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E Math (IGCSE)

H1 Math

H2 Math

IB AA HL Math

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PSLE

PSLE Foundation Mathematics

<p>The diagram below shows a sketch of the graph of $y = \sin^2 x$ for&nbsp;$0 \le x \le \frac{\pi}{2}$.</p>
<p>Rectangles each of width $\frac{\pi}{2n}$ are drawn under the curve for&nbsp;$0 \le x \le \frac{\pi}{2}$.</p>
<div class="graph-block" data-graph-config="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" data-width="400" data-height="300" style="width:400px;height:300px;max-width:100%;display:block;margin:0 auto;"></div>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Show that&nbsp;$A$, the total area of all the rectangles, is given by $a \sum_{k=1}^{n-1} \sin^2 \frac{k\pi}{2n}$, where&nbsp;$a$ is to be determined.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Find the exact value of $\displaystyle \lim_{n\to\infty} a \sum_{k=1}^{n-1} \sin^2 \frac{k\pi}{2n}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Hence, find the value of $\int_{0}^{1} \sin^{-1} \sqrt{y} \ \mathrm{d}y$.<span class="inline-marks">[2]</span></div></li></ol>

The diagram below shows a sketch of the graph of $y = \sin^2 x$ for $0 \le x \le \frac{\pi}{2}$.

Rectangles each of width $\frac{\pi}{2n}$ are drawn under the curve for $0 \le x \le \frac{\pi}{2}$.

Show that $A$, the total area of all the rectangles, is given by $a \sum_{k=1}^{n-1} \sin^2 \frac{k\pi}{2n}$, where $a$ is to be determined.

Limit of an expression

Integrating Powers of Sine and Cosine

Integrable Forms of Trigonometric Functions

Area Enclosed by a Curve and the y-axis

Riemann Sums

Sigma Notation

2024 YIJC P1 Q2

Applications of Integration

Techniques of Integration

Sequence and Series

<ol class="abc-list"><li><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760753909486-xr3u9j011k.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA3NTM5MDk0ODYteHIzdTlqMDExay5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzODUzODE5LCJleHAiOjE3OTM4NTQxMTl9.QpKd5xLE9EHBODTXaBR-Z6irkflMKoy1J762pL4i8gc" alt="List item image" class="aligncenter custom-width" style="max-width: min(400px, 100%); height: auto;" /></div><div class="list-item-paragraph">The diagram shows part of the graph of $y={{x}^{2}}+2$ from $x=1$ to $x=2$. The area under the curve in this interval may be approximated by the total area of $n$ rectangles, $A$, as shown. The width of each rectangle is $\frac{1}{n}$.</div><div class="list-item-paragraph">Given that&nbsp;$\underset{r=1}{\overset{n}{\mathop \sum }}\,{{r}^{2}}=\frac{n}{6}\left( n+1 \right)\left( 2n+1 \right)$, show that&nbsp;$A=\frac{13}{3}+\frac{3}{2n}+\frac{1}{6{{n}^{2}}}$.</div><div class="list-item-paragraph">Deduce the exact value of $\mathop{\displaystyle\displaystyle \displaystyle\int }_{1}^{2}\left( 2+{{x}^{2}} \right)\,\mathrm{d}x$, justifying your answer. <span class="inline-marks">[6]</span></div></li><li><div class="list-item-paragraph">Region $R$ is bounded by the curve $y=\ln x$, $x=1$, and $y=\ln 5$.</div><ol class="i-list"><li><div class="list-item-paragraph">Find the area of $R$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Find the exact volume of the solid generated when&nbsp;$R$ is rotated through $4$ right angles about the $y$-axis.<span class="inline-marks">[4]</span></div></li></ol></li></ol>

Integration by Parts

Volume of Revolution about the y-axis

Summing Polynomial Series

2024 CJC P1 Q9

<ol class="abc-list"><li><div class="list-item-paragraph">The curve $C$ has parametric equations</div><div class="list-item-paragraph" style="text-align: center">$x=\sqrt{3}\cos t$, $y=3\sin t$, for $-\frac{\pi }{2}\le t\le \frac{\pi }{2}.$</div><ol class="i-list"><li><div class="list-item-paragraph">Sketch $C$, indicating clearly the exact coordinates of the axial intercepts.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the exact value of the area of the region bounded by $C$ and the $y$–axis.</div><p style="text-align: right;">[5]</p></li></ol></li></ol>

