<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Find $\displaystyle\int{\left( {{\cot }^{6}}2x+{{\cot }^{4}}2x-\sin \frac{x}{2}\sin \frac{3x}{2} \right)}\mathrm{ d}x$.</span></div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">The diagram shows the graphs of $y={{\mathrm{e}}^{2x}}$ and $y={{\left( x+1 \right)}^{2}}$.</div><div class="list-item-paragraph">$R$ is the finite region bounded by the two curves $y={{\mathrm{e}}^{2x}}$ and $y={{\left( x+1 \right)}^{2}}$.</div><div class="list-item-paragraph">Find the volume of the solid formed when $R$ is rotated through $2\pi $ radians about the $y$-axis, giving your answer correct to $4$ decimal places.</div><p style="text-align: right;">[3]</p><div class="list-item-image"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1761540959364-qtzjyukhxf.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjE1NDA5NTkzNjQtcXR6anl1a2h4Zi5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTAzNDkzLCJleHAiOjE3OTM1MDM3OTN9.t-oGD9J8TsdHPYJh2WqNgA2xJSFTVNjtUGWO2PkQ3uM" alt="List item image" /></div></li><li><div class="list-item-paragraph">A curve $C$ has parametric equations</div><div class="list-item-paragraph" style="text-align: center">$x=2\left( \sin \frac{t}{2}-\cos \frac{t}{2} \right)$, $y=\sin \frac{t}{2}+\cos \frac{t}{2}$, for $-\frac{3\pi }{2}\le t\le -\frac{\pi }{2}$.</div><div class="list-item-paragraph">The line $L$ with equation $y=-2x-5$ meets the curve $C$ at the point $\left( -2,\,-1 \right)$. $S$ is the region enclosed by line $L$, curve $C$and the $x$-axis as shown in the diagram below.</div><div class="list-item-image"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1761541142288-0319z6id82yx.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjE1NDExNDIyODgtMDMxOXo2aWQ4Mnl4LnBuZyIsInNjb3BlIjoiZG93bmxvYWQiLCJpYXQiOjE3OTM1MDM0OTMsImV4cCI6MTc5MzUwMzc5M30.j6KLWCxMlI79012h0bOQi5cLBLyzxmEQwofmiEHMYjw" alt="List item image" /></div><ol class="i-list"><li><div class="list-item-paragraph">Show that the x-intercept of curve $C$ is $-2\sqrt{2}$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the exact area of $S$.</div><p style="text-align: right;">[6]</p></li></ol></li></ol>

Find $\displaystyle\displaystyle \displaystyle\int{\left( {{\cot }^{6}}2x+{{\cot }^{4}}2x-\sin \frac{x}{2}\sin \frac{3x}{2} \right)}\mathrm{ d}x$.

The diagram shows the graphs of $y={{\mathrm{e}}^{2x}}$ and $y={{\left( x+1 \right)}^{2}}$.

$R$ is the finite region bounded by the two curves $y={{\mathrm{e}}^{2x}}$ and $y={{\left( x+1 \right)}^{2}}$.

Find the volume of the solid formed when $R$ is rotated through $2\pi $ radians about the $y$-axis, giving your answer correct to $4$ decimal places.

$x=2\left( \sin \frac{t}{2}-\cos \frac{t}{2} \right)$, $y=\sin \frac{t}{2}+\cos \frac{t}{2}$, for $-\frac{3\pi }{2}\le t\le -\frac{\pi }{2}$.

The line $L$ with equation $y=-2x-5$ meets the curve $C$ at the point $\left( -2,\,-1 \right)$. $S$ is the region enclosed by line $L$, curve $C$and the $x$-axis as shown in the diagram below.

Default solution

$\displaystyle\int\left(\cot^{6}2x+\cot^{4}2x-\sin\dfrac{x}{2}\sin\dfrac{3x}{2}\right)\mathrm{d}x$

$= \displaystyle\int\cot^{4}2x(\cot^{2}2x+1)\,\mathrm{d}x-\int\sin\dfrac{x}{2}\sin\dfrac{3x}{2}\,\mathrm{d}x$

Note: $\cot^{2}2x+1=\csc^{2}2x$ and $\sin\dfrac{x}{2}\sin\dfrac{3x}{2}=-\dfrac{1}{2}[\cos 2x-\cos x]$

