Question

Find the area bounded by the curves y = x/2 and y =√x.

Answer 1

Answer:

$\displaystyle \frac{4}{3}$.

Step-by-step explanation:

Start by finding the intersection of the two curves:

$\left\lbrace\begin{aligned}& y = \frac{x}{2} \\ & y = \sqrt{x} \\ & x \ge 0 \\ \end{aligned}\right.$.

$\displaystyle \frac{x}{2} = \sqrt{x}$ while $x \ge 0$.

$\displaystyle \frac{x^{2}}{4} - x = 0$.

$x = 0$ or $x = 4$.

Therefore, these two curves would intersect at two points: $(0,\, 0)$ and $(4,\, 2)$.

The area bounded between $\displaystyle y = \frac{x}{2}$ and $y = \sqrt{x}$ would be between $x = 0$ and $x = 4$.

Refer to the diagram attached. The graph $y = \sqrt{x}$ is always above the graph of $\displaystyle y = \frac{x}{2}$ over the entire bounded area (except for the two intersections).

Therefore, $\displaystyle \left(\sqrt{x} - \frac{x}{2}\right)$ would represent the vertical distance between the upper and lower curve for any given $x$ over this bounded area (where $0 \le x \le 4$.)

Integrating height over the horizontal variable $x$ over some closed interval would give area. Likewise, the area between the two curves in this question could be found with the following integral:

$\begin{aligned}& \int\limits_{0}^{4} \left(\sqrt{x} - \frac{x}{2}\right)\, dx \\ = \; & \int\limits_{0}^{4} \left(x^{1/2} - \frac{x}{2}\right)\, dx \\ =\; & \left[\frac{2}{3}\, x^{3/2} - \frac{x^{2}}{4}\right]_{x=0}^{x=4} \\ =\; & \frac{2\times 8}{3} - 4 \\ =\; &\frac{4}{3} \end{aligned}$.