Using integration find the area of region bounded by the triangle whose vertices are `( 1, 0), (1, 3) a n d (3, 2)`.

Answer

N/a

Explanation

Let we have the vertices of a `DeltaABC " as "A(-1,1),B(0,5) "as " A(-1,1),B(0,5)" and " (3,2)`.

`:." Equation of AB is "y-1=((5-1)/(0+1))(x+1)`
`rArr y-1=4x+4`
`rArr y=4x+5 …(i)`
and equation of BC is `y-5=((2-5)/(3-0))(x-0)`
`rArr y-5=(-3)/3(x)`
`rArr y=5-x`
Similarly, equation of AC is `y-1=((2-1)/(3+1))(x+1)`
`rArr y-1=1/4(x+1)`
`rArr 4y=x+5`
`:. " Area of shaded region "=int_(-1)^(0)(y_(1)-y_(2))dx+int_(0)^(3)(y_(1)-y_(2))dx`
`=int_(-1)^(9)[4x+5-(x-5)/4]dx+int_(0)^(3)[5-x-(x+5)/4]dx`
`=[(4x^(2))/2+5x-x^(2)/8-(5x)/4]_(-1)^(0)+[5x-(x^(2))/2-x^(2)/8-(5x)/4]_(0)^(3)`
`=[0-(4. 1/2+5(-1)-1/8+5/4)]+[(15-9/2-9/8-15/4)-0]`
`=[-2+5-1/8-5/4+15-9/2-9/8-15/4]`
`=18+((1-10-36-9-30)/8)`
`=18+(-84/8)=18-21/2=15/2" sq units"`