1/(⌈(a+b+1)))×[a×⌈a)⌈(b+1)−b⌈(b)⌈(a+1)] =(1/((a+b)!))×[a(a−1)!b!−b(b−1)!a!] =(1/((a+b)!))[a!b!−b!a!]=0](https://www.tinkutara.com/question/Q54834.png)
$$\left.\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{7}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} {dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{7}}} {dx} \\ $$$${I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{{a}} \right)^{\frac{\mathrm{1}}{{b}}} {dx} \\ $$$${x}^{{a}} ={sin}^{\mathrm{2}} \alpha\:\:{x}=\left({sin}\alpha\right)^{\frac{\mathrm{2}}{{a}}} \\ $$$${dx}=\frac{\mathrm{2}}{{a}}×\left({sin}\alpha\right)^{\frac{\mathrm{2}}{{a}}−\mathrm{1}} \left({cos}\alpha\right){d}\alpha \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−{sin}^{\mathrm{2}} \alpha\right)^{\frac{\mathrm{1}}{{b}}} ×\frac{\mathrm{2}}{{a}}×\left({sin}\alpha\right)^{\frac{\mathrm{2}}{{a}}−\mathrm{1}} \left({cos}\alpha\right){d}\alpha \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({cos}\alpha\right)^{\frac{\mathrm{2}}{{b}}} ×\frac{\mathrm{2}}{{a}}×\left({sin}\alpha\right)^{\frac{\mathrm{2}}{{a}}−\mathrm{1}} \left({cos}\alpha\right){d}\alpha \\ $$$$\frac{\mathrm{2}}{{a}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left({sin}\alpha\right)^{\frac{\mathrm{2}}{{a}}−\mathrm{1}} ×\left({cos}\alpha\right)^{\frac{\mathrm{2}}{{b}}+\mathrm{1}} {d}\alpha \\ $$$$\frac{\mathrm{1}}{{a}}×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\alpha\right)^{\mathrm{2}×\frac{\mathrm{1}}{{a}}−\mathrm{1}} \left({cos}\alpha\right)^{\mathrm{2}\left(\frac{\mathrm{1}}{{b}}+\mathrm{1}\right)−\mathrm{1}} {d}\alpha \\ $$$${formula} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\alpha\right)^{\mathrm{2}{p}−\mathrm{1}} \left({cos}\alpha\right)^{\mathrm{2}{q}−\mathrm{1}} {d}\alpha \\ $$$$\frac{\left.\lceil\left({p}\right)\lceil{q}\right)}{\lceil\left({p}+{q}\right)} \\ $$$${so}\:\frac{\mathrm{1}}{{a}}×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\alpha\right)^{\mathrm{2}×\frac{\mathrm{1}}{{a}}−\mathrm{1}} \left({cos}\alpha\right)^{\mathrm{2}\left(\frac{\mathrm{1}}{{b}}+\mathrm{1}\right)−\mathrm{1}} {d}\alpha \\ $$$$=\frac{\mathrm{1}}{{a}}×\frac{\lceil\left(\frac{\mathrm{1}}{{a}}\right)\lceil\left(\frac{\mathrm{1}}{{b}}+\mathrm{1}\right)}{\lceil\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\mathrm{1}\right)} \\ $$$${so}\:\:{I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{7}}×\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{7}}\right)\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}\right)}{\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{7}}+\mathrm{1}\right)}\:\left[{a}=\mathrm{7}\:\:\:{b}=\mathrm{4}\right] \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}×\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\lceil\left(\frac{\mathrm{1}}{\mathrm{7}}+\mathrm{1}\right)}{\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{7}}+\mathrm{1}\right)}\:\left[{a}=\mathrm{4}\:\:\:{b}=\mathrm{7}\right] \\ $$$${required}\:{ans}={I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$$${formula}\:\lceil\left({n}+\mathrm{1}\right)={n}! \\ $$$${pls}\:{check}\:{upto}\:{this}\:{step}… \\ $$$$ \\ $$$${I}_{\mathrm{1}} −{I}_{\mathrm{2}} =\frac{\mathrm{1}}{\lceil\left(\frac{\mathrm{11}}{\mathrm{28}}+\mathrm{1}\right)}\left[\frac{\mathrm{1}}{\mathrm{7}}×\lceil\left(\frac{\mathrm{1}}{\mathrm{7}}\right)×\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{4}}×\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}\right)×\lceil\left(\frac{\mathrm{1}}{\mathrm{7}}+\mathrm{1}\right)\right] \\ $$$${let}\:{a}=\frac{\mathrm{1}}{\mathrm{7}}\:\:\:{b}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left.{ans}=\frac{\mathrm{1}}{\lceil\left({a}+{b}+\mathrm{1}\right)}×\left[{a}×\lceil{a}\right)\lceil\left({b}+\mathrm{1}\right)−{b}\lceil\left({b}\right)\lceil\left({a}+\mathrm{1}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\left({a}+{b}\right)!}×\left[{a}\left({a}−\mathrm{1}\right)!{b}!−{b}\left({b}−\mathrm{1}\right)!{a}!\right] \\ $$$$=\frac{\mathrm{1}}{\left({a}+{b}\right)!}\left[{a}!{b}!−{b}!{a}!\right]=\mathrm{0} \\ $$
Commented by mr W last updated on 12/Feb/19
![f(x)=(1−x^7 )^(1/4) =y ⇒x=(1−y^4 )^(1/7) ⇒f^(−1) (x)=(1−x^4 )^(1/7) =g(x) ⇒g(x) is inverse function of f(x), i.e. g(f(x))=x I_2 =∫_0 ^1 g(y)dy let y=f(x) ⇒g(y)=g(f(x))=x dy=f′(x)dx ⇒I_2 =∫_0 ^1 g(y)dy=∫_(g(0)) ^(g(1)) xf′(x)dx=∫_1 ^0 xf′(x)dx=−∫_0 ^1 xf′(x)dx I_1 =∫_0 ^1 f(x)dx I=I_1 −I_2 =∫_0 ^1 f(x)dx+∫_0 ^1 xf′(x)dx =∫_0 ^1 [f(x)+xf′(x)]dx =[xf(x)]_0 ^1 =0](https://www.tinkutara.com/question/Q54839.png)
$${f}\left({x}\right)=\left(\mathrm{1}−{x}^{\mathrm{7}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} ={y} \\ $$$$\Rightarrow{x}=\left(\mathrm{1}−{y}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{7}}} ={g}\left({x}\right) \\ $$$$\Rightarrow{g}\left({x}\right)\:{is}\:{inverse}\:{function}\:{of}\:{f}\left({x}\right),\:{i}.{e}. \\ $$$${g}\left({f}\left({x}\right)\right)={x} \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} {g}\left({y}\right){dy} \\ $$$${let}\:{y}={f}\left({x}\right) \\ $$$$\Rightarrow{g}\left({y}\right)={g}\left({f}\left({x}\right)\right)={x} \\ $$$${dy}={f}'\left({x}\right){dx} \\ $$$$\Rightarrow{I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} {g}\left({y}\right){dy}=\int_{{g}\left(\mathrm{0}\right)} ^{{g}\left(\mathrm{1}\right)} {xf}'\left({x}\right){dx}=\int_{\mathrm{1}} ^{\mathrm{0}} {xf}'\left({x}\right){dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} {xf}'\left({x}\right){dx} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx} \\ $$$${I}={I}_{\mathrm{1}} −{I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {xf}'\left({x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left[{f}\left({x}\right)+{xf}'\left({x}\right)\right]{dx} \\ $$$$=\left[{xf}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{0} \\ $$
