Question

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

Answer 1

`\displaystyle (4)/(3)`.

Explanation:

Start by finding the intersection of the two curves:

`\left\lbrace\begin{aligned}& y = (x)/(2) \\ & y = √(x) \\ & x \ge 0 \\ \end{aligned}\right.`.

`\displaystyle (x)/(2) = √(x)` while
`x \ge 0`.

`\displaystyle (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 = (x)/(2)` and
`y = √(x)` would be between
`x = 0` and
`x = 4`.

Refer to the diagram attached. The graph
`y = √(x)` is always above the graph of
`\displaystyle y = (x)/(2)` over the entire bounded area (except for the two intersections).

Therefore,
`\displaystyle \left(√(x) - (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(√(x) - (x)/(2)\right)\, dx \\ = \; & \int\limits_(0)^(4) \left(x^(1/2) - (x)/(2)\right)\, dx \\ =\; & \left[(2)/(3)\, x^(3/2) - (x^(2))/(4)\right]_(x=0)^(x=4) \\ =\; & (2* 8)/(3) - 4 \\ =\; &(4)/(3) \end{aligned}`.