1-0-1-1-x-7-1-4-1-x-4-1-7-dx-2-If-f-x-x-3-3x-4-then-the-value-of-1-1-f-x-dx-0-4-f-1-x-dx-3-pi-pi-1-cosx-cos2x-cos13x-1-sinx-si

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Question Number 54826 by rahul 19 last updated on 13/Feb/19

$$1){\int }_0^1({(1-x^7)}^{\frac{1}{4}}-{(1-x^4)}^{\frac{1}{7}})dx=?$$$$2)Iff\left(x\right)=x^3+3x+4thenthevalueof$$$${\int }_{-1}^{1}f\left(x\right)dx+{\int }_0^4{f}^{-1}\left(x\right)dx=?$$$$3){\int }_{-\pi }^{\pi }(1+cosx+cos2x+\dots .+cos13x)(1+sinx+\dots +sin13x)dx=?$$$$4){\int }_0^2(\sqrt{x^3+1}+{(x^2+2x)}^{\frac{1}{3}})dx=?$$$$(Thistimefunc.arenotinverseofeach$$$$other,right?)$$

Commented by maxmathsup by imad last updated on 12/Feb/19

$$2)wehave{\int }_{-1}^{1}f\left(x\right)dx={\int }_{-1}^1(x^3+3x+4)dx={[\frac{x^4}{4}+\frac{3}{2}{x}^2+4x]}_{-1}^1$$$$=\frac{1}{4}+\frac{3}{2}+4-\frac{1}{4}-\frac{3}{2}+4=8letfind{f}^{-1}\left(x\right)$$$$f\left(x\right)=y\iff x^3+3x+4=y\iff x^3+3x+4-y=0changementx=u+vgive$$$$u^3+3uv(u+v)+v^3+3(u+v)+4-y=0\Rightarrow$$$$u^3+v^3+4-y+(u+v)(3uv+3)=0\Rightarrow u^3+v^3=y-4anduv=-1\Rightarrow$$$$so{u}^{3}and{v}^{3}aresolutionoftheequation{X}^2-(y-4)X-1=0\Rightarrow$$$$\mathrm{\Delta }={(y-4)}^2+4>0\Rightarrow X_1=\frac{y-4+\sqrt{{(y-4)}^2+4}}{2}and$$$$X_2=\frac{y-4-\sqrt{{(y-4)}^2)4}}{2}\Rightarrow x{=}^3\sqrt{X_1}+\sqrt{X_2}$$$$\Rightarrow f^{-1}\left(x\right)={\left(\frac{x-4+\sqrt{{(x-4)}^2+4}}{2}\right)}^{\frac{1}{3}}+{\left(\frac{x-4-\sqrt{{(x-4)}^2+4}}{2}\right)}^{\frac{1}{3}}\Rightarrow$$$${\int }_0^4{f}^{-1}\left(x\right)dx=\frac{1}{\left({}^3\sqrt{2}\right)}{\int }_0^4{(x-4+\sqrt{{(x-4)}^2+4})}^{\frac{1}{3}}dx+\frac{1}{\left({}^3\sqrt{2}\right)}{\int }_0^4{(x-4-\sqrt{{(x-4)}^2+4})}^{\frac{1}{3}}dx$$$$letI={\int }_0^4{(x-4+\sqrt{{(x-4)}^2+4})}^{\frac{1}{3}}dxchang.x-4=2sh\left(t\right)give$$$$I={\int }_{-argsh\left(2\right)}^0{(2sh\left(t\right)+2ch\left(t\right))}^{\frac{1}{3}}2ch\left(t\right)dt$$$$=2\left({}^3\sqrt{2}\right){\int }_{-ln(2+\sqrt{5})}^0(sh\left(t\right)+ch{\left(t\right)}^{\frac{1}{3}}ch\left(t\right)dt\dots .becontinued\dots$$$$letA={\int }_0^4{f}^{-1}\left(x\right)dxchangement{f}^{-1}\left(x\right)=tgivex=f\left(t\right)\Rightarrow$$$$A={\int }_{f^{-1}\left(0\right)}^{f^{-1}\left(4\right)}t{f}^{{}^{\prime }}\left(t\right)dt={\left[tf\left(t\right)\right]}_{f^{-1}\left(0\right)}^{f^{-1}\left(4\right)}-{\int }_{f^{-1}\left(0\right)}^{f^{-1}\left(4\right)}f\left(t\right)dt$$$$but{f}^{-1}\left(4\right)=1-1=0and{f}^{-1}\left(0\right)={\left(\frac{-4+\sqrt{20}}{2}\right)}^{\frac{1}{3}}+{\left(\frac{-4-\sqrt{20}}{2}\right)}^{\frac{1}{3}}$$$$={(-2+\sqrt{5})}^{\frac{1}{3}}-{(2+\sqrt{5})}^{\frac{1}{3}}={\alpha }_0\Rightarrow$$$$A={\left[tf\left(t\right)\right]}_{{\alpha }_0}^0-{\int }_{{\alpha }_0}^{0}f\left(t\right)dt=-{\alpha }_{0}f\left({\alpha }_0\right)+{\int }_0^{{\alpha }_0}(t^3+3t+4)dt$$$$=-{\alpha }_{0}f\left({\alpha }_0\right)+\frac{{\alpha }_0^3}{3}+\frac{3}{2}{\alpha }_0^2+4{\alpha }_0$$$$sothevalueof{\int }_0^4{f}^{-1}\left(x\right)dxisknown.$$

