Question

Using integration, find the area of the triangular region, the equations of whose sides are y = 2x + 1, y = 3x+ 1 and x = 4.

Answer 1

Solving the given equations

The point of intersection of the three lines are A(0, 1), B(4, 13) and C(4, 9).

We need to find the area of ABC

Area under line AB = area OABCL

\[ \Rightarrow\text{ Area OABCL }= \int_0^4 \left( 3x + 1 \right) dx ...............\left[\text{ Equation of BC is }y = 3x + 1 \text{ and }x \text{ moves from A, }x = 0\text{ to }B, x = 4 \right] \]
\[ = \left[ 3\frac{x^2}{2} + x \right]_0^4 \]
\[ = \left[ 3\frac{4^2}{2} + 4 \right]\]
\[ = 24 + 4 = 28 \text{ sq . units }\]
Area under line BC = Area OACL
\[ \Rightarrow\text{ Area OACL }= \int_0^4 \left( 2x + 1 \right)dx ................\left[\text{ Equation of BC is }y = 2x + 1\text{ and }x \text{ moves from A, }x = 0\text{ to }C, x = 4 \right] \]
\[ = \left[ 2\frac{x^2}{2} + x \right]_0^4 \]
\[ = 16 + 4 = 20\text{ sq . units }\]
\[ \therefore\text{ Area }\Delta \text{ ABC } \hspace{0.167em} = \text{ Area OABCL - Area OACL }\]
\[ \Rightarrow\text{ Area }\Delta\text{ ABC }= 28 - 20 = 8\text{ sq . units }\]
\[ \therefore\text{ Area of triangle formed by the three given lines = 8 sq . units }\]