| $ \displaystyle \int_{a}^{b}{{\left[ {f(x)-g(x)} \right]}}\ dx$ |
Where $ \displaystyle f(x)$ is the curve of upper boundry and $ \displaystyle g(x)$ is the curve of lower boundry.
ဆရာႀကီး ေဒါက္တာ ေရႊေက်ာ္ ၏ Drill For Exam Blog မွ Area under curve ပုစာၦမ်ား ျဖစ္ပါသည္။
Problem (1)
(i) Show that, for $ \displaystyle x≥0, 9x^2−x^3≤108.$
(ii) Find the area of the shaded region bounded by the curve, the line $ \displaystyle AB$ and the $ \displaystyle y$-axis.
Show/Hide Solution
$ \displaystyle \begin{array}{l}\ \ \ \ \text{Line : }y-2x+18\Rightarrow y=2x-18\\\\\ \ \ \ \text{Curve : }y=9{{x}^{2}}-{{x}^{3}}\end{array}$
$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=18x-3{{x}^{2}}=3x(6-x)$
$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=0\ \text{when}\ 3x(6-x)=0$
$ \displaystyle {\therefore \ \ x=0\ \text{or }x=6}$
$ \displaystyle \therefore \ \ \text{When }x=0, y=0$
$ \displaystyle {\therefore \ \ \text{When }x=6,\ }$
$ \displaystyle {\ \ \ \ y=9({{6}^{2}})-{{{(6)}}^{3}}=108}$
$ \displaystyle \ \ \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=-6x$
$ \displaystyle \therefore \ \ \text{When }x=6,\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=-36<0$
$ \displaystyle \begin{array}{l}\therefore \ \ y=108\ \text{is a maximum value}\text{.}\\\\\therefore \ \ \text{For}\ x\ge 0,\ y\le 0.\\\\\ \ \ \ \text{According to the diagram, }\\\ \ \ \ \text{the line and curve intersect when }y=0.\\\\\therefore \ \ 2x-18=0\Rightarrow x=9.\\\\\ \ \ \ \text{Let the curve of upper boundry be }{{y}_{1}}\text{ and }\\\ \ \ \ \text{that of the lower boundry be }{{y}_{2}}\text{, and }\\\ \ \ \ \text{let the area of the shaded region be }A.\\\\\therefore \ \ {{y}_{1}}=9{{x}^{2}}-{{x}^{3}},\ {{y}_{2}}=2x-18\ \operatorname{and}\end{array}$
$ \displaystyle \ \ \ A=\int_{0}^{9}{{\left( {{{y}_{1}}-{{y}_{2}}} \right)}}\ dx$
$ \displaystyle \ \ \ \ \ \ \ =\int_{0}^{9}{{\left( {9{{x}^{2}}-{{x}^{3}}-2x+18} \right)}}\ dx$
$ \displaystyle \ \ \ \ \ \ \ =\left. {\frac{9}{3}{{x}^{3}}-\frac{{{{x}^{4}}}}{4}-\frac{{2{{x}^{2}}}}{2}+18x} \right]_{0}^{9}$
$ \displaystyle \ \ \ \ \ \ \ =\left. {3{{x}^{3}}-\frac{{{{x}^{4}}}}{4}-{{x}^{2}}+18x} \right]_{0}^{9}$
$ \displaystyle \ \ \ \ \ \ \ =3{{(9)}^{3}}-\frac{{{{{(9)}}^{4}}}}{4}-{{(9)}^{2}}+18(9)$
$ \displaystyle \ \ \ \ \ \ \ =627.75$
$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=18x-3{{x}^{2}}=3x(6-x)$
$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=0\ \text{when}\ 3x(6-x)=0$
$ \displaystyle {\therefore \ \ x=0\ \text{or }x=6}$
$ \displaystyle \therefore \ \ \text{When }x=0, y=0$
$ \displaystyle {\therefore \ \ \text{When }x=6,\ }$
$ \displaystyle {\ \ \ \ y=9({{6}^{2}})-{{{(6)}}^{3}}=108}$
$ \displaystyle \ \ \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=-6x$
$ \displaystyle \therefore \ \ \text{When }x=6,\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=-36<0$
$ \displaystyle \begin{array}{l}\therefore \ \ y=108\ \text{is a maximum value}\text{.}\\\\\therefore \ \ \text{For}\ x\ge 0,\ y\le 0.\\\\\ \ \ \ \text{According to the diagram, }\\\ \ \ \ \text{the line and curve intersect when }y=0.\\\\\therefore \ \ 2x-18=0\Rightarrow x=9.\\\\\ \ \ \ \text{Let the curve of upper boundry be }{{y}_{1}}\text{ and }\\\ \ \ \ \text{that of the lower boundry be }{{y}_{2}}\text{, and }\\\ \ \ \ \text{let the area of the shaded region be }A.\\\\\therefore \ \ {{y}_{1}}=9{{x}^{2}}-{{x}^{3}},\ {{y}_{2}}=2x-18\ \operatorname{and}\end{array}$
$ \displaystyle \ \ \ A=\int_{0}^{9}{{\left( {{{y}_{1}}-{{y}_{2}}} \right)}}\ dx$
$ \displaystyle \ \ \ \ \ \ \ =\int_{0}^{9}{{\left( {9{{x}^{2}}-{{x}^{3}}-2x+18} \right)}}\ dx$
$ \displaystyle \ \ \ \ \ \ \ =\left. {\frac{9}{3}{{x}^{3}}-\frac{{{{x}^{4}}}}{4}-\frac{{2{{x}^{2}}}}{2}+18x} \right]_{0}^{9}$
$ \displaystyle \ \ \ \ \ \ \ =\left. {3{{x}^{3}}-\frac{{{{x}^{4}}}}{4}-{{x}^{2}}+18x} \right]_{0}^{9}$
$ \displaystyle \ \ \ \ \ \ \ =3{{(9)}^{3}}-\frac{{{{{(9)}}^{4}}}}{4}-{{(9)}^{2}}+18(9)$
$ \displaystyle \ \ \ \ \ \ \ =627.75$
Problem (2)
(i) Find the coordinates of $ \displaystyle P.$
(ii) Find the area of the shaded region bounded by the curve, the normal and the $ \displaystyle y$-axis.
Show/Hide Solution
$ \displaystyle \ \ \ \text{Curve : }y=2\sin 3x,$
$ \displaystyle \ \ \ \text{When }x=\frac{\pi }{9},$
$ \displaystyle \ \ \ y=2\sin 3\left( {\frac{\pi }{9}} \right)$
$ \displaystyle \ \ \ y=2\sin 3\left( {\frac{\pi }{9}} \right)$
$ \displaystyle \ \ \ \ \ =\sqrt{3}$
$ \displaystyle \ \ \ \frac{{dy}}{{dx}}=6\cos 3x$
$ \displaystyle \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{x=\frac{\pi }{9}}}}=6\cos 3\left( {\frac{\pi }{9}} \right)=3$
$ \displaystyle \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{x=\frac{\pi }{9}}}}=6\cos 3\left( {\frac{\pi }{9}} \right)=3$
$ \displaystyle \ \ \ y=\sqrt{3}-\frac{1}{3}\left( {x-\frac{\pi }{9}} \right)$
$ \displaystyle \ \ \ y=\sqrt{3}+\frac{\pi }{{27}}-\frac{x}{3}$
$ \displaystyle \ \ \ \text{When the normal cuts the }y\text{-axis,}\ x=0.$
$ \displaystyle \therefore \ y=\sqrt{3}+\frac{\pi }{{27}}$
$ \displaystyle \therefore \ P=(0,\sqrt{3}+\frac{\pi }{{27}})=(0,1.85)$
$ \displaystyle \ \ \ \text{Let the area of the shaded region be }A.$
$ \displaystyle \therefore A=\left. {\left( {\sqrt{3}+\frac{\pi }{{27}}} \right)x-\frac{{{{x}^{2}}}}{6}+\frac{2}{3}\cos 3x} \right]_{0}^{{\frac{\pi }{9}}}$
$ \displaystyle \therefore A=\left( {\sqrt{3}+\frac{\pi }{{27}}} \right)\frac{\pi }{9}-\frac{{{{\pi }^{2}}}}{{486}}+\frac{2}{3}\cos 3\left( {\frac{\pi }{9}} \right)-\frac{2}{3}\cos (0)$
$ \displaystyle \therefore A=\frac{{\sqrt{3}\pi }}{9}+\frac{{{{\pi }^{2}}}}{{243}}-\frac{{{{\pi }^{2}}}}{{486}}+\frac{1}{3}-\frac{2}{3}=2.912$
$ \displaystyle \ \ \ \text{When }x=\frac{\pi }{9},$
$ \displaystyle \ \ \ y=2\sin 3\left( {\frac{\pi }{9}} \right)$
$ \displaystyle \ \ \ y=2\sin 3\left( {\frac{\pi }{9}} \right)$
$ \displaystyle \ \ \ \ \ =\sqrt{3}$
$ \displaystyle \ \ \ \frac{{dy}}{{dx}}=6\cos 3x$
