Question

Find the area under the curve y = \[\sqrt{6x + 4}\] above x-axis from x = 0 to x = 2. Draw a sketch of curve also.

Answer 1

\[y = \sqrt{6x + 4}\text{ represents a parabola, with vertex }V( - \frac{2}{3}, 0)\text{ and symmetrical about }x -\text{ axis }\]
\[x = 0\text{ is the }y -\text{ axis } .\text{ The curve cuts it at }A(0, 2 )\text{ and }A'(0, - 2)\]
\[x = \text{ 2 is a line parallel to } y - \text{ axis, cutting the } x - \text{ axis at }C(2, 0)\]
\[\text{ The enclosed area of the curve between }x = 0 \text{ and }x = 2\text{ and above }x - \text{ axis }=\text{ area OABC }\]
\[\text{ Consider, a vertical strip of length }= \left| y \right|\text{ and width }= dx \]
\[\text{ Area of approximating rectangle }= \left| y \right| dx\]
\[\text{ The approximating rectangle moves from }x = 0\text{ to } x = 2 \]
\[ \Rightarrow\text{ area OABC }= \int_0^2 \left| y \right| dx\]
\[ \Rightarrow A = \int_0^2 y dx ..................\left[ As, y > 0 , \left| y \right| = y \right]\]
\[ \Rightarrow A = \int_0^2 \sqrt{6x + 4} dx \]
\[ \Rightarrow A = \int_0^2 \left( 6x + 4 \right)^\frac{1}{2} dx\]
\[ \Rightarrow A = \frac{1}{6} \left[ \frac{\left( 6x + 4 \right)^\frac{3}{2}}{\frac{3}{2}} \right]_0^2 \]
\[ \Rightarrow A = \frac{2}{18}\left[ {16}^{{}^\frac{3}{2}} - 4^\frac{3}{2} \right]\]
\[ \Rightarrow A = \frac{2}{18}\left[ 4^3 - 2^3 \right]\]
\[ \Rightarrow A = \frac{2}{18}\left[ 64 - 8 \right] = \frac{2}{18} \times 56 = \frac{56}{9} \text{ sq . units }\]
\[ \therefore\text{ Enclosed area }= \frac{56}{9} \text{ sq . units }\]