Calculus - The definite integral – Areas between curves

Find the area of the region enclosed by the curves $f(x) = x^2$ and $g(x) = \sqrt{x}$.

1. Find points of intersection: $x^2 = \sqrt{x}$ $x^4 = x$ $x^4 - x = 0$ $x(x^3 - 1) = 0$ $x = 0$ and $x = 1$.

2. Determine which function is upper on $[0, 1]$. For $x = 0.25$, $\sqrt{0.25} = 0.5$ and $0.25^2 = 0.0625$. So $g(x) = \sqrt{x}$ is the upper function.

3. Set up the integral: $A = \int_{0}^{1} (\sqrt{x} - x^2) \, dx$ $A = \int_{0}^{1} (x^{1/2} - x^2) \, dx$ $A = \left[ \frac{2}{3}x^{3/2} - \frac{1}{3}x^3 \right]_{0}^{1}$ $A = (\frac{2}{3}(1)^{3/2} - \frac{1}{3}(1)^3) - (0)$ $A = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \text{ units}^2$

We first identify the boundaries by finding where the functions meet. Then, we integrate the difference (top function minus bottom function) over those boundaries.

Find the area bounded by $y = x^2 - 4$ and the $x$-axis.

1. The $x$-axis is the line $y = 0$. Find intersections: $x^2 - 4 = 0 \implies x = 2, x = -2$

2. On $[-2, 2]$, the $x$-axis ($y=0$) is above the parabola $y = x^2 - 4$.

3. Set up the integral: $A = \int_{-2}^{2} (0 - (x^2 - 4)) \, dx$ $A = \int_{-2}^{2} (4 - x^2) \, dx$ $A = \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2}$ $A = (4(2) - \frac{2^3}{3}) - (4(-2) - \frac{(-2)^3}{3})$ $A = (8 - \frac{8}{3}) - (-8 + \frac{8}{3})$ $A = \frac{16}{3} - (-\frac{16}{3}) = \frac{32}{3} \text{ units}^2$

Since the region is below the $x$-axis, the function $y=0$ is the upper boundary. Subtracting the curve from zero ensures a positive area result.