Resources for Areas by Integration
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Questions
22
With Worked SolutionClick Here -
Video Tutorials
1
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HSC Questions
17
With Worked SolutionClick Here
Areas by Integration Theory
![Areas by integration involve areas between functions for both \(x\) and \(y\) axis orientation\\ \begin{multicols}{2} \textbf{Example 1}\\%11001 The shaded area between the curves \(y = 4 - {x^2}\) and \(y = {x^2}- 2x\) is equal to? \\ \begin{center} \begin{tikzpicture}[ declare function={a(\x)=4-(\x)^2;}, declare function={b(\x)=(\x)^2-2*\x;}, ] \def \domain{-2.8:2.8} \def \xmax{4} \def \xmin{-4} \def \ymax{6} \def \ymin{-4} \def \xlabel{x} \def \ylabel{y} \begin{axis}[ axis lines=middle, axis line style={Stealth-Stealth,very thick}, grid=both, %major, ylabel = $\ylabel$, xlabel = $\xlabel$, width=3in, height=2.5in, ymin=\ymin, ymax=\ymax, xmin=\xmin, xmax=\xmax, minor x tick num=1, minor y tick num=1, axis line style = thick, major tick style = thick, minor tick style = thick, xtick distance = 1, xlabel style={right}, ytick distance = 2, ylabel style={above}, x grid style={thin, opacity=0.8}, y grid style={thin, opacity=0.8}, axis on top=false, xtick={-6,-4,...,6}, ytick={-6,-4,-2,...,8}, % extra x ticks={0}, extra x tick style={xticklabel style={anchor=north east}} ] %FUNCTION \draw [draw=black,thin, opacity=0.5] (\xmin,\ymin) rectangle (\xmax,\ymax); \addplot[name path=a, ultra thick, latex-latex, samples=300, smooth, domain=\domain, red] {a(x)} node [pos=0.9, left, red, font=\small] {}; \addplot[name path=b, ultra thick, latex-latex, samples=300, smooth, domain=-1.5:3.5, blue] {b(x)} node [pos=0.9, left, red, font=\small] {}; \addplot[color=yellow!60,opacity=0.6] fill between[of=a and b, soft clip={domain=-1:2}]; \end{axis} \end{tikzpicture} %\includegraphics[width=0.3\textwidth]{81107a85-ac88-407e-b7d6-a5221f96828c} \end{center} \textbf{Example 1 solution}\\ $\begin{aligned} &\text{Solving simultaneously}\\ &y=x^{2}-2 x,\, y=4-x^{2}\\ &\begin{aligned} x^{2}-2 x&=4-x^{2} \\ 2 x^{2}-2 x-4&=0 \\ x^{2}-x-2&=0 \\ (x-2)(x+1)=0 & \rightarrow x=2 \text { or }-1\end{aligned} \\ &\begin{aligned}\therefore \text { Area }&=\int_{-1}^{2}\left(4-x^{2}\right)-\left(x^{2}-2 x\right) d x\\ &=\int_{-1}^{2}-2 x^{2}+2 x+4 d x \\ &=\left[-\frac{2 x^{3}}{3}+x^{2}+4 x\right]_{-1}^{2}\\ &=\left[\left(-\frac{16}{3}+4+8\right)-\left(\frac{2}{3}+1-4\right)\right] \\ &=9 u^{2} \end{aligned} \end{aligned}$\\ \columnbreak \textbf{Example 2}\\ %11003 The shaded area between \(y = \sqrt x \), \(y = - 2x + 3\) and the y-axis is equal to?\\ \begin{center} \begin{tikzpicture}[ declare function={a(\x)=sqrt(\x);}, declare function={b(\x)=-2*\x+3;}, ] \def \domain{0:1.8} \def \xmax{2} \def \xmin{-1} \def \ymax{4} \def \ymin{-1} \def \xlabel{x} \def \ylabel{y} \begin{axis}[ axis lines=middle, axis line style={Stealth-Stealth,very thick}, grid=both, %major, ylabel = $\ylabel$, xlabel = $\xlabel$, width=3in, height=2.5in, ymin=\ymin, ymax=\ymax, xmin=\xmin, xmax=\xmax, minor x tick num=1, minor y tick num=1, axis line style = thick, major tick style = thick, minor tick style = thick, xtick distance = 1, xlabel style={right}, ytick distance = 2, ylabel style={above}, x grid style={thin, opacity=0.8}, y grid style={thin, opacity=0.8}, axis on top=false, xtick={-1,-0.5,0,0.5,...,2}, ytick={-1,0,1,...,4}, % extra x ticks={0}, extra x tick style={xticklabel style={anchor=north east}} ] %FUNCTION \draw [draw=black,thin, opacity=0.5] (\xmin,\ymin) rectangle (\xmax,\ymax); \addplot[name path=a, ultra thick, -latex, samples=300, smooth, domain=\domain, red] {a(x)} node [pos=0.9, left, red, font=\small] {}; \addplot[name path=b, ultra thick, latex-latex, samples=300, smooth, domain=-0.4:1.8, blue] {b(x)} node [pos=0.9, left, red, font=\small] {}; \addplot[color=yellow!60,opacity=0.6] fill between[of=a and b,soft clip={domain=-0.01:1}]; \end{axis} \end{tikzpicture} %\includegraphics[width=0.3\textwidth]{3198c105-c329-4be1-b7c2-1600828b9867} \end{center} \textbf{Example 2 solution}\\ $\begin{aligned} &\text{Solving simultaneously: } y=\sqrt{x}, y=-2 x+3\\ &\begin{array}{l} y=\sqrt{x} \rightarrow x=y^{2} \rightarrow y=-2 y^{2}+3 \rightarrow 2 y^{2}+y-3=0 \\ (2 y+3)(y-1)=0=y=-1.5 \text { or } y=1 \\ y=-2 x+3 \rightarrow 2 x=3-y \rightarrow x=\frac{3}{2}-\frac{1}{2} y \end{array}\\ &\begin{aligned}\therefore \text { Area }&=\int_{1}^{3} \frac{3}{2}-\frac{1}{2} y d y+\int_{0}^{1} y^{2} d y\\ &=\left[\frac{3}{2} y-\frac{1}{2} \frac{y^{2}}{2}\right]_{1}^{3}+\left[\frac{y^{3}}{3}\right]_{0}^{1} \\ &=\left[\left(\frac{9}{2}-\frac{9}{4}\right)-\left(\frac{3}{2}-\frac{1}{4}\right)\right]+\left[\frac{1}{3}-0\right]\\ &=1 \frac{1}{3} u^{2} \end{aligned} \end{aligned}$\\ \end{multicols}](/media/qflhgsve/4871.png)