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प्रश्न
Using the method of integration, find the area of the region bounded by the lines 3x − 2y + 1 = 0, 2x + 3y − 21 = 0 and x − 5y + 9 = 0
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उत्तर
`3"x" - 2"y" + 1 = 0 ⇒ "y"_1 =((3"x"+1))/2` ...........(i)
`2"x" - 3"y" - 21 = 0 ⇒ "y"_2 =((21-2"x"))/3` .....(ii)
`"x" - 5"y" + 9 = 0 ⇒ "y"_3 = (("x"+9))/5` ......(iii)
Point of intersection of (i) and (ii) is A(3, 5)
Point of intersection of (ii) and (iii) is B(6, 3) and
Point of intersection of (iii) and (i) is C(1, 2).
Therefore, the area of the region bounded=`int_1^3"y"_1."dx"+int_3^6"y"_2."dx"-int_1^6"y"_3". dx"`
`=int_3^1((3"x"+1))/2."dx"+int_3^6((21-2"x"))/3."dx"-int_1^6(("x"+9))/5."dx"`
`=1/2((3"x"^2)/2+"x")_1^3+1/3(21"x"-"x"^2)_3^6-1/5("x"^2/2+9"x")_1^6`
`1/2[14]+1/3[36]-1/5[125/2]`
`= 7+12-12.5`
`= 6.5 "sq.units"`