$= \displaystyle\int(\cot 2x)^{4}\csc^{2}2x\,\mathrm{d}x+\dfrac{1}{2}\int(\cos 2x-\cos x)\,\mathrm{d}x$

Let $u=\cot 2x$, $\mathrm{d}u=-2\csc^{2}2x\,\mathrm{d}x$:

$= -\dfrac{1}{2}\displaystyle\int u^{4}\,\mathrm{d}u+\dfrac{1}{2}\left[\dfrac{\sin 2x}{2}-\sin x\right]+C$

$= -\dfrac{1}{10}\cot^{5}2x+\dfrac{1}{4}\sin 2x-\dfrac{1}{2}\sin x+C$

$y_{1}=(x+1)^{2} \Rightarrow x_{1}=-1-\sqrt{y}$ (left branch, $x<-1$)

$y_{2}=\mathrm{e}^{2x} \Rightarrow x_{2}=\dfrac{1}{2}\ln y$

Intersection at $(-1.2785,\,0.07754)$

Required volume:

$= \pi\displaystyle\int_{0.07754}^{1}\left(\dfrac{1}{2}\ln y\right)^{2}\mathrm{d}y+\pi\int_{0}^{0.07754}(-1-\sqrt{y})^{2}\,\mathrm{d}y-\pi\int_{0}^{1}(-1+\sqrt{y})^{2}\,\mathrm{d}y$

$= 0.55928\ldots$

$= 0.5593$ units$^{3}$ (4 d.p.)

Primary Solution

Area bounded by 2 curves

Applications of Integration

H2 Math

2020 ASRJC P2 Q4

At $x$-intercept, $y=0$:

$\sin\dfrac{t}{2}+\cos\dfrac{t}{2}=0$

$\tan\dfrac{t}{2}=-1$

Basic angle $\alpha=\dfrac{\pi}{4}$. Since $-\dfrac{3\pi}{4}\leq\dfrac{t}{2}\leq-\dfrac{\pi}{4}$:

$\dfrac{t}{2}=-\dfrac{\pi}{4}$, so $t=-\dfrac{\pi}{2}$

$x=2\left(\sin\left(-\dfrac{\pi}{4}\right)-\cos\left(-\dfrac{\pi}{4}\right)\right)=2\left(-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}\right)=-\dfrac{4}{\sqrt{2}}=-2\sqrt{2}$ (shown)

Area of $S$ = Area under curve from $x=-2\sqrt{2}$ to $x=-2$ minus triangle area.

At $(-2,-1)$: $t=-\pi$. At $(-2\sqrt{2},0)$: $t=-\dfrac{\pi}{2}$.

$\dfrac{\mathrm{d}x}{\mathrm{d}t}=\cos\dfrac{t}{2}+\sin\dfrac{t}{2}$

Triangle (vertices $(-2\sqrt{2},0),(-2,0),(-2,-1)$): area $=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot 1$

Wait actually vertices at $(-5/2,0), (-2,0), (-2,-1)$... The triangle is $\frac{1}{2}(\frac{1}{2})(1)=\frac{1}{4}$

Parametric area:

$\left|\displaystyle\int_{-\pi}^{-\frac{\pi}{2}}\left(\sin\dfrac{t}{2}+\cos\dfrac{t}{2}\right)\left(\cos\dfrac{t}{2}+\sin\dfrac{t}{2}\right)\mathrm{d}t\right|$

$= \displaystyle\int_{-\pi}^{-\frac{\pi}{2}}\left(\sin^{2}\dfrac{t}{2}+2\sin\dfrac{t}{2}\cos\dfrac{t}{2}+\cos^{2}\dfrac{t}{2}\right)\mathrm{d}t$

$= \displaystyle\int_{-\pi}^{-\frac{\pi}{2}}(1+\sin t)\,\mathrm{d}t$

$= \left[t-\cos t\right]_{-\pi}^{-\frac{\pi}{2}}$

$= \left(-\dfrac{\pi}{2}-\cos\left(-\dfrac{\pi}{2}\right)\right)-\left(-\pi-\cos(-\pi)\right)$

$= \left(-\dfrac{\pi}{2}-0\right)-(-\pi+1)$

$= \dfrac{\pi}{2}-1$

Area of $S = \dfrac{\pi}{2}-1-\dfrac{1}{4}=\dfrac{\pi}{2}-\dfrac{5}{4}$ units$^{2}$

2020 ASRJC P2 Q4

2020 ASRJC P2 Q4

Video Solution:

Default solution