CBSE Class 12 · 2012 · Area Under CurvesLong Answer

Using the method of integration, find the area of the region bounded by the following lines:

3 x - y - 3 = 0

2 x + y - 12 = 0

and

x - 2 y - 1 = 0

Answer: $10$ square units

Solution

Solving the lines in pairs gives the vertices

(1, 0)

(3, 6)

and

(5, 2)

.
Writing

y

for each line (line 1:

y = 3 x - 3

, line 2:

y = 12 - 2 x

, line 3:

y = \frac{x - 1}{2}

) with the lowest line

x - 2 y - 1 = 0

as the lower edge:

A = \int_{1}^{3} (3 x - 3) d x + \int_{3}^{5} (12 - 2 x) d x - \int_{1}^{5} \frac{x - 1}{2} d x .

= 12 - 6 - 16 + 24 - 6 + 2 = 10.

Hence the area of the triangular region is

10

square units.

CBSE Class 12 Board · 2012 · Subjective