Question

Sketch the graph of y = \[\sqrt{x + 1}\]  in [0, 4] and determine the area of the region enclosed by the curve, the x-axis and the lines x = 0, x = 4.

Answer 1

\[y = \sqrt{x + 1} \text{ in }\left[ 0, 4 \right] \text{ represents a curve which is part of a parabola }\]
\[x = 4\text{ represents a line parallel to }y - \text{ axis and cutting }x - \text{ axis at }(4, 0)\]
\[\text{ Enclosed area bound by the curve and lines }x = 0\text{ and }x = 4\text{ is OABCO}\]
\[\text{ Consider a vertical strip of lenght }= \left| y \right| \text{ and width }= dx\]
\[ \therefore\text{ Area of approximating rectangle }= \left| y \right| dx\]
\[\text{ The approximating rectangle moves from }x = 0\text{ to } x = 4\]
\[ \Rightarrow\text{ A = Area OABCO }= \int_0^4 \left| y \right| dx\]
\[ \Rightarrow A = \int_0^4 y dx ..................\left[ y > 0 \Rightarrow \left| y \right| = y \right]\]
\[ \Rightarrow A = \int_0^4 \sqrt{x + 1} dx\]
\[ \Rightarrow A = \int_0^4 \left( x + 1 \right)^\frac{1}{2} dx\]
\[ \Rightarrow A = \left[ \frac{\left( x + 1 \right)^\frac{3}{2}}{\frac{3}{2}} \right]_0^4 \]
\[ \Rightarrow A = \frac{2}{3}\left( 5^{{}^\frac{3}{2}} - 1 \right)\text{ sq . units }\]
\[ \therefore\text{ Enclosed area between the curve and given lines }= \frac{2}{3}\left( 5^{{}^\frac{3}{2}} - 1 \right)\text{ sq . units }\]