Atanov's formula (PDF)

Atanov’s formula for area of a parabolic segment
Given a parabola 𝒚 = −𝒙𝟐 + 𝒃𝒙 + 𝒄, which is passed through by a line 𝒚 = 𝒌𝒙 + 𝒎 at points (𝒙𝟎 ; 𝒚𝟎 ) and
(𝒙𝟏 ; 𝒚𝟏 ) (Fig. 1). Then the area of parabolic segment is given by

𝑨=

(∆𝒙)𝟑
𝟔

, where ∆𝒙 = 𝒙𝟏 − 𝒙𝟎

Proof.
𝒙𝟏

𝒙𝟏

𝒙𝟏

𝑨 = ∫ (−𝒙𝟐 + 𝒃𝒙 + 𝒄) 𝒅𝒙 − ∫ (𝒌𝒙 + 𝒎) 𝒅𝒙 = ∫ (−𝒙𝟐 + (𝒃 − 𝒌)𝒙 + (𝒄 − 𝒎)) 𝒅𝒙
𝒙𝟎

𝒙𝟎

𝒙𝟎

where equation 𝒚 = −𝒙𝟐 + (𝒃 − 𝒌)𝒙 + (𝒄 − 𝒎) is a parabola, which intersects x-axis at points (𝒙𝟎 ; 𝟎) and
(𝒙𝟏 ; 𝟎) (Fig. 2). Thus, we see that the original area is equal to the area of a segment bounded by this
parabola and the x-axis .

If we shift the origin to point (𝒙𝟎 ; 𝟎) (Fig. 3), parabola will be represented by the equation 𝒚 = −𝒙(𝒙 − ∆𝒙).

And we receive
∆𝒙

∆𝒙

∆𝒙

𝒙𝟑 ∆𝒙 ∙ 𝒙𝟐
𝑨 = ∫ −𝒙(𝒙 − ∆𝒙) 𝒅𝒙 = ∫ (−𝒙 + ∆𝒙 ∙ 𝒙) 𝒅𝒙 = [− +
]
𝟑
𝟐
𝟎
𝟐

𝟎

𝟎

∆𝒙𝟑 ∆𝒙 ∙ ∆𝒙𝟐 ∆𝒙𝟑
𝑨=−
+
=
𝟑
𝟐
𝟔
It's quite easy to see that in more common case when parabola is represented by the equation
𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄, coefficient 𝒂 appears in the formula:
|𝒂|∆𝒙𝟑
𝑨=
𝟔