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

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