Commented by rahul 19 last updated on 12/Feb/19
$${Sir},{I}\:'{m}\:{new}\:{to}\:{this}\:{function}! \\ $$
Commented by rahul 19 last updated on 12/Feb/19
Commented by rahul 19 last updated on 12/Feb/19
$${this}\:{is}\:{sol}^{{n}} {given}.\:{Pl}\:{explain}\:{how} \\ $$$$\:\int_{\mathrm{0}} ^{\mathrm{1}} {g}\left({y}\right){dy}\:=\:\int_{\mathrm{1}} ^{\mathrm{0}} {xf}\:'\left({x}\right){dx}\:? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 12/Feb/19
$${excellent}\:{sir}… \\ $$
Commented by rahul 19 last updated on 13/Feb/19
$${Thank}\:{you}\:{sir}! \\ $$
Answered by mr W last updated on 12/Feb/19
![Q(2) f(x)=x^3 +3x+4 I_1 =∫_(−1) ^1 f(x)dx=∫_(−1) ^0 f(x)dx+∫_0 ^1 f(x)dx I_2 =∫_0 ^4 f^(−1) (x)dx let x=f(t)⇒t=f^(−1) (x) dx=f′(t)dt I_2 =∫_(f^(−1) (0)) ^(f^(−1) (4)) tf′(t)dt=∫_(−1) ^0 tf′(t)dt=∫_(−1) ^0 xf′(x)dx I=I_1 +I_2 =∫_(−1) ^0 f(x)dx+∫_0 ^1 f(x)dx+∫_(−1) ^0 xf′(x)dx =∫_(−1) ^0 [f(x)+xf′(x)]dx+∫_0 ^1 f(x)dx =[xf(x)]_(−1) ^0 +∫_0 ^1 f(x)dx =0+∫_0 ^1 f(x)dx =∫_0 ^1 f(x)dx =∫_0 ^1 (x^3 +3x+4)dx =[(x^4 /4)+((3x^2 )/2)+4x]_0 ^1 =(1/4)+(3/2)+4 =((23)/4) ⇒∫_(−1) ^( 1) f(x)dx + ∫_0 ^( 4) f^( −1) (x)dx =((23)/4)](https://www.tinkutara.com/question/Q54842.png)
$${Q}\left(\mathrm{2}\right) \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{4} \\ $$$${I}_{\mathrm{1}} =\int_{−\mathrm{1}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\int_{−\mathrm{1}} ^{\mathrm{0}} {f}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx} \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{4}} {f}^{−\mathrm{1}} \left({x}\right){dx} \\ $$$${let}\:{x}={f}\left({t}\right)\Rightarrow{t}={f}^{−\mathrm{1}} \left({x}\right) \\ $$$${dx}={f}'\left({t}\right){dt} \\ $$$${I}_{\mathrm{2}} =\int_{{f}^{−\mathrm{1}} \left(\mathrm{0}\right)} ^{{f}^{−\mathrm{1}} \left(\mathrm{4}\right)} {tf}'\left({t}\right){dt}=\int_{−\mathrm{1}} ^{\mathrm{0}} {tf}'\left({t}\right){dt}=\int_{−\mathrm{1}} ^{\mathrm{0}} {xf}'\left({x}\right){dx} \\ $$$${I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} =\int_{−\mathrm{1}} ^{\mathrm{0}} {f}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}+\int_{−\mathrm{1}} ^{\mathrm{0}} {xf}'\left({x}\right){dx} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} \left[{f}\left({x}\right)+{xf}'\left({x}\right)\right]{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx} \\ $$$$=\left[{xf}\left({x}\right)\right]_{−\mathrm{1}} ^{\mathrm{0}} +\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx} \\ $$$$=\mathrm{0}+\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{4}\right){dx} \\ $$$$=\left[\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{4}{x}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{4} \\ $$$$=\frac{\mathrm{23}}{\mathrm{4}} \\ $$$$\:\Rightarrow\int_{−\mathrm{1}} ^{\:\mathrm{1}} {f}\left({x}\right){dx}\:+\:\int_{\mathrm{0}} ^{\:\mathrm{4}} {f}^{\:−\mathrm{1}} \left({x}\right){dx}\:=\frac{\mathrm{23}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 12/Feb/19
