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Question

Mathematics Question on applications of integrals

The area (in s units) bounded by the curves y=x,2yx+3=0,xy = \sqrt{x} , 2y - x + 3 = 0, x-axis and lying in the first quadrant is

A

3636

B

1818

C

274 \frac{27}{4}

D

99

Answer

99

Explanation

Solution

Solving y=xy = \sqrt{x} with 2yx+3=0,2y - x + 3 = 0, we have 2xx+3=0(x3)(x+1)=02 \sqrt{x} - x + 3 = 0 \, \Rightarrow \, ( \sqrt{x} - 3)(\sqrt{x} +1) = 0
x=1,9\therefore \, \, \, x = 1, 9
Area =03[(2y+3)y2]dy = \int\limits_{0}^{3} \left[\left(2y+3\right)-y^{2}\right]dy
=y2+3yy3303= y^2 + 3y - \frac{y^3}{3}|^3_0
=9+99=9= 9 + 9 - 9 = 9