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Area bounded by y²=x³ and y=x

Short answer

The curves intersect at (0,0) and (1,1). For 0≤x≤1, y runs from x^(3/2) to x. The double integral gives area = ∫₀¹ (x - x^(3/2)) dx = 1/10 square unit.

Find, by double integration, the area of the region bounded by y² = x³ and y = x.

Curves. y² = x³ (a cuspidal curve; its upper branch is y = x^(3/2)) and the line y = x.

Intersection

Put y = x in y² = x³: x² = x³, so x²(1 − x) = 0, giving x = 0 and x = 1. Points: (0,0) and (1,1).

For 0 < x < 1, x^(3/2) < x, so the curve lies below the line and the region lies between y = x^(3/2) and y = x.

y
1 |           *(1,1)
  |        . /
  |      ./ .      line y = x
  |    ./ .        curve y² = x³
  |  _/ .
  |_/_______ x
 0          1

Area by double integration

A = ∬ dy dx = ∫₀¹ ∫_{x^{3/2}}^{x} dy dx
  = ∫₀¹ ( x − x^{3/2} ) dx
  = [ x²/2 − (2/5) x^{5/2} ]₀¹
  = 1/2 − 2/5
  = 1/10

Answer: Area = 1/10 square unit.

✓ Answer checked by Claude Opus 5.5 (high effort)

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