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Question

Find the area bounded in the first quadrant between the ellipse x216+y29=1 and the line 3x+4y=12 is

A
3π sq. units
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B
(3π6) sq. units
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C
(3π2) sq. units
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D
(3π4) sq. units
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Solution

The correct option is B (3π6) sq. units
x216+y29=1
y=3416x2
and 3x+4y=12
y=123x4
Area bounded in first quadrant=40(y1y2)dx
=40(3416x2123x4)dx
=34[x216x2+8sin1x4]4014[12x3x22]40
=34(8π2)14[4824]
=(3π6) sq.units

998506_1039877_ans_c8c75ba80df04e95b5fca976f8db9c9c.png

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