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Question 13 of 42

Q.Find the area of the region bounded by the parabola y=4−x2y = 4 - x^2, xx-axis and the lines x=0x = 0, x=2x = 2.

Puducherry TnboardTamil Nadu HSC (DGE) Commerce Board 2020Subjective· 2mImportance★★★★★
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On [0,2][0, 2], y=4−x2≥0y = 4 - x^2 \ge 0, so the required area is ∫02(4−x2) dx=[4x−x33]02=163\int_0^2 (4 - x^2)\,dx = \left[4x - \frac{x^3}{3}\right]_0^2 = \frac{16}{3} sq.units.

Step 1 — Confirm the curve lies above the xx-axis. For x∈[0,2]x \in [0, 2], 4−x2≥04 - x^2 \ge 0 (it is 44 at x=0x = 0 and 00 at x=2x = 2), so the area is a straightforward definite integral.

Step 2 — Set up the integral.

Area=∫02(4−x2) dx.\text{Area} = \int_0^2 (4 - x^2)\,dx.

Step 3 — Integrate and evaluate. …

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