Area Between Curves

$x^2 - 2x = x + 4 \Rightarrow x^2 - 3x - 4 = 0 \Rightarrow (x - 4)(x + 1) = 0$, so $x = -1$ and $x = 4$.

At $x = 0$: the line gives $4$, the parabola $0$. The line is on top.

$\displaystyle\int_{-1}^{4}\big[(x + 4) - (x^2 - 2x)\big]\,dx = \int_{-1}^{4}(-x^2 + 3x + 4)\,dx$.

$\Big[-\tfrac{x^3}{3} + \tfrac{3x^2}{2} + 4x\Big]_{-1}^{4} = \tfrac{56}{3} - \left(-\tfrac{13}{6}\right) = \tfrac{125}{6}$.

$\sqrt{x} = \dfrac{x^2}{8} \Rightarrow x = 0$ or $x^{3/2} = 8$, so $x = 0$ and $x = 4$.

At $x = 1$: $\sqrt{1} = 1 > \tfrac{1}{8}$. The root is on top.

$\displaystyle\int_0^4\left(\sqrt{x} - \frac{x^2}{8}\right)\,dx$.

$\Big[\tfrac{2}{3}x^{3/2} - \tfrac{x^3}{24}\Big]_0^4 = \tfrac{16}{3} - \tfrac{8}{3} = \tfrac{8}{3}$.

$\sin x = \cos x$ at $x = \tfrac{\pi}{4}$. The interval $\left[0, \tfrac{\pi}{2}\right]$ is given.

On $\left[0, \tfrac{\pi}{4}\right]$, $\cos x$ is on top; on $\left[\tfrac{\pi}{4}, \tfrac{\pi}{2}\right]$, $\sin x$ is. Split at $\tfrac{\pi}{4}$.

$\displaystyle\int_0^{\pi/4}(\cos x - \sin x)\,dx + \int_{\pi/4}^{\pi/2}(\sin x - \cos x)\,dx$.

$\big[\sin x + \cos x\big]_0^{\pi/4} = \sqrt{2} - 1$ and $\big[-\cos x - \sin x\big]_{\pi/4}^{\pi/2} = \sqrt{2} - 1$. Total $2\sqrt{2} - 2$.

$x = y^2$ and $x = y + 2$.

$y^2 = y + 2 \Rightarrow (y - 2)(y + 1) = 0$, so $y = -1$ and $y = 2$.

At $y = 0$: the line gives $x = 2$, the parabola $x = 0$. The line is on the right.

$\displaystyle\int_{-1}^{2}(y + 2 - y^2)\,dy = \Big[\tfrac{y^2}{2} + 2y - \tfrac{y^3}{3}\Big]_{-1}^{2} = \tfrac{10}{3} + \tfrac{7}{6} = \tfrac{9}{2}$.

The curve is $x = 4 - y^2$; the $y$-axis is $x = 0$.

$4 - y^2 = 0 \Rightarrow y = \pm 2$.

At $y = 0$: $x = 4 > 0$. The curve is on the right.

$\displaystyle\int_{-2}^{2}(4 - y^2)\,dy = \Big[4y - \tfrac{y^3}{3}\Big]_{-2}^{2} = \tfrac{32}{3}$.

$y = \log x \Rightarrow x = e^{y}$. The $y$-axis is $x = 0$.

The limits are given: $y = 0$ to $y = 1$.

$e^{y} > 0$, so the curve is on the right.

$\displaystyle\int_0^1 (e^{y} - 0)\,dy = e - 1$.

The region lies under both curves and above the $x$-axis, with a corner on each pair.

$\sqrt{x} = 2 - x$ at $x = 1$, the point $(1, 1)$. $\sqrt{x} = 0$ at $x = 0$; $2 - x = 0$ at $x = 2$.

The top is $\sqrt{x}$ on $[0, 1]$ and $2 - x$ on $[1, 2]$; the bottom is always $y = 0$.

$\displaystyle\int_0^1 \sqrt{x}\,dx + \int_1^2 (2 - x)\,dx = \tfrac{2}{3} + \tfrac{1}{2} = \tfrac{7}{6}$.

A cap: below the parabola $y = 2 - x^2$, above the V of $y = |x|$. Both are even, so the region is symmetric.

For $x \ge 0$: $x = 2 - x^2 \Rightarrow (x + 2)(x - 1) = 0$, so $x = 1$; by symmetry, $x = -1$ too.

The bottom changes at $x = 0$ ($-x$ on the left, $x$ on the right). Double the right half.

$2\displaystyle\int_0^1 (2 - x^2 - x)\,dx = 2\left(2 - \tfrac{1}{3} - \tfrac{1}{2}\right) = \tfrac{7}{3}$.

$\Delta x = \dfrac{2 - 0}{4} = \dfrac{1}{2}$.

Left endpoints: $0, \tfrac{1}{2}, 1, \tfrac{3}{2}$.

Height $(x^2 + 1) - x$: $\ 1,\ \tfrac{3}{4},\ 1,\ \tfrac{7}{4}$.

$\tfrac{1}{2}\left(1 + \tfrac{3}{4} + 1 + \tfrac{7}{4}\right) = \tfrac{1}{2} \cdot \tfrac{9}{2} = \tfrac{9}{4}$.

$\Delta x = \dfrac{2 - 0}{4} = \dfrac{1}{2}$.

Right endpoints: $\tfrac{1}{2}, 1, \tfrac{3}{2}, 2$.

Height $(8 - x) - x^2$: $\ \tfrac{29}{4},\ 6,\ \tfrac{17}{4},\ 2$.

$\tfrac{1}{2}\left(\tfrac{29}{4} + 6 + \tfrac{17}{4} + 2\right) = \tfrac{1}{2} \cdot \tfrac{39}{2} = \tfrac{39}{4}$.