Area Between Curves (x-functions) - Calculus 2

1. The curves $y = \sqrt{x}$ and $y = x^2$ intersect at two points. The area between these curves is:
   Answer: $\frac{1}{3}$ To find intersections, solve $\sqrt{x} = x^2$. Squaring both sides: $x = x^4$, so $x^4 - x = 0$, giving $x(x^3 - 1) = 0$, thus $x(x-1)(x^2+x+1) = 0$. Since we need $x \geq 0$ for $\sqrt{x}$ to be defined, and $x^2 + x + 1 > 0$ for all real $x$, the intersections are at $x = 0$ and $x = 1$. To determine which function is larger in $(0,1)$, test $x = \frac{1}{4}$: $\sqrt{\frac{1}{4}} = \frac{1}{2}$ and $(\frac{1}{4})^2 = \frac{1}{16}$. Since $\frac{1}{2} > \frac{1}{16}$, we have $\sqrt{x} > x^2$ on $(0,1)$. Area = $\int_0^1 (\sqrt{x} - x^2) dx = \int_0^1 (x^{1/2} - x^2) dx = \left[\frac{x^{3/2}}{3/2} - \frac{x^3}{3}\right]_0^1 = \left[\frac{2x^{3/2}}{3} - \frac{x^3}{3}\right]_0^1 = \frac{2}{3} - \frac{1}{3} - 0 = \frac{1}{3}$.
2. The area between $y = \frac{1}{x^2}$ and $y = \frac{4}{x^2}$ from $x = 1$ to $x = 2$ is:
   Answer: $\frac{3}{2}$ Since $\frac{4}{x^2} > \frac{1}{x^2}$ for all $x > 0$, the area is $\int_1^2 \left(\frac{4}{x^2} - \frac{1}{x^2}\right) dx = \int_1^2 \frac{3}{x^2} dx = \int_1^2 3x^{-2} dx = 3 \int_1^2 x^{-2} dx = 3[-x^{-1}]_1^2 = 3\left[-\frac{1}{x}\right]_1^2 = 3\left(-\frac{1}{2} - \left(-\frac{1}{1}\right)\right) = 3\left(-\frac{1}{2} + 1\right) = 3 \cdot \frac{1}{2} = \frac{3}{2}$.
3. Find the area of the finite region bounded by $y = x^3 - x$ and $y = 3x$.
   Answer: $8$ To find intersections, solve $x^3 - x = 3x$, which gives $x^3 - 4x = 0$, so $x(x^2 - 4) = 0$, thus $x(x-2)(x+2) = 0$. The curves intersect at $x = -2, 0, 2$. To determine which function is larger: For $x \in (-2, 0)$, test $x = -1$: $y_1 = (-1)^3 - (-1) = -1 + 1 = 0$ and $y_2 = 3(-1) = -3$, so $x^3 - x > 3x$. For $x \in (0, 2)$, test $x = 1$: $y_1 = 1^3 - 1 = 0$ and $y_2 = 3(1) = 3$, so $3x > x^3 - x$. Area = $\int_{-2}^0 (x^3 - x - 3x) dx + \int_0^2 (3x - (x^3 - x)) dx = \int_{-2}^0 (x^3 - 4x) dx + \int_0^2 (-x^3 + 4x) dx = \left[\frac{x^4}{4} - 2x^2\right]_{-2}^0 + \left[-\frac{x^4}{4} + 2x^2\right]_0^2 = (0 - (4 - 8)) + ((-4 + 8) - 0) = 4 + 4 = 8$.

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