S=2(∫x-x/4)dx+∫(1/x-x/4)dx=2(S1+S2)=
первый интеграл по х от 0 до 1, второй от 1 до 2
S1=(3/4)*(x^2/2)=3x^2/8=3/8
S2=lnx-(1/4)*x^2/2=lnx-x^2/8=ln2-0.5-(ln1-1/8)=ln2-0.5+1/8=ln2-3/8
S=2(3/8+ln2-3/8)=2ln2
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Answers & Comments
S=2(∫x-x/4)dx+∫(1/x-x/4)dx=2(S1+S2)=
первый интеграл по х от 0 до 1, второй от 1 до 2
S1=(3/4)*(x^2/2)=3x^2/8=3/8
S2=lnx-(1/4)*x^2/2=lnx-x^2/8=ln2-0.5-(ln1-1/8)=ln2-0.5+1/8=ln2-3/8
S=2(3/8+ln2-3/8)=2ln2