Application of Integrals

Maths · Class 12

Lesson 5 of 7 · 7 min

Parabolas and symmetry

NCERT §8.1–8.2 (standard forms)

The flower bed is shaped by a parabolic hedge, y = x², and a straight path along y = 4 (in metres). The region between them is the bed.

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In short

Standard parabolas such as y = x², y² = 4ax and x² = 4ay are symmetric about an axis. Work out the area on one side of the axis and double it.

Worked example: the region between y = x² and the line y = 4. With horizontal strips, each strip runs from x = −√y to x = √y, length 2√y, so A = ∫₀⁴ 2√y dy = 32/3.

The same region with vertical strips: the strip height is 4 − x² for x from −2 to 2, so A = ∫₋₂² (4 − x²) dx = 16 − 16/3 = 32/3.

A useful check: the region is 2/3 of the 4 × 4 rectangle around it, 2/3 × 16 = 32/3. A parabolic segment cut off by a chord perpendicular to the axis always fills 2/3 of the rectangle drawn around it.

Worked example: the region between y² = 4ax and its latus rectum x = a. The upper half is y = 2√(ax), so A = 2∫₀ᵃ 2√(ax) dx = 8a²/3. For y² = 4x (a = 1) the area is 8/3.

Symmetry also works for odd functions: the region under y = x³ from −1 to 1 has area 2 × 1/4 = 1/2, though ∫₋₁¹ x³ dx = 0.

Parabolas and symmetry | Application of Integrals | Lumi Learn