![certainly one can also do like this: I_1 =∫_(−1) ^1 f(x)dx=∫_(−1) ^1 (x^3 +3x+4)dx =[(x^4 /4)+((3x^2 )/2)+4x]_(−1) ^1 =8 I_2 =∫_(f^(−1) (0)) ^(f^(−1) (4)) tf′(t)dt=∫_(−1) ^0 tf′(t)dt=∫_(−1) ^0 t(3t^2 +3)dt =3∫_(−1) ^0 (t^3 +t)dt =3[(t^4 /4)+(t^2 /2)]_(−1) ^0 =−3((1/4)+(1/2)) =−(9/4) ⇒I=8−(9/4)=((23)/4)](https://www.tinkutara.com/question/Q54843.png)
$${certainly}\:{one}\:{can}\:{also}\:{do}\:{like}\:{this}: \\ $$$${I}_{\mathrm{1}} =\int_{−\mathrm{1}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\int_{−\mathrm{1}} ^{\mathrm{1}} \left({x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{4}\right){dx} \\ $$$$=\left[\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{4}{x}\right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$=\mathrm{8} \\ $$$${I}_{\mathrm{2}} =\int_{{f}^{−\mathrm{1}} \left(\mathrm{0}\right)} ^{{f}^{−\mathrm{1}} \left(\mathrm{4}\right)} {tf}'\left({t}\right){dt}=\int_{−\mathrm{1}} ^{\mathrm{0}} {tf}'\left({t}\right){dt}=\int_{−\mathrm{1}} ^{\mathrm{0}} {t}\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}\right){dt} \\ $$$$=\mathrm{3}\int_{−\mathrm{1}} ^{\mathrm{0}} \left({t}^{\mathrm{3}} +{t}\right){dt} \\ $$$$=\mathrm{3}\left[\frac{{t}^{\mathrm{4}} }{\mathrm{4}}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{−\mathrm{1}} ^{\mathrm{0}} \\ $$$$=−\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=−\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Rightarrow{I}=\mathrm{8}−\frac{\mathrm{9}}{\mathrm{4}}=\frac{\mathrm{23}}{\mathrm{4}} \\ $$
Commented by rahul 19 last updated on 13/Feb/19
$${yes}\:,\:\mathrm{2}{nd}\:{method}\:{works}\:{only}\:{when}\:{fun}^{{c}} \\ $$$${is}\:{easy}\:{to}\:{integrate}! \\ $$
Answered by mr W last updated on 12/Feb/19
$${Q}\left(\mathrm{3}\right) \\ $$$${I}=\int_{−\pi} ^{\:\pi} \left(\mathrm{1}+{cosx}+{cos}\mathrm{2}{x}+….+{cos}\mathrm{13}{x}\right)\left(\mathrm{1}+{sinx}+…+{sin}\mathrm{13}{x}\right){dx} \\ $$$$=\int_{−\pi} ^{\:\pi} \left(\mathrm{1}+{cosx}+{cos}\mathrm{2}{x}+….+{cos}\mathrm{13}{x}\right){dx}+\int_{−\pi} ^{\pi} \underset{−−−−−−−−{even}−−−−−−−−−} {\left(\mathrm{1}+{cosx}+{cos}\mathrm{2}{x}+….+{cos}\mathrm{13}{x}\right)}\left(\underset{−−−−−−{odd}−−−−−} {{sinx}+…+{sin}\mathrm{13}{x}}\right){dx} \\ $$$$=\int_{−\pi} ^{\:\pi} \left(\mathrm{1}+{cosx}+{cos}\mathrm{2}{x}+….+{cos}\mathrm{13}{x}\right){dx}+\mathrm{0} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\:\pi} \left(\mathrm{1}+{cosx}+{cos}\mathrm{2}{x}+….+{cos}\mathrm{13}{x}\right){dx} \\ $$$$=\mathrm{2}\left({x}+\mathrm{sin}\:{x}+\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}+…+\frac{\mathrm{sin}\:\mathrm{13}{x}}{\mathrm{13}}\right)_{\mathrm{0}} ^{\pi} \\ $$$$=\mathrm{2}\pi \\ $$
Commented by rahul 19 last updated on 13/Feb/19
thank you sir!
