Question 1155132
{{{drawing( 600, 600, -10, 20, -10, 20,
green(line(5,20,5,-20)), locate(1,1.2,A),circle(1,3,.12),locate(1,3,p(1,3)),
 graph( 600, 600, -10, 20, -10, 20, 3x,3x, 3/x^2)) }}}


the area {{{3x}}} from {{{0}}} to {{{1}}} is a right triangle , length of its legs are {{{1}}} and {{{3}}}

=> the area is: {{{3*1/2=3/2=1.5 }}}

then,take integral of {{{3/x^2}}} from {{{1}}} to {{{5}}}


{{{int(1->5)}}}{{{ (3/x^2)*dx)}}}...take the constant out

{{{3*int(1->5)}}}{{{ (1/x^2)*dx)}}}.......{{{1/x^2=x^-2}}}

{{{3*int(1->5)}}}{{{ (x^-2)*dx)}}}

{{{3*(4/5)}}}

{{{12/5)}}}

{{{2.4)}}}


total area: {{{1.5+2.4=3.9 }}}