Commented by maxmathsup by imad last updated on 12/Feb/19

$$ingeneralletfind{\int }_a^b{f}^{-1}\left(x\right)dxchang.f^{-1}\left(x\right)=tgivex=f\left(t\right)\Rightarrow$$$${\int }_a^b{f}^{-1}\left(x\right)dx={\int }_{f^{-1}\left(a\right)}^{f^{-1}\left(b\right)}t{f}^{{}^{\prime }}\left(t\right)dt={\left[tf\left(t\right)\right]}_{f^{-1}\left(a\right)}^{f^{-1}\left(b\right)}-{\int }_{f^{-1}\left(a\right)}^{f^{-1}\left(b\right)}f\left(t\right)dt$$$$={bf}^{-1}\left(b\right)-{af}^{-1}\left(a\right)-{\int }_{f^{-1}\left(a\right)}^{f^{-1}\left(b\right)}f\left(t\right)dt.$$

Commented by kaivan.ahmadi last updated on 12/Feb/19

$$\mathrm{y}={\mathrm{f}}^{-1}\left(\mathrm{x}\right)\Rightarrow \mathrm{f}\left(\mathrm{y}\right)=\mathrm{x}\Rightarrow {\mathrm{y}}^3+3\mathrm{y}+4=\mathrm{x},\mathrm{dx}=({3\mathrm{y}}^2+3)\mathrm{dy}$$$$\mathrm{if}\mathrm{x}=0\Rightarrow \mathrm{y}=-1$$$$\mathrm{if}\mathrm{x}=4\Rightarrow \mathrm{y}=0\mathrm{or}-3$$$${\int }_0^4{\mathrm{f}}^{-1}\left(\mathrm{x}\right)\mathrm{dx}={\int }_{-1}^0\mathrm{y}({3\mathrm{y}}^2+3)\mathrm{dy}={[\frac{{3\mathrm{y}}^4}{4}+\frac{{3\mathrm{y}}^2}{2}]}_{-1}^0=-(\frac{3}{4}+\frac{3}{2})=-\frac{9}{4}$$$$\mathrm{is}\mathrm{it}\mathrm{true}?$$$$$$

