Question

Make a rough sketch of the graph of the function y = 4 − x², 0 ≤ x ≤ 2 and determine the area enclosed by the curve, the x-axis and the lines x = 0 and x = 2.

Answer 1

$$\begin{gathered} y=4-x^2,0\le x\le 2\mathrm{represents}\mathrm{a}\mathrm{half}\mathrm{parabola}\mathrm{with}\mathrm{vetex}\mathrm{at}(4,0) \\ x=2\mathrm{represents}\mathrm{a}\mathrm{line}\mathrm{parallel}\mathrm{to}\mathit{}y-\mathrm{axis}\mathrm{and}\mathrm{cutting}\mathit{}x-\mathrm{axis}\mathrm{at}(2,0) \\ \mathrm{In}\mathrm{quadrant}\mathrm{OABO},\mathrm{consider}\mathrm{a}\mathrm{vertical}\mathrm{strip}\mathrm{of}\mathrm{length}=\left|y\right|,\mathrm{width}=dx \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{approximating}\mathrm{rectangle}=\left|y\right|dx \\ \mathrm{The}\mathrm{approximating}\mathrm{rectangle}\mathrm{moves}\mathrm{from}x=0\mathrm{to}x=2 \\ \Rightarrow A=\mathrm{Area}\mathrm{OABO}={\int }_0^2\left|y\right|dx \\ \Rightarrow A={\int }_0^{2}ydx\left[\mathrm{As},y>o,\left|y\right|=y\right] \\ \Rightarrow A={\int }_0^2\left(4-x^2\right)dx \\ \Rightarrow A={\left[4x-\frac{x^3}{3}\right]}_0^2 \\ \Rightarrow A=8-\frac{8}{3} \\ \Rightarrow A=\frac{16}{3}\mathrm{sq}.\mathrm{units} \\ \therefore \mathrm{The}\mathrm{area}\mathrm{enclosed}\mathrm{by}\mathrm{the}\mathrm{curve}\mathrm{and}x-\mathrm{axis}\mathrm{and}\mathrm{given}\mathrm{lines}=\frac{16}{3}\mathrm{sq}.\mathrm{units} \end{gathered}$$