Question

Using integration, find the area of the region bounded by triangle whose vertices are

(1,0) , (2,2) and (3,1) .​

Answer 1

Given : vertices of triangle (1,0) , (2,2) and (3,1) .​

To Find : area of the region bounded by triangle Using integration,

Solution:

A = ( 1,0)

B = ( 2, 2)

C = ( 3, 1)

Draw BD & CE perpendicular on x axis

D = ( 2, 0)  , E = ( 3, 0)

Area of ΔABC =  Area of ΔABD + area of trapzium BCED - Area of ΔACE

AB = y - 0 = ((2 - 0)/(2 - 1) )(x - 1)  =  2(x - 1)

Area of ΔABD  =

$\int\limits^2_1 {2(x-1)} \, dx$

= 2 (x²/2 - x)  limits 1 to 2

= 2 ( 2²/2 - 2 - 1²/2  + 1)

= 2(1/2)

=1

BC , y = 4 - x

area of trapezium BCED

  $\int\limits^3_2 {4-x} \, dx$

= 3/2

AC = (1/2)(x - 1)

Area of ΔACE  =

$\int\limits^2_1 {\frac{1}{2} (x-1)} \, dx$

= 1

Area of triangle = 1 + 3/2 - 1 = 3/2 sq units

Area of triangle

= (1/2) | 1 ( 2 - 1)  + 2( 1 - 0) + 3(0 - 2) |

= (1/2) | 1 + 2 - 6|

= (1/2) | - 3|

= 3/2  sq units

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