Commented by maxmathsup by imad last updated on 12/Feb/19

$$3)letsimlifyA\left(x\right)=(1+cosx+cos\left(2x\right)+\dots +cos\left(3x\right))(1+sinx+sin\left(2x\right)+\dots +sin\left(13x\right))$$$$let{S}_n\left(x\right)=\sum _{k=0}^{n}cos\left(kx\right)and{W}_n\left(x\right)=\sum _{k=0}^{n}sin\left(kx\right)\Rightarrow S_n+{iW}_n$$$$=\sum _{k=0}^n{\left(e^{ix}\right)}^k=\frac{1-{\left(e^{ix}\right)}^{n+1}}{1-e^{ix}}=\frac{1-e^{i(n+1)x}}{1-cosx-isinx}$$$$=\frac{1-cos(n+1)x-isin(n+1)x}{1-cosx-isinx}=\frac{2{sin}^2\frac{(n+1)x}{2}-2isin\left(\frac{(n+1)x}{2}\right)cos\left(\frac{(n+1)x}{2}\right)}{2{sin}^2\left(\frac{x}{2}\right)-2isin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right)}$$$$=\frac{-isin\left(\frac{(n+1)x}{2}\right)e^{i\frac{(n+1)x}{2}}}{-isin\left(\frac{x}{2}\right)e^{\frac{ix}{2}}}=\frac{sin\left(\frac{(n+1)x}{2}\right)e^{i\frac{nx}{2}}}{sin\left(\frac{x}{2}\right)}$$$$=\frac{sin\left(\frac{(n+1)x}{2}\right)}{sin\left(\frac{x}{2}\right)}\{cos\left(\frac{nx}{2}\right)+isin\left(\frac{nx}{2}\right)\}\Rightarrow$$$$S_n\left(x\right)=\frac{sin\left(\frac{(n+1)x}{2}\right)cos\left(\frac{nx}{2}\right)}{sin\left(\frac{x}{2}\right)}and{W}_n\left(x\right)=\frac{sin\left(\frac{(n+1)x}{2}\right)sin\left(\frac{nx}{2}\right)}{sin\left(\frac{x}{2}\right)}\Rightarrow$$$$1+cosx+cos\left(2x\right)+\dots cos\left(13x\right)=\frac{sin\left(7x\right)cos\left(\frac{13x}{2}\right)}{sin\left(\frac{x}{2}\right)}and$$$$1+sinx+sin\left(2x\right)+\dots +sin\left(13x\right)=1+\frac{sin\left(7x\right)sin\left(\frac{13x}{2}\right)}{sin\left(\frac{x}{2}\right)}\Rightarrow$$$$(\dots )\times (\dots )=\frac{sin\left(7x\right)cos\left(\frac{13x}{2}\right)}{sin\left(\frac{x}{2}\right)}+\frac{{sin}^2\left(7x\right)sin\left(13x\right)}{2{sin}^2\left(\frac{x}{2}\right)}\Rightarrow$$$${\int }_{-\pi }^{\pi }A\left(x\right)dx={\int }_{-\pi }^{\pi }\frac{sin\left(7x\right)cos\left(\frac{13x}{2}\right)}{sin\left(\frac{x}{2}\right)}dx+{\int }_{-\pi }^{\pi }\frac{{sin}^2\left(7x\right)sin\left(13x\right)}{2{sin}^2\left(\frac{x}{2}\right)}dx$$$${\int }_{-\pi }^{\pi }\frac{sin\left(7x\right)cos\left(\frac{13x}{2}\right)}{sin\left(\frac{x}{2}\right)}dx=2{\int }_0^{\pi }\frac{sin\left(7x\right)cos\left(\frac{13x}{2}\right)}{sin\left(\frac{x}{2}\right)}dx\left(evenfunction\right)$$$${\int }_{-\pi }^{\pi }\frac{{sin}^2\left(7x\right)sin\left(13x\right)}{2{sin}^2\left(\frac{x}{2}\right)}dx=0\left(oddfunction\right)\dots becontinued\dots .$$$$$$

