<p>The region bounded by the axes and the curve $y=\cos x$ from $x=0$ to $x=\frac{1}{2}\pi $ is divided into two parts, of areas ${{A}_{1}}$ and ${{A}_{2}}$, by the curve $y=\sin x$ (see diagram). Prove that ${{A}_{2}}=\left( \sqrt{2} \right){{A}_{1}}$.</p><p style="text-align: right;">[6]</p>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/pasted/1762623426071-w10ut3dj6z.png" width="400" class="aligncenter" /></figure>
<p>The two curves meet at $P$. The line through $P$parallel to the $x$-axis meets the $y$-axis at $Q$. The region $OPQ$, bounded by the arc $OP$ and the lines $PQ$ and $QO$, is rotated through $4$ right angles about the $y$-axis to form a solid of revolution of volume $V$. It is given that</p>
<p style="text-align: center">$C=\pi \displaystyle\int\nolimits_{0}^{\frac{1}{\sqrt{2}}}{{{\left( {{\sin }^{-1}}y \right)}^{2}}}\mathrm{d}y$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">By substituting $u={{\sin }^{-1}}y$, show that $V=\pi \displaystyle\int\nolimits_{0}^{\frac{1}{4}\pi }{{{u}^{2}}\cos u\,\mathrm{d}u}$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Show that $\frac{\mathrm{d}}{\mathrm{d}u}\left( {{u}^{2}}\sin u+2u\cos u-2\sin u \right)={{u}^{2}}\cos u$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Hence find the exact value of $V$.</div><p style="text-align: right;">[2]</p></li></ol>

The region bounded by the axes and the curve $y=\cos x$ from $x=0$ to $x=\frac{1}{2}\pi $ is divided into two parts, of areas ${{A}_{1}}$ and ${{A}_{2}}$, by the curve $y=\sin x$ (see diagram). Prove that ${{A}_{2}}=\left( \sqrt{2} \right){{A}_{1}}$.

The two curves meet at $P$. The line through $P$parallel to the $x$-axis meets the $y$-axis at $Q$. The region $OPQ$, bounded by the arc $OP$ and the lines $PQ$ and $QO$, is rotated through $4$ right angles about the $y$-axis to form a solid of revolution of volume $V$. It is given that

$C=\pi \displaystyle\displaystyle \displaystyle\int\nolimits_{0}^{\frac{1}{\sqrt{2}}}{{{\left( {{\sin }^{-1}}y \right)}^{2}}}\mathrm{d}y$.

By substituting $u={{\sin }^{-1}}y$, show that $V=\pi \displaystyle\displaystyle \displaystyle\int\nolimits_{0}^{\frac{1}{4}\pi }{{{u}^{2}}\cos u\,\mathrm{d}u}$.

Show that $\frac{\mathrm{d}}{\mathrm{d}u}\left( {{u}^{2}}\sin u+2u\cos u-2\sin u \right)={{u}^{2}}\cos u$.

Applications of Intergration

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

N2003 P1 Q14

\[A_1 = \displaystyle\int\nolimits_0^{\frac{\pi}{4}} (\cos x - \sin x) \, \mathrm{d}x = \left[ \sin x + \cos x \right]_0^{\frac{\pi}{4}}\]

\[= \left( \sin\frac{\pi}{4} + \cos\frac{\pi}{4} \right) - (\sin 0 + \cos 0)\]

\[= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - (0 + 1) = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1\]

\[A_2 = \displaystyle\int\nolimits_0^{\frac{\pi}{4}} \sin x \, \mathrm{d}x + \displaystyle\int\nolimits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \, \mathrm{d}x\]

\[= \left[ -\cos x \right]_0^{\frac{\pi}{4}} + \left[ \sin x \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}\]

\[= -\left( \cos\frac{\pi}{4} - \cos 0 \right) + \left( \sin\frac{\pi}{2} - \sin\frac{\pi}{4} \right)\]

\[= -\left( \frac{1}{\sqrt{2}} - 1 \right) + \left( 1 - \frac{1}{\sqrt{2}} \right)\]

\[= 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2} = \sqrt{2}(\sqrt{2} - 1)\]

Let $u = \sin^{-1} y$. Differentiating w.r.t. $y$:

\[\frac{\mathrm{d}u}{\mathrm{d}y} = \frac{1}{\sqrt{1-y^2}} \Rightarrow dy = \sqrt{1-y^2} \, du\]

When $y = \frac{1}{\sqrt{2}}$, $u = \sin^{-1}\frac{1}{\sqrt{2}} = \frac{\pi}{4}$. When $y = 0$, $u = 0$.

\[V = \pi \displaystyle\int\nolimits_0^{\frac{1}{\sqrt{2}}} x^2 \, \mathrm{d}y = \pi \displaystyle\int\nolimits_0^{\frac{1}{\sqrt{2}}} (\sin^{-1} y)^2 \, \mathrm{d}y\]

\[= \pi \displaystyle\int\nolimits_0^{\frac{\pi}{4}} u^2 \sqrt{1-y^2} \, du\]

Since $y = \sin u$, $\sqrt{1-y^2} = \sqrt{1-\sin^2 u} = \cos u$

\[V = \pi \displaystyle\int\nolimits_0^{\frac{\pi}{4}} u^2 \cos u \, du\]

\[\frac{\mathrm{d}}{\mathrm{d}u}(u^2 \sin u + 2u \cos u - 2 \sin u)\]

\[= u^2 \cos u + 2u \sin u + 2 \cos u - 2u \sin u - 2 \cos u\]

From (ii): $\displaystyle\int u^2 \cos u \, du = u^2 \sin u + 2u \cos u - 2 \sin u + C$

\[= \pi \left[ u^2 \sin u + 2u \cos u - 2 \sin u \right]_0^{\frac{\pi}{4}}\]

\[= \pi \left[ \left(\frac{\pi}{4}\right)^2 \sin\frac{\pi}{4} + 2\left(\frac{\pi}{4}\right) \cos\frac{\pi}{4} - 2\sin\frac{\pi}{4} - (0 + 0 - 0) \right]\]

\[= \pi \left[ \frac{\pi^2}{16} \cdot \frac{1}{\sqrt{2}} + \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} - 2 \cdot \frac{1}{\sqrt{2}} \right]\]

\[= \frac{\pi}{\sqrt{2}} \left( \frac{\pi^2}{16} + \frac{\pi}{2} - 2 \right) \text{ units}^3\]

N2003 P1 Q14

N2003 P1 Q14

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