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in Integrals calculus by (41.7k points)

The area (in square units) bounded by the curves y = x,  2y - x + 3 = 0, x-axis, and lying in the first quadrant is

(A) 36 

(B) 18 

(C) 27/4 

(D) 9

1 Answer

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Answer is (D) 9

First solving the equations, we have 

2x = x - 3   (1) 

Squaring on both sides of Eq. (1), we get 

4x = x2 - 6x + 9 ⇒ x2 - 10x + 9 ⇒ x = 9, x = 1

Since x = 1 intersects the parabola below the x-axis, this point is extraneous.

So, for x = 9 we have, y = 3. 

Therefore, the required area under the curve (see Fig. 24.33) is

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