Commented by Meritguide1234 last updated on 13/Feb/19

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Feb/19

$$1){\int }_0^1{(1-x^7)}^{\frac{1}{4}}dx-{\int }_0^1{(1-x^4)}^{\frac{1}{7}}dx$$$$I_1-I_2$$$${\int }_0^1{(1-x^a)}^{\frac{1}{b}}dx$$$$x^a={sin}^2\alpha x={\left(sin\alpha \right)}^{\frac{2}{a}}$$$$dx=\frac{2}{a}\times {\left(sin\alpha \right)}^{\frac{2}{a}-1}\left(cos\alpha \right)d\alpha$$$${\int }_0^{\frac{\pi }{2}}{(1-{sin}^2\alpha )}^{\frac{1}{b}}\times \frac{2}{a}\times {\left(sin\alpha \right)}^{\frac{2}{a}-1}\left(cos\alpha \right)d\alpha$$$${\int }_0^{\frac{\pi }{2}}{\left(cos\alpha \right)}^{\frac{2}{b}}\times \frac{2}{a}\times {\left(sin\alpha \right)}^{\frac{2}{a}-1}\left(cos\alpha \right)d\alpha$$$$\frac{2}{a}{\int }_0^{\frac{\pi }{2}}{\left(sin\alpha \right)}^{\frac{2}{a}-1}\times {\left(cos\alpha \right)}^{\frac{2}{b}+1}d\alpha$$$$\frac{1}{a}\times 2{\int }_0^{\frac{\pi }{2}}{\left(sin\alpha \right)}^{2\times \frac{1}{a}-1}{\left(cos\alpha \right)}^{2(\frac{1}{b}+1)-1}d\alpha$$$$formula$$$$2{\int }_0^{\frac{\pi }{2}}{\left(sin\alpha \right)}^{2p-1}{\left(cos\alpha \right)}^{2q-1}d\alpha$$$$\frac{\lceil \left(p\right)\lceil q)}{\lceil (p+q)}$$$$so\frac{1}{a}\times 2{\int }_0^{\frac{\pi }{2}}{\left(sin\alpha \right)}^{2\times \frac{1}{a}-1}{\left(cos\alpha \right)}^{2(\frac{1}{b}+1)-1}d\alpha$$$$=\frac{1}{a}\times \frac{\lceil \left(\frac{1}{a}\right)\lceil (\frac{1}{b}+1)}{\lceil (\frac{1}{a}+\frac{1}{b}+1)}$$$$so{I}_1=\frac{1}{7}\times \frac{\lceil \left(\frac{1}{7}\right)\lceil (\frac{1}{4}+1)}{\lceil (\frac{1}{4}+\frac{1}{7}+1)}[a=7b=4]$$$$I_2=\frac{1}{4}\times \frac{\lceil \left(\frac{1}{4}\right)\lceil (\frac{1}{7}+1)}{\lceil (\frac{1}{4}+\frac{1}{7}+1)}[a=4b=7]$$$$requiredans=I_1-I_2$$$$formula\lceil (n+1)=n!$$$$plscheckuptothisstep\dots$$$$$$$$I_1-I_2=\frac{1}{\lceil (\frac{11}{28}+1)}[\frac{1}{7}\times \lceil \left(\frac{1}{7}\right)\times \lceil (\frac{1}{4}+1)-\frac{1}{4}\times \lceil \left(\frac{1}{4}\right)\times \lceil (\frac{1}{7}+1)]$$$$leta=\frac{1}{7}b=\frac{1}{4}$$$$ans=\frac{1}{\lceil (a+b+1)}\times [a\times \lceil a)\lceil (b+1)-b\lceil \left(b\right)\lceil (a+1)]$$$$=\frac{1}{(a+b)!}\times [a(a-1)!b!-b(b-1)!a!]$$$$=\frac{1}{(a+b)!}[a!b!-b!a!]=0$$

