Question

Using integration, find the area of the region bounded by the line y − 1 = x, the x − axis and the ordinates x= −2 and x = 3.

Answer 1

\[\because y - 1 = x\text{ is a straight line cutting x - axis at D( - 1, 0) and y - axis at A( 0, 1) }\]
\[\text{ And, }x = - 2\text{ and }x = 3 \text{ are straight lines parallel to y - axis }\]
\[\text{ Also, since }\left| y \right| = - \left( 1 + x \right),\text{ for } x \leq - 1\]
\[\text{ And }\left| y \right| = \left( 1 + x \right), x > - 1\]
\[ \therefore\text{ Area of region bound by line }y - 1 = x , x \text{ axis and the ordinates } x = - 2\text{ and }x = 3\text{ is }\]
\[\text{ Area A = area FED + area DOA + area OABC }\]
\[ \Rightarrow A = \int_{- 2}^3 \left| y \right| dx\]
\[ \Rightarrow A = \int_{- 2}^{- 1} \left| y \right| dx + \int_{- 1}^0 \left| y \right| dx + \int_0^3 \left| y \right| dx\]
\[ \Rightarrow A = \int_{- 2}^{- 1} - \left( 1 + x \right) dx + \int_{- 1}^0 \left( 1 + x \right) dx + \int_0^3 \left( 1 + x \right) dx\]
\[ \Rightarrow A = - \left[ x + \frac{x^2}{2} \right]_{- 2}^{- 1} + \left[ x + \frac{x^2}{2} \right]_{- 1}^0 + \left[ x + \frac{x^2}{2} \right]_0^3 \]
\[ \Rightarrow A = \left[ 1 - \frac{1}{2} - 2 + \frac{4}{2} \right] + \left[ 0 + 1 - \frac{1}{2} \right] + \left[ 3 + \frac{9}{2} \right]\]
\[ \Rightarrow A = - 1 + \frac{3}{2} + 1 - \frac{1}{2} + 3 + \frac{9}{2} = 3 + \frac{3 - 1 + 9}{2} = 3 + \frac{11}{2} = \frac{17}{2}\text{ sq . units }\]
\[ \therefore \text{Area of the bound region }= \frac{17}{2}\text{ sq . units }\]