Question:
The area (in s units) bounded by the curves $y = \sqrt{x} , 2y - x + 3 = 0, x$-axis and lying in the first quadrant is
The area (in s units) bounded by the curves $y = \sqrt{x} , 2y - x + 3 = 0, x$-axis and lying in the first quadrant is
Updated On: Sep 3, 2024
- $36$
- $18$
- $ \frac{27}{4}$
- $9$
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The Correct Option is D
Solution and Explanation
Solving $y = \sqrt{x}$ with $2y - x + 3 = 0,$ we have $2 \sqrt{x} - x + 3 = 0 \, \Rightarrow \, ( \sqrt{x} - 3)(\sqrt{x} +1) = 0$
$\therefore \, \, \, x = 1, 9$
Area $ = \int\limits_{0}^{3} \left[\left(2y+3\right)-y^{2}\right]dy$
$ = y^2 + 3y - \frac{y^3}{3}|^3_0$
$ = 9 + 9 - 9 = 9$
$\therefore \, \, \, x = 1, 9$
Area $ = \int\limits_{0}^{3} \left[\left(2y+3\right)-y^{2}\right]dy$
$ = y^2 + 3y - \frac{y^3}{3}|^3_0$
$ = 9 + 9 - 9 = 9$
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Concepts Used:
Applications of Integrals
There are distinct applications of integrals, out of which some are as follows:
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Integrals are used to find:
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In Physics
Integrals are used to find:
- Centre of gravity
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