Commented by mr W last updated on 12/Feb/19

$$f\left(x\right)={(1-x^7)}^{\frac{1}{4}}=y$$$$\Rightarrow x={(1-y^4)}^{\frac{1}{7}}$$$$\Rightarrow f^{-1}\left(x\right)={(1-x^4)}^{\frac{1}{7}}=g\left(x\right)$$$$\Rightarrow g\left(x\right)isinversefunctionoff\left(x\right),i.e.$$$$g\left(f\left(x\right)\right)=x$$$$I_2={\int }_0^{1}g\left(y\right)dy$$$$lety=f\left(x\right)$$$$\Rightarrow g\left(y\right)=g\left(f\left(x\right)\right)=x$$$$dy=f^{\prime }\left(x\right)dx$$$$\Rightarrow I_2={\int }_0^{1}g\left(y\right)dy={\int }_{g\left(0\right)}^{g\left(1\right)}{xf}^{\prime }\left(x\right)dx={\int }_1^0{xf}^{\prime }\left(x\right)dx=-{\int }_0^1{xf}^{\prime }\left(x\right)dx$$$$I_1={\int }_0^{1}f\left(x\right)dx$$$$I=I_1-I_2={\int }_0^{1}f\left(x\right)dx+{\int }_0^1{xf}^{\prime }\left(x\right)dx$$$$={\int }_0^1[f\left(x\right)+{xf}^{\prime }\left(x\right)]dx$$$$={\left[xf\left(x\right)\right]}_0^1$$$$=0$$

Commented by rahul 19 last updated on 12/Feb/19

$$Sir,I{}^{\prime }mnewtothisfunction!$$

Commented by rahul 19 last updated on 12/Feb/19

Commented by rahul 19 last updated on 12/Feb/19

$$thisis{sol}^{n}given.Plexplainhow$$$${\int }_0^{1}g\left(y\right)dy={\int }_1^{0}xf{}^{\prime }\left(x\right)dx?$$

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Feb/19

$$excellentsir\dots$$

Commented by rahul 19 last updated on 13/Feb/19

$$Thankyousir!$$

Answered by mr W last updated on 12/Feb/19

$$Q\left(2\right)$$$$f\left(x\right)=x^3+3x+4$$$$I_1={\int }_{-1}^{1}f\left(x\right)dx={\int }_{-1}^{0}f\left(x\right)dx+{\int }_0^{1}f\left(x\right)dx$$$$I_2={\int }_0^4{f}^{-1}\left(x\right)dx$$$$letx=f\left(t\right)\Rightarrow t=f^{-1}\left(x\right)$$$$dx=f^{\prime }\left(t\right)dt$$$$I_2={\int }_{f^{-1}\left(0\right)}^{f^{-1}\left(4\right)}{tf}^{\prime }\left(t\right)dt={\int }_{-1}^0{tf}^{\prime }\left(t\right)dt={\int }_{-1}^0{xf}^{\prime }\left(x\right)dx$$$$I=I_1+I_2={\int }_{-1}^{0}f\left(x\right)dx+{\int }_0^{1}f\left(x\right)dx+{\int }_{-1}^0{xf}^{\prime }\left(x\right)dx$$$$={\int }_{-1}^0[f\left(x\right)+{xf}^{\prime }\left(x\right)]dx+{\int }_0^{1}f\left(x\right)dx$$$$={\left[xf\left(x\right)\right]}_{-1}^0+{\int }_0^{1}f\left(x\right)dx$$$$=0+{\int }_0^{1}f\left(x\right)dx$$$$={\int }_0^{1}f\left(x\right)dx$$$$={\int }_0^1(x^3+3x+4)dx$$$$={[\frac{x^4}{4}+\frac{3x^2}{2}+4x]}_0^1$$$$=\frac{1}{4}+\frac{3}{2}+4$$$$=\frac{23}{4}$$$$\Rightarrow {\int }_{-1}^{1}f\left(x\right)dx+{\int }_0^4{f}^{-1}\left(x\right)dx=\frac{23}{4}$$