Sketching Parametric Curves

2013 VJC P1 Q10

Graphing Techniques

<p>The graph of $y={{\mathrm{e}}^{x}}$ is shown in the diagram. Rectangles, each of width $\frac{1}{n}$, where $n$ is an Integer, are drawn under the curve.</p>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760334739323-9a3kqahz2p.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjAzMzQ3MzkzMjMtOWEza3FhaHoycC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzODUzODE5LCJleHAiOjE3OTM4NTQxMTl9.5l2_WuQl4OaU_7XjtLycdPs8FpU96qNAU4AIaK1jDF4" width="400" class="aligncenter" /></figure>
<ol class="i-list"><li><div class="list-item-paragraph">The region ${{A}_{1}}$ is bounded by the $x$- axis, the $y$- axis, the line $x=1$ and the curve $y={{\mathrm{e}}^{x}}$.</div><div class="list-item-paragraph">The area of ${{A}_{1}}$ may be appropriated by the total area of all the $n$ rectangles.</div><div class="list-item-paragraph">Given that $1+x+{{x}^{2}}+{{x}^{3}}+...+{{x}^{n}}=\frac{1-{{x}^{n+1}}}{1-x}$, show that the area of ${{A}_{1}}$ is approximately $\frac{\mathrm{e}-1}{n\left( {{\mathrm{e}}^{\frac{1}{n}}}-1 \right)}$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">The region ${{A}_{2}}$ is bounded by the $y$- axis, the line $y=\mathrm{e}$ and the curve $y={{\mathrm{e}}^{x}}$. Using <strong>(i)</strong>, explain how the area of ${{A}_{2}}$ can be approximated and state, with a reason, whether this area is an underestimation or an overestimation.</div><p style="text-align: right;">[2]</p></li></ol>

The graph of $y={{\mathrm{e}}^{x}}$ is shown in the diagram. Rectangles, each of width $\frac{1}{n}$, where $n$ is an Integer, are drawn under the curve.

Application of Integration [Rectangles under Graph]

<p>The diagram below shows the curve $C$ with equation given by</p>
<p style="text-align: center">$x={{\mathrm{e}}^{-t}}\cos 2t$, $y=\sin t$,</p>
<p>where $\frac{\pi }{2}\le t\le \pi $.</p>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760263952557-sl2i03a42y.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjAyNjM5NTI1NTctc2wyaTAzYTQyeS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzODUzODE5LCJleHAiOjE3OTM4NTQxMTl9.QuWvi1wXCeh0KGTf_1vCmsrdfGf0CetZVeXVaG52B14" width="400" class="aligncenter" /></figure>
<p>Write down the exact values of $a$ and $b$. By considering the integral $\displaystyle\int_{0}^{a}{x\,\mathrm{d}y}$, show that the exact area of region $R$ is $\frac{3}{10}\left[ \left( \sqrt{2} \right){{\mathrm{e}}^{-\frac{3\pi }{4}}}-{{\mathrm{e}}^{-\pi }} \right]$.</p><p style="text-align: right;">[7]</p>

The diagram below shows the curve $C$ with equation given by

$x={{\mathrm{e}}^{-t}}\cos 2t$, $y=\sin t$,

Write down the exact values of $a$ and $b$. By considering the integral $\displaystyle\displaystyle \displaystyle\int_{0}^{a}{x\,\mathrm{d}y}$, show that the exact area of region $R$ is $\frac{3}{10}\left[ \left( \sqrt{2} \right){{\mathrm{e}}^{-\frac{3\pi }{4}}}-{{\mathrm{e}}^{-\pi }} \right]$.

Trigonometric Integration Using Product-to-Sum Formulae

Cyclic Integration by Parts

2008 NJC Promo Q13

<p>$O$ is the origin and $A$ is the point on the curve $y=\tan x$ where&nbsp;$x=\frac{1}{3}\pi $.</p>
<ol class="i-list"><li data-list-item-id="item-1760258240689-tv6a2hdwi" data-list-item-path="item-1760258240689-tv6a2hdwi"><div class="list-item-paragraph">Calculate the area of the region $R$ enclosed by the arc $OA$, the $x$-axis and the line $x=\frac{1}{3}\pi $, giving your answer in an exact form.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760258241371-0i8gvcif7" data-list-item-path="item-1760258241371-0i8gvcif7"><div class="list-item-paragraph">The region $S$ is enclosed by the arc $OA$, the $y$-axis and the line $y=\sqrt{3}$. Find the volume of the solid of revolution formed when&nbsp;$S$ is rotated through $360{}^\circ $ about the $x$-axis, giving your answer in an exact form.<span class="inline-marks">[6]</span></div></li><li data-list-item-id="item-1760258242151-34dc7fqo5" data-list-item-path="item-1760258242151-34dc7fqo5"><div class="list-item-paragraph"><p>Find $\int_{0}^{\sqrt{3}} \tan^{-1} y\, \mathrm{d}y$.</p><span class="inline-marks">[3]</span></div></li></ol>
<p style="text-align: center"></p>

$O$ is the origin and $A$ is the point on the curve $y=\tan x$ where $x=\frac{1}{3}\pi $.

