Question

Find the area enclosed by the curve 2x + 3y = 12, lines x = 1 and x = 5.​

Answer 1

Given : 2x + 3y = 12, lines x = 1 and x = 5.

To Find : area enclosed

Solution:

2x + 3y = 12

x = 1 and x = 5

2x + 3y = 12 and x = 1

=> 2 + 3y = 12

=> y = 10/3

2x + 3y = 12 and x = 5

=> 10 + 3y = 12

=> y = 2/3

x = 1 and x = 5 does not intersects

Hence we do not get any closed figure

if We find area using x axis Then we get trapezium with parallel side length

= 10/3 and 2/3

and height = 5 - 1 = 4

Area = (1/2) ( 10/3 + 2/3) * 4

= 8 sq units

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Answer 2

Given:

• 2x + 3y = 12
• x = 1
• x = 5

To find:

Area enclosed by the curve with the x axis ?

Calculation:

We will use definite integral to solve this question.

• First of all, refer to the graph.

• The line 2x + 3y = 12 is being intersected by x=1 and x=5.

• So, out upper and lower limits of integration will be 5 and 1 respectively.

$\sf \: 2x + 3y = 12 \\ \implies \sf 3y = 12 - 2x \\ \sf \implies y = 4 - ( \dfrac{2}{3} )x$

Let area be A :

$\displaystyle \sf \therefore \: A = \int_{1}^{5} \: y \: dx$

$\displaystyle \sf \implies \: A = \int_{1}^{5} \: \bigg[ 4 - ( \dfrac{2}{3} )x\bigg] \: dx$

$\displaystyle \sf \implies \: A =\: \bigg[ 4x\bigg]_{1}^{5} - \bigg[ \dfrac{ {x}^{2} }{3} \bigg]_{1}^{5}$

$\displaystyle \sf \implies \: A =\: \bigg[ 20 - 4\bigg] - \bigg[ \dfrac{ {5}^{2} - 1}{3} \bigg]$

$\displaystyle \sf \implies \: A =\: \bigg[ 16\bigg] - \bigg[ \dfrac{ 25 - 1}{3} \bigg]$

$\displaystyle \sf \implies \: A =\: \bigg[ 16\bigg] - \bigg[ \dfrac{ 24}{3} \bigg]$

$\displaystyle \sf \implies \: A =\: 16 - 8$

$\displaystyle \sf \implies \: A =\: 8 \: units$

So, area is 8 units.