CBSE Class 12 · 2011 · Area Under CurvesLong Answer

Using integration find the area of the triangular region whose sides have equations

y = 2 x + 1

y = 3 x + 1

and

x = 4

Answer: $8$ square units

Solution

The sides meet at

(0, 1)

(lines

y = 2 x + 1, y = 3 x + 1

), at

(4, 9)

(

y = 2 x + 1, x = 4

) and at

(4, 13)

(

y = 3 x + 1, x = 4

).
For

0 \leq x \leq 4

the upper edge is

y = 3 x + 1

and the lower edge is

y = 2 x + 1

A = \int_{0}^{4} [(3 x + 1) - (2 x + 1)] d x = \int_{0}^{4} x d x = [\frac{x ^{2}}{2}]_{0}^{4} = 8.

Hence the area of the triangular region is

8

square units.

CBSE Class 12 Board · 2011 · Subjective