Question

Draw the rough sketch of y² + 1 = x, x ≤ 2. Find the area enclosed by the curve and the line x = 2.

Answer 1

\[y^2 + 1 = x, x \leq 2\text{ is a parabola with vertex }(1, 0)\text{ and symmetrical about + ve side of }x - \text{ axis }\]
\[x = 2\text{ is a line parallel to }y -\text{ axis and cutting }x - \text{ axis at }(2, 0)\]
\[\text{ Consider, a vertical section of height }= \left| y \right| \text{ and width } = dx \text{ in the first quadrant }\]
\[ \Rightarrow\text{ area of corresponding rectangle }= \left| y \right| dx\]
\[\text{ Since, the corresponding rectangle is moving from }x = 1\text{ to }x = 2\text{ and the curve being symmetrical }\]
\[ \therefore\text{ A = area ABCA }= 2 \times\text{ area ABDA }\]
\[ \Rightarrow A = 2 \int_1^2 \left| y \right| dx \]
\[ \Rightarrow A = 2 \int_1^2 y dx ................\left[ \left| y \right| = y \text{ as }y > 0 \right]\]
\[ \Rightarrow A = 2 \int_1^2 \sqrt{x - 1} dx ..............\left[ y^2 + 1 = x \Rightarrow y = \sqrt{x - 1} \right] \]
\[ \Rightarrow A = 2 \left[ \frac{\left( x - 1 \right)^\frac{3}{2}}{\frac{3}{2}} \right]_1^2 \]
\[ \Rightarrow A = \frac{4}{3}\left( 1^\frac{3}{2} - 0 \right)\]
\[ \Rightarrow A = \frac{4}{3}\text{ sq . units }\]
\[ \therefore\text{ Enclosed area }= \frac{4}{3}\text{ sq . units }\]