Area Under Curves — CBSE Class 12 · 2009 Board Question with Solution

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

2 x + y = 4

3 x - 2 y = 6

and

x - 3 y + 5 = 0

Answer: $\frac{7}{2}$ square units

Solution

Find the vertices by taking the lines in pairs:

2 x + y = 4

and

x - 3 y + 5 = 0

meet at

(1, 2)

;

2 x + y = 4

and

3 x - 2 y = 6

meet at

(2, 0)

;

3 x - 2 y = 6

and

x - 3 y + 5 = 0

meet at

(4, 3)

.
Writing

y

for each line and integrating over

x \in [1, 4]

(upper edge

x - 3 y + 5 = 0

, i.e.

y = \frac{x + 5}{3}

; lower edges

2 x + y = 4

for

1 \leq x \leq 2

and

3 x - 2 y = 6

for

2 \leq x \leq 4

A = \int_{1}^{4} \frac{x + 5}{3} d x - \int_{1}^{2} (4 - 2 x) d x - \int_{2}^{4} \frac{3 x - 6}{2} d x .

= \frac{1}{3} \cdot \frac{45}{2} - 1 - 3 = \frac{15}{2} - 4 = \frac{7}{2} .

Hence the area of the triangular region is

\frac{7}{2}

square units.

CBSE Class 12 Board · 2009 · Subjective