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Question
Find the area of the smaller region bounded by the curves `x^2/25 + y^2/16` = 1 and `x/5 + y/4` = 1, using integration.
Solution
`x^2/25 + y^2/16` = 1 and `x/5 + y/4` = 1
`\implies` y = `4/5 sqrt(25 - x^2)`
The points of intersection of the given curve and line are A(0, 4) and B(5, 0).
Area of shaded region = Area of ellipse in I quadrant – Area of triangle ΔOAB
= `int_0^5 (4/5 sqrt(25 - x^2))dx - 1/2 xx 5 xx 4`
= `4/5 [x/2 sqrt(25 - x^2) + 25/2 sin^-1 x/5]_0^5 - 10 ...[∵ int (sqrt(a^2 - x^2))dx = x/2 sqrt(a^2 - x^2) + a^2/2 sin^-1 x/a]`
= `4/5 [0 + 25/2 sin^-1 1] - 10`
= `10 xx π/2 - 10`
= (5π – 10) sq. units.
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