Archimedes found a way to calculate the area of parabolic segment by exhaustion

Question

Archimedes found a way to calculate the area of parabolic segment by exhaustion
method. Let us consider a parabolic segment given by a curve y
2 = x, 0 ? x ? a.
Denote A = (a, ?a) and B = (b, ??b) where 0 < b ? a. Archimedes constructed a
triangle ABC having AB for its base and takes the point C = (c, ?c) such that the
tangent at C to the parabola was parallel to the line segment AB. Let the area of
triangle ABC be denoted by ?. In each of the two smaller segment cut off by the
chords AC and CB, Archimedes similarly constructed inscribed triangles ADC and
CEB. From the properties of the parabola, he demonstrated that each of the two new
triangles had an area equal to 1/8?. Hence the area of ADC and CEB equaled 1/4?.
Next four more triangles were constructed with vertices on the parabola and bases on
the new chords AD, DC, CE and EB Each of these four triangles had an area equal
to 1/8
that of the triangle ADC, or equal to 1/8?. Repeating this process. Then the
area of the parabolic section is approximated to be
?

1 +
1
4
+
1
4
2
+ ... +
1
4
n
+ ...
.
By using the above method, find the area of the parabolic segment y^2 = x, 0 ? x ? a.
Verify your result by integral in Calculus.

I already know the AREA PARABOLIC SEGMENT=(4/3) * AREA OF TRIANGLE

Explanation / Answer

http://www.intmath.com/blog/archimedes-and-the-area-of-a-parabolic-segment/1652

This is where you can get your answer and more explantion if needed.

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