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Question

Area enclosed by curve y39y+x=0 and Y - axis is -

A
92
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B
9
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C
812
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D
81
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Solution

The correct option is D 812
The given equation of curve can be written as x=f(y)=9yy3. Now to calculate the area we need to find the boundaries of this curve i.e ordinates or the point where this curve is meeting Y - axis.
f(y)=y(9y2)
f(y)=y.(3+y)(3y)
So, the points where f(y) is meeting y - axis are y = -3, y = 0 & y = 3.
Important thing to note here is that the function is changing its signs.
I.e. from y = -3 to y = 0 f(y) is negative.
& from y = 0 to y = 3 f(y) is positive.
Since, the function in negative in the interval (-3, 0 ) we’ll take absolute value of it. Because we are interested in the area enclosed and not the algebraic sum of area.
Let the area enclosed be A.
A=03|f(y)|dy+30f(y)dyA=03|9yy3|dy+30(9yy3)dyA=03(9yy3)dy+309yy3dyA=[9y22y44]03+[9y22y44]30A=[00812+814]+[812814]A=81812A=812

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