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  • Using integration, find the area of the region bounded by the lines and <img alt="x + y = 12." src="https://learn.careers360.c

Using integration, find the area of the region bounded by the lines x -y = 0, 3x - y = 0 and x + y = 12.

 

Answers (1)

Given:

Equations

x -y = 0 ...(i)

3x - y = 0 ...(ii)

Equation (i) - Equation (ii)

x - y - 3x + y = 0 

x = 0 , y = 0 

Intesrsect point O(0,0)

Put value  of y from equation(i) in equation (iii)

x + y = 12...(iii)

2x = 12

x = 6 =>  y = 6 

Intesrsect point B(6,6)

Put value  of y from equation(ii) in equation (iii)

x + 3x = 12

4x = 16 => x = 3

y = 9 

Intesrsect point A(3,9)

 

$Required area $ = $Area of triangle AOB$ \\\\ = \int_0^3 3x dx+ \int_3^6 (12-x) dx - \int_0^6 x dx \\\\ = \frac{3}{2} \left [ x^2 \right ]_0^3 + 12\left [ x \right ]_3^6 - \frac{1}{2} \left [ x^2 \right ]_3^6 - \frac{1}{2} \left [ x^2 \right ]_0^6 \\\\ = \frac{27}{2} + 36 - \frac{27}{2} - 18 \\\\ = 18 \ square \ units

Posted by

Safeer PP

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