Volume of Revolution about the x-axis

Area Bounded by Curves

N2002 P1 Q14

<p>The curve $C$ has parametric equations</p>
<p style="text-align: center">$x={{t}^{3}}-t,y={{\mathrm{e}}^{t}},\mathrm{for }t\ge -1.$</p>
<ol class="i-list"><li><div class="list-item-paragraph">Sketch $C$, labelling all axial intercept(s).</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the equation of the tangent to the curve when $t=1$, leaving your answer in terms of $e$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">The normal to the curve $C$ at point $P\left( {{p}^{3}}-p,{{\mathrm{e}}^{p}} \right)$ is parallel to the $x$-axis and intersects the $y$-axis at point $S\left( 0,s \right)$, where$s<1$. Find the exact value of $p$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the area of the region bounded by the curve $C$ and the $y$-axis.</div><p style="text-align: right;">[3]</p></li></ol>

$x={{t}^{3}}-t,y={{\mathrm{e}}^{t}},\mathrm{for }t\ge -1.$

Tangents and Normals to Parametric Curves

Differentiation of Parametric Equations

Tangents and Normals Parallel to the Axes

Derivatives of Exponential Functions

2013 NJC P1 P8

Applications of Differentiation

Techniques of Differentiation

<p>The shaded region $R$ is bounded by the curves $x=-3{{\mathrm{e}}^{-2y}}$, $x=\frac{1}{2}-{{\mathrm{e}}^{2y}}$ and the $x$-axis is shown in the diagram below.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1759578872337-ih5zdfrf96a.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NTk1Nzg4NzIzMzctaWg1emRmcmY5NmEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5Mzg1MzgxOSwiZXhwIjoxNzkzODU0MTE5fQ.qPILy32HEKG9HsLy8JrGLVtf4rgIXINt5l77eO7mZe4" style="max-width: min(500px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-1759578848964-3ayqswf9n" data-list-item-path="item-1759578848964-3ayqswf9n"><div class="list-item-paragraph">Find the exact region of the region $R$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-1759578849954-04zccermh" data-list-item-path="item-1759578849954-04zccermh"><div class="list-item-paragraph">Find the volume of the solid or revolution formed when&nbsp;$R$ is rotated through $4$ right angles about the $y$-axis, giving your answer correct to $2$ decimal places.<span class="inline-marks">[2]</span></div></li></ol>

The shaded region $R$ is bounded by the curves $x=-3{{\mathrm{e}}^{-2y}}$, $x=\frac{1}{2}-{{\mathrm{e}}^{2y}}$ and the $x$-axis is shown in the diagram below.

Find the volume of the solid or revolution formed when $R$ is rotated through $4$ right angles about the $y$-axis, giving your answer correct to $2$ decimal places.

Area bounded by 2 curves

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<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">A curve has parametric equations</span></div><div class="list-item-paragraph" style="text-align: center">$x=\sqrt{2}\cos t$, $y=3\sin t$, for $-\frac{\pi }{2}\le t\le \frac{\pi }{2}$.</div><div class="list-item-paragraph">Find the exact area bounded by the curve, the $y$-axis and the line $x=1$.</div><p style="text-align: right;">[5]</p></li><li><div class="list-item-content"><span class="list-item-text">The curve $C$ has an equation ${{y}^{2}}-2y-x=-2$. The region $S$ is bounded by the curve $C$ and the line $y=x-2$. Find the volume of the solid formed when the region $S$ is rotated about the $y$-axis through four right angles.</span></div><p style="text-align: right;">[3]</p></li></ol>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1759485021427-lluxbtjv3b.jpg?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NTk0ODUwMjE0MjctbGx1eGJ0anYzYi5qcGciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzODUzODE5LCJleHAiOjE3OTM4NTQxMTl9.Xutw4pUXjbOCr1OuXyl3lMuvflsurT3IS7Nr_3DJD2c" width="400" class="aligncenter" /></figure>

The curve $C$ has an equation ${{y}^{2}}-2y-x=-2$. The region $S$ is bounded by the curve $C$ and the line $y=x-2$. Find the volume of the solid formed when the region $S$ is rotated about the $y$-axis through four right angles.

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<div class="graph-block" data-graph-config="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" data-width="600" data-height="201" style="width:600px;height:201px;max-width:100%;display:block;margin:0 auto;"></div>
<p>The diagram shows the curve with equation $y={{x}^{2}}$. The area of the region bounded by the curve, the lines $x=1$, $x=2$ and the $x$-axis is equal to the area of the region bounded by the curve, the lines $y=a$, $y=4$ and the $y$-axis, where&nbsp;$a&lt;4$. Find the value of $a$.<span class="inline-marks">[4]</span></p>

The diagram shows the curve with equation $y={{x}^{2}}$. The area of the region bounded by the curve, the lines $x=1$, $x=2$ and the $x$-axis is equal to the area of the region bounded by the curve, the lines $y=a$, $y=4$ and the $y$-axis, where $a<4$. Find the value of $a$.

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