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2007
P2 Q4
Topical
Areas & Volumes
07 P2 Q4
2007 H2 Mathematics Paper 2 Question 4
Definite Integrals: Areas and Volumes
Answers
(i)
∫
0
5
π
3
sin
2
x
d
x
=
5
6
π
+
1
8
3
{\displaystyle \int_0^{\frac{5 \pi}{3}} \sin^2 x \; \mathrm{d}x} \allowbreak {\displaystyle = \frac{5}{6} \pi + \frac{1}{8} \sqrt{3}}
∫
0
3
5
π
sin
2
x
d
x
=
6
5
π
+
8
1
3
∫
0
5
π
3
cos
2
x
d
x
=
5
6
π
−
1
8
3
{\displaystyle \int_0^{\frac{5 \pi}{3}} \cos^2 x \; \mathrm{d}x} \allowbreak {\displaystyle = \frac{5}{6} \pi - \frac{1}{8} \sqrt{3}}
∫
0
3
5
π
cos
2
x
d
x
=
6
5
π
−
8
1
3
(iia)
(
π
−
2
)
units
2
{(\pi - 2) \textrm{ units}^2}
(
π
−
2
)
units
2
(iib)
5.391
units
3
{5.391 \textrm{ units}^3}
5.391
units
3
Full solutions
(i)
=
∫
0
5
π
3
sin
2
x
d
x
=
∫
0
5
π
3
1
−
cos
2
x
2
d
x
=
[
x
2
−
sin
2
x
4
]
0
5
π
3
=
1
2
(
5
π
3
)
−
1
4
(
sin
10
π
3
−
sin
0
)
=
5
6
π
+
1
8
3
■
\begin{align*} & = \int_0^\frac{5 \pi}{3} \sin^2 x \; \mathrm{d}x \\ & = \int_0^\frac{5 \pi}{3} \frac{1-\cos 2x}{2} \; \mathrm{d}x \\ & = \left[ \frac{x}{2} - \frac{\sin 2x}{4} \right]_0^{\frac{5 \pi}{3}}\\ & = \frac{1}{2} \left( \frac{5 \pi}{3} \right) - \frac{1}{4} \left( \sin \frac{10\pi}{3} - \sin 0 \right) \\ & = \frac{5}{6} \pi + \frac{1}{8} \sqrt{3} \; \blacksquare \end{align*}
=
∫
0
3
5
π
sin
2
x
d
x
=
∫
0
3
5
π
2
1
−
cos
2
x
d
x
=
[
2
x
−
4
sin
2
x
]
0
3
5
π
=
2
1
(
3
5
π
)
−
4
1
(
sin
3
10
π
−
sin
0
)
=
6
5
π
+
8
1
3
■
=
∫
0
5
π
3
cos
2
x
d
x
=
∫
0
5
π
3
1
−
sin
2
x
d
x
=
[
x
]
0
5
π
3
−
(
5
6
π
+
1
8
3
)
=
5
π
3
−
(
5
6
π
+
1
8
3
)
=
5
6
π
−
1
8
3
■
\begin{align*} & = \int_0^\frac{5 \pi}{3} \cos^2 x \; \mathrm{d}x \\ & = \int_0^\frac{5 \pi}{3} 1 - \sin^2 x \; \mathrm{d}x \\ & = \Big[ x \Big]_0^\frac{5 \pi}{3} - \left( \frac{5}{6} \pi + \frac{1}{8} \sqrt{3} \right) \\ & = \frac{5 \pi}{3} - \left( \frac{5}{6} \pi + \frac{1}{8} \sqrt{3} \right) \\ & = \frac{5}{6} \pi - \frac{1}{8} \sqrt{3} \; \blacksquare \end{align*}
=
∫
0
3
5
π
cos
2
x
d
x
=
∫
0
3
5
π
1
−
sin
2
x
d
x
=
[
x
]
0
3
5
π
−
(
6
5
π
+
8
1
3
)
=
3
5
π
−
(
6
5
π
+
8
1
3
)
=
6
5
π
−
8
1
3
■
(iia)
∫
0
1
2
π
x
2
sin
x
d
x
=
[
x
2
(
−
cos
x
)
]
0
1
2
π
−
∫
0
1
2
π
2
x
(
−
cos
x
)
d
x
=
0
−
0
+
∫
0
1
2
π
2
x
cos
x
d
x
=
[
2
x
sin
x
]
0
1
2
π
−
∫
0
1
2
π
2
sin
x
d
x
=
2
(
1
2
π
)
−
0
+
[
2
cos
x
]
0
1
2
π
=
π
+
2
cos
1
2
π
−
2
cos
0
=
(
π
−
2
)
units
2
■
\begin{align*} & \int_0^{\frac{1}{2}\pi} x^{2} \sin x \; \mathrm{d}x \\ & = \Big[ x^2 (- \cos x) \Big]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} 2 x (- \cos x) \; \mathrm{d}x \\ & = 0 - 0 + \int_0^{\frac{1}{2}\pi} 2 x \cos x \; \mathrm{d}x \\ & = \Big[ 2x \sin x \Big]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} 2 \sin x \; \mathrm{d}x \\ & = 2 \left( \frac{1}{2}\pi \right) - 0 + \Big[ 2 \cos x \Big]_0^{\frac{1}{2}\pi} \\ & = \pi + 2 \cos \frac{1}{2}\pi - 2\cos 0 \\ & = (\pi - 2) \textrm{ units}^2 \; \blacksquare \\ \end{align*}
∫
0
2
1
π
x
2
sin
x
d
x
=
[
x
2
(
−
cos
x
)
]
0
2
1
π
−
∫
0
2
1
π
2
x
(
−
cos
x
)
d
x
=
0
−
0
+
∫
0
2
1
π
2
x
cos
x
d
x
=
[
2
x
sin
x
]
0
2
1
π
−
∫
0
2
1
π
2
sin
x
d
x
=
2
(
2
1
π
)
−
0
+
[
2
cos
x
]
0
2
1
π
=
π
+
2
cos
2
1
π
−
2
cos
0
=
(
π
−
2
)
units
2
■
(iib)
Volume of revolution
=
π
∫
0
1
2
π
(
x
2
sin
x
)
2
d
x
=
5.391
units
3
(3 dp)
■
\begin{align*} & \textrm{Volume of revolution} \\ & = \pi \int_0^{\frac{1}{2}\pi} \left( x^2 \sin x \right)^2 \; \mathrm{d}x \\ & = 5.391 \textrm{ units}^3 \textrm{ (3 dp)} \; \blacksquare \\ \end{align*}
Volume of revolution
=
π
∫
0
2
1
π
(
x
2
sin
x
)
2
d
x
=
5.391
units
3
(3 dp)
■
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