The area of the region bounded by curves y=x^2 and line y=4 is: Option 1: 48/3 sq. units Option 2: 16/3 sq. units Option 3: 32/3 sq. units Option 4: 8/3 sq. units

Answer: 32/3 sq. units. Solution: The curve meets the line at x=±2, and by symmetry the area is 2_0^2(4-x^2)dx. This equals 2[4x-x^3/3]_0^2=2(8-8/3)=32/3 square units.

Applied Mathematics

Calculus

Medium

applied

The area of the region bounded by curves $y=x^2$ and line $y=4$ is:

$\frac{48}{3}$ sq. units

$\frac{16}{3}$ sq. units

$\frac{32}{3}$ sq. units

$\frac{8}{3}$ sq. units

✅ Correct Option: 3

The curve meets the line at $x=\pm2$ , and by symmetry the area is $2\int_0^2(4-x^2)dx$ .

This equals $2\left[4x-\frac{x^3}{3}\right]_0^2=2\left(8-\frac{8}{3}\right)=\frac{32}{3}$ square units.

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Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The area of the region bounded by curves $y=x^2$ and line $y=4$ is: