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

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