$ \displaystyle \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{x=\frac{\pi }{9}}}}=6\cos 3\left( {\frac{\pi }{9}} \right)=3$
$ \displaystyle \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{x=\frac{\pi }{9}}}}=6\cos 3\left( {\frac{\pi }{9}} \right)=3$
$ \displaystyle \ \ \ y=\sqrt{3}-\frac{1}{3}\left( {x-\frac{\pi }{9}} \right)$
$ \displaystyle \ \ \ y=\sqrt{3}+\frac{\pi }{{27}}-\frac{x}{3}$
$ \displaystyle \ \ \ \text{When the normal cuts the }y\text{-axis,}\ x=0.$
$ \displaystyle \therefore \ y=\sqrt{3}+\frac{\pi }{{27}}$
$ \displaystyle \therefore \ P=(0,\sqrt{3}+\frac{\pi }{{27}})=(0,1.85)$
$ \displaystyle \ \ \ \text{Let the area of the shaded region be }A.$
$ \displaystyle \therefore A=\left. {\left( {\sqrt{3}+\frac{\pi }{{27}}} \right)x-\frac{{{{x}^{2}}}}{6}+\frac{2}{3}\cos 3x} \right]_{0}^{{\frac{\pi }{9}}}$
$ \displaystyle \therefore A=\left( {\sqrt{3}+\frac{\pi }{{27}}} \right)\frac{\pi }{9}-\frac{{{{\pi }^{2}}}}{{486}}+\frac{2}{3}\cos 3\left( {\frac{\pi }{9}} \right)-\frac{2}{3}\cos (0)$
$ \displaystyle \therefore A=\frac{{\sqrt{3}\pi }}{9}+\frac{{{{\pi }^{2}}}}{{243}}-\frac{{{{\pi }^{2}}}}{{486}}+\frac{1}{3}-\frac{2}{3}=2.912$
Problem (3)
The diagram shows part of the curve $ \displaystyle y=\sin \frac{1}{2}x$. The tangent to the curve at the point $ \displaystyle P\left( {\frac{{3\pi }}{2},\frac{{\sqrt{2}}}{2}} \right)$ cuts the $ \displaystyle x$-axis at the point $ \displaystyle Q$. (i) Find the coordinates of $ \displaystyle Q$. (ii) Find the area of the shaded region bounded by the curve, the tangent and the $ \displaystyle x$-axis.
Show/Hide Solution
$ \displaystyle \ \ \ \ y=\sin \frac{1}{2}x$
$ \displaystyle \ \ \ \ y=0\Rightarrow \sin \frac{1}{2}x=0$
$ \displaystyle \therefore \ \ \frac{1}{2}x=0\ \text{or}\ \frac{1}{2}x=\pi \ \ (\text{for first half cycle)}$
$ \displaystyle \therefore \ \ x=0\ \text{or}\ x=2\pi $
$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=\frac{1}{2}\cos \frac{1}{2}x$
$ \displaystyle \ \ \ \ \text{At }P\left( {\frac{{3\pi }}{2},\frac{{\sqrt{2}}}{2}} \right),\frac{{dy}}{{dx}}=\frac{1}{2}\cos \frac{{3\pi }}{4}=-\frac{{\sqrt{2}}}{4}$
$ \displaystyle \therefore \ \ \text{Equation of tangent at }P\ \text{is}$
$ \displaystyle \ \ \ \ y=\frac{{\sqrt{2}}}{2}-\frac{{\sqrt{2}}}{4}\left( {x-\frac{{3\pi }}{2}} \right)$
$ \displaystyle \therefore \ \ y=\frac{{\sqrt{2}}}{2}+\frac{{3\sqrt{2}\pi }}{8}-\frac{{\sqrt{2}}}{4}x$
$ \displaystyle \ \ \ \ \text{When the tangent cuts the }x\text{-axis, }y=0.$
$ \displaystyle \therefore \ \ \frac{{\sqrt{2}}}{2}+\frac{{3\sqrt{2}\pi }}{8}-\frac{{\sqrt{2}}}{4}x=0$
$ \displaystyle \therefore \ \ x=2+\frac{{3\pi }}{2}$
$ \displaystyle \therefore \ \ Q=\left( {2+\frac{{3\pi }}{2},0} \right)=(6.71,0)$
$ \displaystyle \ \ \ \ \text{Area of }\Delta PQR=\frac{1}{2}\cdot PQ\cdot PR$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}(2)\left( {\frac{{\sqrt{2}}}{2}} \right)$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sqrt{2}}}{2}$
