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Determine the area under the curve y = √(a2 – x2) included between the lines x = 0 and x = a.

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= √(a2 – x2)

Squaring both sides

⇒ y2 = a2 - x2

⇒ x2 + y2 = a2

This is equation of circle having center as (0, 0) and radius a

Now in = √(a2 – x2) -a ≤ x ≤ a and y ≥ 0 which means x and y both positive or x negative and y positive hence the curve = √(a2 – x2) has to be above X-axis in 1st and 2nd quadrant

x = 0 is equation of Y-axis and x = a is a line parallel to Y-axis passing through (a, 0)

Using uv rule of integration where u and v are functions of x

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