
Να βρείτε το εμβαδό του χωρίου που περικλείεται απο την
τον xx' , και τις ευθείες x = 0 και x = 1Απάντηση:
Συντονιστής: R BORIS

τον xx' , και τις ευθείες x = 0 και x = 1
τον
, και τις ευθείες
και 
έτσι για κάθε
θα ισχύει ότι 
![\displaystyle{E = \int_{\,0}^{\,1} {\left( {\int_{\,0}^{\,x} {\frac{{{t^4}}}{{{t^6} + 1}}dt} } \right)} dx = \int_{\,0}^{\,1} {\left( {x'\int_{\,0}^{\,x} {\frac{{{t^4}}}{{{t^6} + 1}}dt} } \right)} dx = \left[ {x\int_{\,0}^{\,x} {\frac{{{t^4}}}{{{t^6} + 1}}dt} } \right]_0^1 - \int_{\,0}^{\,1} {\left( {x{{\left( {\int_{\,0}^{\,x} {\frac{{{t^4}}}{{{t^6} + 1}}dt} } \right)}^\prime }} \right)} dx} \displaystyle{E = \int_{\,0}^{\,1} {\left( {\int_{\,0}^{\,x} {\frac{{{t^4}}}{{{t^6} + 1}}dt} } \right)} dx = \int_{\,0}^{\,1} {\left( {x'\int_{\,0}^{\,x} {\frac{{{t^4}}}{{{t^6} + 1}}dt} } \right)} dx = \left[ {x\int_{\,0}^{\,x} {\frac{{{t^4}}}{{{t^6} + 1}}dt} } \right]_0^1 - \int_{\,0}^{\,1} {\left( {x{{\left( {\int_{\,0}^{\,x} {\frac{{{t^4}}}{{{t^6} + 1}}dt} } \right)}^\prime }} \right)} dx}](/forum/ext/geomar/texintegr/latexrender/pictures/6ae9fe19750dfc97af058064d1eac878.png)

![\displaystyle{ = f\left( 1 \right) - \frac{1}{6}\left[ {\ln \left( {{x^6} + 1} \right)} \right]_0^1 = f\left( 1 \right) - \frac{1}{6}\ln \left( 2 \right) = \ln {e^{f\left( 1 \right)}} - \ln \sqrt[6]{2} = \ln \frac{{{e^{f\left( 1 \right)}}}}{{\sqrt[6]{2}}}} \displaystyle{ = f\left( 1 \right) - \frac{1}{6}\left[ {\ln \left( {{x^6} + 1} \right)} \right]_0^1 = f\left( 1 \right) - \frac{1}{6}\ln \left( 2 \right) = \ln {e^{f\left( 1 \right)}} - \ln \sqrt[6]{2} = \ln \frac{{{e^{f\left( 1 \right)}}}}{{\sqrt[6]{2}}}}](/forum/ext/geomar/texintegr/latexrender/pictures/4ad4c8d1164a6d662649de61b6f297f2.png)
, οπότε με ολοκλήρωση κατά παράγοντες έχουμε ![\displaystyle{E=\int_{0}^{1}(x)^{\prime}\left(\int_{0}^{x}\frac{t^4}{t^6+1}dt \right)dx=\left[x\int_{0}^{x}\frac{t^4}{t^6+1}dt \right]_{0}^{1}-\int_{0}^{1}x\frac{x^4}{x^6+1}dx=f(1)-\int_{0}^{1}\frac{x^5}{x^6+1}dx=f(1)-\frac{1}{6}[\ln (x^6+1)]_{0}^{1}=f(1)-\frac{1}{6}\ln 2.} \displaystyle{E=\int_{0}^{1}(x)^{\prime}\left(\int_{0}^{x}\frac{t^4}{t^6+1}dt \right)dx=\left[x\int_{0}^{x}\frac{t^4}{t^6+1}dt \right]_{0}^{1}-\int_{0}^{1}x\frac{x^4}{x^6+1}dx=f(1)-\int_{0}^{1}\frac{x^5}{x^6+1}dx=f(1)-\frac{1}{6}[\ln (x^6+1)]_{0}^{1}=f(1)-\frac{1}{6}\ln 2.}](/forum/ext/geomar/texintegr/latexrender/pictures/33948c56ad4d7e6443f8a18255579654.png)
Μέλη σε αυτήν τη Δ. Συζήτηση: Δεν υπάρχουν εγγεγραμμένα μέλη και 2 επισκέπτες