[math] \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow} \newcommand{\ddxof}[1]{\frac{d #1}{d x}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\dydx}{\ddxof y} \newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}} \newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}} \newcommand{\dist}{\mathrm{distance}} \newcommand{\arccot}{\mathrm{arccot\:}} \newcommand{\arccsc}{\mathrm{arccsc\:}} \newcommand{\arcsec}{\mathrm{arcsec\:}} \newcommand{\arctanh}{\mathrm{arctanh\:}} \newcommand{\arcsinh}{\mathrm{arcsinh\:}} \newcommand{\arccosh}{\mathrm{arccosh\:}} \newcommand{\sech}{\mathrm{sech\:}} \newcommand{\csch}{\mathrm{csch\:}} \newcommand{\conj}[1]{\overline{#1}} \newcommand{\mathds}{\mathbb} [/math]

• If [math]f(y) = -y^2 + y + 2[/math], sketch the region bounded by the curve [math]x = f(y)[/math], the [math]y[/math]-axis, and the lines [math]y = 0[/math] and [math]y = 1[/math]. Find its area.
• Find the area bounded by the curve [math]x = -y^2 + y + 2[/math] and the [math]y[/math]-axis.
• The equation [math]x + y^2 = 4[/math] can be solved for [math]x[/math] as a function of [math]y[/math], or for [math]y[/math] as plus or minus a function of [math]x[/math]. Sketch the region in the first quadrant bounded by the curve [math]x + y^2 = 4[/math], and find its area first by integrating a function of [math]y[/math] and then by integrating a function of [math]x[/math].