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  • The area (in square units) bounded by the curves <img alt="y=\sqrt{x},\; 2y-x+3=0,x-axis," src="https://learn.careers360.com/latex-image/?%5Cdpi%7B100%7D%20y%3D%5Csqrt%7Bx%7D%2C%5C%3B%202y-x&pl

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

Option: 1

\frac{27}{4}


Option: 2

9


Option: 3

36


Option: 4

18


Answers (1)

best_answer

As learnt in concept

Area along x axis -

Let y_{1}= f_{1}(x)\, and \, y_{2}= f_{2}(x) be two curve then area bounded between the curves and the lines

x = a and x = b is

\left | \int_{a}^{b} \Delta y\, dx\right |= \left | \int_{a}^{b}\left ( y_{2}-y_{1} \right ) dx\right |

 

- wherein

Where \Delta y= f_{2}\left ( x \right )-f_{1}(x)

 

 Area=\int_{0}^{9}\sqrt{x}dx-\int_{3}^{9}\frac{(x-3)}{2}dx\\ \left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_{0}^{9}-\frac{1}{2}\left [ \frac{x^{2}}{2} \right ]_{3}^{9}+\frac{3}{2}(x)_{3}^{9}\\ 18-\frac{1}{2}\times 36+\frac{3}{2}\times 6\\ = 9

Posted by

Rishi

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