Commented by mr W last updated on 12/Feb/19

$$certainlyonecanalsodolikethis:$$$$I_1={\int }_{-1}^{1}f\left(x\right)dx={\int }_{-1}^1(x^3+3x+4)dx$$$$={[\frac{x^4}{4}+\frac{3x^2}{2}+4x]}_{-1}^1$$$$=8$$$$I_2={\int }_{f^{-1}\left(0\right)}^{f^{-1}\left(4\right)}{tf}^{\prime }\left(t\right)dt={\int }_{-1}^0{tf}^{\prime }\left(t\right)dt={\int }_{-1}^{0}t(3t^2+3)dt$$$$=3{\int }_{-1}^0(t^3+t)dt$$$$=3{[\frac{t^4}{4}+\frac{t^2}{2}]}_{-1}^0$$$$=-3(\frac{1}{4}+\frac{1}{2})$$$$=-\frac{9}{4}$$$$\Rightarrow I=8-\frac{9}{4}=\frac{23}{4}$$

Commented by rahul 19 last updated on 13/Feb/19

$$yes,2ndmethodworksonlywhen{fun}^c$$$$iseasytointegrate!$$

Answered by mr W last updated on 12/Feb/19

$$Q\left(3\right)$$$$I={\int }_{-\pi }^{\pi }(1+cosx+cos2x+\dots .+cos13x)(1+sinx+\dots +sin13x)dx$$$$={\int }_{-\pi }^{\pi }(1+cosx+cos2x+\dots .+cos13x)dx+{\int }_{-\pi }^{\pi }\underset{--------even---------}{(1+cosx+cos2x+\dots .+cos13x)}\left(\underset{------odd-----}{sinx+\dots +sin13x}\right)dx$$$$={\int }_{-\pi }^{\pi }(1+cosx+cos2x+\dots .+cos13x)dx+0$$$$=2{\int }_0^{\pi }(1+cosx+cos2x+\dots .+cos13x)dx$$$$=2{(x+\sin x+\frac{\sin 2x}{2}+\dots +\frac{\sin 13x}{13})}_0^{\pi }$$$$=2\pi$$

Commented by rahul 19 last updated on 13/Feb/19

thank you sir!

Answered by mr W last updated on 13/Feb/19

$$Q\left(4\right)$$$$let{I}_2={\int }_0^2{(x^2+2x)}^{\frac{1}{3}}dx={\int }_0^2{[{(x+1)}^2-1]}^{\frac{1}{3}}d(x+1)$$$$={\int }_1^3{(t^2-1)}^{\frac{1}{3}}dt$$$$={\int }_1^3{(x^2-1)}^{\frac{1}{3}}dx={\int }_1^{3}g\left(x\right)dx$$$$withg\left(x\right)={(x^2-1)}^{\frac{1}{3}}$$$$$$$$$$$$letf\left(x\right)=\sqrt{x^3+1}=y$$$$x={(y^2-1)}^{\frac{1}{3}}$$$$f^{-1}\left(x\right)={(x^2-1)}^{\frac{1}{3}}=g\left(x\right)$$$$$$$$let{I}_1={\int }_0^2\sqrt{x^3+1}dx={\int }_0^{2}f\left(x\right)dx$$$$letx=g\left(t\right)$$$$f\left(x\right)=f\left(g\left(t\right)\right)=t$$$$dx=g^{\prime }\left(t\right)dt$$$$I_1={\int }_0^{2}f\left(x\right)dx={\int }_{f\left(0\right)}^{f\left(2\right)}{tg}^{\prime }\left(t\right)dt={\int }_1^3{tg}^{\prime }\left(t\right)dt$$$$={\int }_1^3{xg}^{\prime }\left(x\right)dx$$$$$$$$I=I_1+I_2={\int }_1^3{xg}^{\prime }\left(x\right)dx+{\int }_1^{3}g\left(x\right)dx$$$$={\int }_1^3[g\left(x\right)+{xg}^{\prime }\left(x\right)]dx$$$$={\left[xg\left(x\right)\right]}_1^3$$$$=3{(3^2-1)}^{\frac{1}{3}}-1{(1^2-1)}^{\frac{1}{3}}$$$$=3\times 8^{\frac{1}{3}}$$$$=3\times 2$$$$=6$$$$\Rightarrow {\int }_0^2(\sqrt{x^3+1}+{(x^2+2x)}^{\frac{1}{3}})dx=6$$

Commented by rahul 19 last updated on 14/Feb/19

$$Thankssir!$$

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