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NDA 2023 Maths · Mathematics · Definite Integrals · medium

Interpreting as an area, the value of the definite integral ∫0^3√(9-x^2 dx, taken between the limits 0 and 3 of the radius-3 curve, is

Single correct option

Interpreting as an area, the value of the definite integral ∫039−x2 dx\displaystyle\int_0^3\sqrt{9-x^2}\,dx, taken between the limits 00 and 33 of the radius-33 curve, is
  1. A. 9π2\dfrac{9\pi}2
  2. B. 9π4\dfrac{9\pi}4
  3. C. 9π8\dfrac{9\pi}8
  4. D. 9π9\pi
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Answer: (B) 9π4\dfrac{9\pi}4

The integrand is the upper half of the circle x2+y2=9x^2+y^2=9, and the limits 00 to 33 cover one quarter of it. The area is 14π⋅32=9π4\dfrac14\pi\cdot3^2=\dfrac{9\pi}4. The value 9π2\dfrac{9\pi}{2} is the half-circle (−3-3 to 33).
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