$ \displaystyle \ \ \ \ \text{Area of yellow region}=\int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\left( {\sin \frac{1}{2}x} \right)dx}}$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left. {-2\cos \frac{1}{2}x} \right]_{{\frac{{3\pi }}{2}}}^{{2\pi }}$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-2\left( {\cos \pi -\cos \frac{{3\pi }}{4}} \right)$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-2\left( {-1+\frac{{\sqrt{2}}}{2}} \right)$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2-\sqrt{2}$
$ \displaystyle \ \ \ \ \text{Let the required area be }A.$
$ \displaystyle \therefore \ \ A=\text{Area of }\Delta PQR-\text{Area of yellow region}$
$ \displaystyle \therefore \ \ A=\frac{{\sqrt{2}}}{2}-2+\sqrt{2}=\frac{{3\sqrt{2}}}{2}-2$
$ \displaystyle \ \ \ \ y=0\Rightarrow \sin \frac{1}{2}x=0$
$ \displaystyle \therefore \ \ \frac{1}{2}x=0\ \text{or}\ \frac{1}{2}x=\pi \ \ (\text{for first half cycle)}$
$ \displaystyle \therefore \ \ x=0\ \text{or}\ x=2\pi $
$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=\frac{1}{2}\cos \frac{1}{2}x$
$ \displaystyle \ \ \ \ \text{At }P\left( {\frac{{3\pi }}{2},\frac{{\sqrt{2}}}{2}} \right),\frac{{dy}}{{dx}}=\frac{1}{2}\cos \frac{{3\pi }}{4}=-\frac{{\sqrt{2}}}{4}$
$ \displaystyle \therefore \ \ \text{Equation of tangent at }P\ \text{is}$
$ \displaystyle \ \ \ \ y=\frac{{\sqrt{2}}}{2}-\frac{{\sqrt{2}}}{4}\left( {x-\frac{{3\pi }}{2}} \right)$
$ \displaystyle \therefore \ \ y=\frac{{\sqrt{2}}}{2}+\frac{{3\sqrt{2}\pi }}{8}-\frac{{\sqrt{2}}}{4}x$
$ \displaystyle \ \ \ \ \text{When the tangent cuts the }x\text{-axis, }y=0.$
$ \displaystyle \therefore \ \ \frac{{\sqrt{2}}}{2}+\frac{{3\sqrt{2}\pi }}{8}-\frac{{\sqrt{2}}}{4}x=0$
$ \displaystyle \therefore \ \ x=2+\frac{{3\pi }}{2}$
$ \displaystyle \therefore \ \ Q=\left( {2+\frac{{3\pi }}{2},0} \right)=(6.71,0)$
$ \displaystyle \ \ \ \ \text{Area of }\Delta PQR=\frac{1}{2}\cdot PQ\cdot PR$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}(2)\left( {\frac{{\sqrt{2}}}{2}} \right)$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\sqrt{2}}}{2}$
$ \displaystyle \ \ \ \ \text{Area of yellow region}=\int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\left( {\sin \frac{1}{2}x} \right)dx}}$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left. {-2\cos \frac{1}{2}x} \right]_{{\frac{{3\pi }}{2}}}^{{2\pi }}$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-2\left( {\cos \pi -\cos \frac{{3\pi }}{4}} \right)$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-2\left( {-1+\frac{{\sqrt{2}}}{2}} \right)$
$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2-\sqrt{2}$
$ \displaystyle \ \ \ \ \text{Let the required area be }A.$
$ \displaystyle \therefore \ \ A=\text{Area of }\Delta PQR-\text{Area of yellow region}$
$ \displaystyle \therefore \ \ A=\frac{{\sqrt{2}}}{2}-2+\sqrt{2}=\frac{{3\sqrt{2}}}{2}-2$
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