Question 749494
The function is {{{y=2*ln(x)}}}
Its graph crosses the x-axis at the point where {{{y=0}}}
{{{system(y=2*ln(x),y=0)}}} --> {{{2*ln(x)=0}}} --> {{{ln(x)=0}}} --> {{{x=1}}}
The x-coordinate of point P is {{{x=t}}}.
The slope of the tangent at point P is the value of the derivative at that point.
y'={{{2/x}}}, so the slope of the tangent at {{{x=t}}} is {{{2/t}}}.
Since the line {{{alpha}}} tangent at P passes through the origin, its equation must be
{{{y=(2/t)x}}}
At point P, with {{{x=t}}}, {{{y=(2/t)t}}} --> {{{y=2}}}
Since point P is on the graph of {{{y=2*ln(x)}}}, its y-coordinate is {{{y=2*ln(t)}}}
So {{{system(y=2,y=2*ln(t))}}} --> {{{2=2*ln(t)}}} --> {{{ln(t)=1}}} --> {{{system(highlight(A=1),t=e)}}} and P is (e,2).
 
Now we can find the equation of {{{alpha}}}:
{{{system(y=(2/t)x,t=e)}}} --> {{{y=(2/e)x}}} --> {{{highlight(B=2)
 
Since the slope of line {{{alpha}}} is {{{2/e}}},
line {{{m}}} perpendicular to {{{alpha}}} must have a slope of {{{-1/(2/e)=-e/2}}}.
As {{{m}}} passes through P(e,2) its equation is
{{{y-2=(-e/2)(x-e)}}} --> {{{y-2=(-e/2)x+e^2/2}}} --> {{{y=(-e/2)x+e^2/2+2}}} --> {{{y=-e*x/2+e^2/2+2}}}
So {{{highlight(C=2)}}}, {{{highlight(D=2)}}}, and {{{highlight(E=2)}}}
The line {{{m}}} crosses the x-axis at the point where {{{y=0}}}
{{{system(y=-e*x/2+e^2/2+2,y=0)}}} --> {{{-e*x/2+e^2/2+2=0}}} --> {{{-e*x+e^2+4=0}}} --> {{{e^2+4=ex}}} --> {{{x=(e^2+4)/e}}} --> {{{x=e+4/e}}}
 
The area {{{S}}} of the region bounded by the curve {{{y=2*ln(x)}}}, the straight line {{{m}}}, and the x-axis is shown below.
{{{graph(300,300,-1,6,-1,6,2*ln(x),-e*x/2+e^2/2+2)}}}
 
{{{S}}} can be can be calculated as the sum of:
the area below {{{y=2*ln(x)}}}, and above the x-axis, between {{{x=1}}} and {{{x=e}}}, {{{int(2*ln(x), dx, 1, e )=2*int(ln(x), dx, 1, e )}}}
plus the area below {{{m}}} between {{{x=e}}} and {{{x=e+4/e}}}, {{{int((-e*x/2+e^2/2+2), dx, e, e+4/e )}}}
 
{{{int((-e*x/2+e^2/2+2), dx, e, e+4/e )}}} is easier than it seems.
It's just the area of the triangle with vertices (e,0), P(e,2), and (e+e/4,0)
Its base is {{{4/e}}}; its height is {{{2}}}, and its area is {{{(1/2)*2*(4/e)=4/e}}}.
 
Since {{{int(ln(x), dx)=2(x*ln(x)-x)}}},
{{{2*int(ln(x), dx, 1, e )=2*((e*ln(e)-e)-(1*ln(1)-1))=2((e*1-e)-(1*0-1))=2((e-e)-(0-1))=2}}}
 
So {{{S=2+4/e}}} and {{{highlight(G=4)}}}