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

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