Answered by mr W last updated on 13/Feb/19
![Q(4) let I_2 =∫_0 ^2 (x^2 +2x)^(1/3) dx=∫_0 ^2 [(x+1)^2 −1]^(1/3) d(x+1) =∫_1 ^3 (t^2 −1)^(1/3) dt =∫_1 ^3 (x^2 −1)^(1/3) dx=∫_1 ^3 g(x)dx with g(x)=(x^2 −1)^(1/3) let f(x)=(√(x^3 +1))=y x=(y^2 −1)^(1/3) f^(−1) (x)=(x^2 −1)^(1/3) =g(x) let I_1 =∫_0 ^2 (√(x^3 +1))dx=∫_0 ^2 f(x)dx let x=g(t) f(x)=f(g(t))=t dx=g′(t)dt I_1 =∫_0 ^2 f(x)dx=∫_(f(0)) ^(f(2)) tg′(t)dt=∫_1 ^3 tg′(t)dt =∫_1 ^3 xg′(x)dx I=I_1 +I_2 =∫_1 ^3 xg′(x)dx+∫_1 ^3 g(x)dx =∫_1 ^3 [g(x)+xg′(x)]dx =[xg(x)]_1 ^3 =3(3^2 −1)^(1/3) −1(1^2 −1)^(1/3) =3×8^(1/3) =3×2 =6 ⇒ ∫_0 ^( 2) ((√(x^3 +1)) + (x^2 +2x)^(1/3) )dx =6](https://www.tinkutara.com/question/Q54865.png)
$${Q}\left(\mathrm{4}\right) \\ $$$${let}\:{I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx}=\int_{\mathrm{0}} ^{\mathrm{2}} \left[\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right]^{\frac{\mathrm{1}}{\mathrm{3}}} {d}\left({x}+\mathrm{1}\right) \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{3}} \left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dt} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{3}} \left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx}=\int_{\mathrm{1}} ^{\mathrm{3}} {g}\left({x}\right){dx} \\ $$$${with}\:{g}\left({x}\right)=\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$ \\ $$$$ \\ $$$${let}\:{f}\left({x}\right)=\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}={y} \\ $$$${x}=\left({y}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ={g}\left({x}\right) \\ $$$$ \\ $$$${let}\:{I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} +\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\mathrm{2}} {f}\left({x}\right){dx} \\ $$$${let}\:{x}={g}\left({t}\right) \\ $$$${f}\left({x}\right)={f}\left({g}\left({t}\right)\right)={t} \\ $$$${dx}={g}'\left({t}\right){dt} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{2}} {f}\left({x}\right){dx}=\int_{{f}\left(\mathrm{0}\right)} ^{{f}\left(\mathrm{2}\right)} {tg}'\left({t}\right){dt}=\int_{\mathrm{1}} ^{\mathrm{3}} {tg}'\left({t}\right){dt} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{3}} {xg}'\left({x}\right){dx} \\ $$$$ \\ $$$${I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} =\int_{\mathrm{1}} ^{\mathrm{3}} {xg}'\left({x}\right){dx}+\int_{\mathrm{1}} ^{\mathrm{3}} {g}\left({x}\right){dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{3}} \left[{g}\left({x}\right)+{xg}'\left({x}\right)\right]{dx} \\ $$$$=\left[{xg}\left({x}\right)\right]_{\mathrm{1}} ^{\mathrm{3}} \\ $$$$=\mathrm{3}\left(\mathrm{3}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1}\left(\mathrm{1}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\mathrm{3}×\mathrm{8}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\mathrm{3}×\mathrm{2} \\ $$$$=\mathrm{6} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\left(\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:+\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\right){dx}\:=\mathrm{6} \\ $$
Commented by rahul 19 last updated on 14/Feb/19
$${Thanks}\:{sir}! \\ $$