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2011
P1 Q5
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Areas & Volumes
11 P1 Q5
2011 H2 Mathematics Paper 1 Question 5
Definite Integrals: Areas and Volumes
Answers
(i)
(ii)
[
0
,
2
]
{\left[ 0, 2 \right]}
[
0
,
2
]
(iii)
a
=
2
+
5
{a = 2 + \sqrt{5}}
a
=
2
+
5
Full solutions
(i)
(ii)
From the graphs, set of values of
x
{x}
x
for which
f
(
∣
x
∣
)
=
∣
f
(
x
)
∣
:
{f\left(\left|x\right|\right)=\left|f(x)\right|:}
f
(
∣
x
∣
)
=
∣
f
(
x
)
∣
:
[
0
,
2
]
■
\left[ 0, 2 \right] \; \blacksquare
[
0
,
2
]
■
(iii)
∫
−
1
1
f
(
∣
x
∣
)
d
x
=
∫
1
a
∣
f
(
x
)
∣
d
x
2
∫
0
1
f
(
x
)
d
x
=
∫
1
2
f
(
x
)
d
x
+
∫
2
a
−
f
(
x
)
d
x
2
∫
0
1
2
−
x
d
x
=
∫
1
2
2
−
x
d
x
+
∫
2
a
x
−
2
d
x
2
[
2
x
−
1
2
x
2
]
0
1
=
[
2
x
−
1
2
x
2
]
1
2
+
[
1
2
x
2
−
2
x
]
2
a
2
(
3
2
)
=
4
−
1
2
(
4
)
−
2
+
1
2
+
1
2
a
2
−
2
a
−
1
2
(
4
)
+
4
\begin{gather*} \int_{-1}^1 f\left(\left|x\right|\right) \; \mathrm{d}x = \int_1^a \left|f(x)\right| \; \mathrm{d}x \\ 2 \int_{0}^1 f(x) \; \mathrm{d}x = \int_1^2 f(x) \; \mathrm{d}x + \int_2^a -f(x) \; \mathrm{d}x \\ 2 \int_{0}^1 2-x \; \mathrm{d}x = \int_1^2 2-x \; \mathrm{d}x + \int_2^a x-2 \; \mathrm{d}x \\ 2 \left[ 2x - \frac{1}{2}x^2 \right]_{0}^1 = \left[ 2x - \frac{1}{2}x^2 \right]_{1}^2 + \left[ \frac{1}{2}x^2 - 2x \right]_{2}^a \\ 2 \left( \frac{3}{2} \right) = 4 - \frac{1}{2} (4) - 2 + \frac{1}{2} + \frac{1}{2}a^2 - 2a -\frac{1}{2}(4) + 4 \\ \end{gather*}
∫
−
1
1
f
(
∣
x
∣
)
d
x
=
∫
1
a
∣
f
(
x
)
∣
d
x
2
∫
0
1
f
(
x
)
d
x
=
∫
1
2
f
(
x
)
d
x
+
∫
2
a
−
f
(
x
)
d
x
2
∫
0
1
2
−
x
d
x
=
∫
1
2
2
−
x
d
x
+
∫
2
a
x
−
2
d
x
2
[
2
x
−
2
1
x
2
]
0
1
=
[
2
x
−
2
1
x
2
]
1
2
+
[
2
1
x
2
−
2
x
]
2
a
2
(
2
3
)
=
4
−
2
1
(
4
)
−
2
+
2
1
+
2
1
a
2
−
2
a
−
2
1
(
4
)
+
4
1
2
a
2
−
2
a
+
5
2
=
3
a
2
−
4
a
−
1
=
0
\begin{align*} \frac{1}{2}a^2 - 2a + \frac{5}{2} &= 3 \\ a^2 -4a - 1 &= 0 \end{align*}
2
1
a
2
−
2
a
+
2
5
a
2
−
4
a
−
1
=
3
=
0
a
=
4
±
4
2
−
4
(
−
1
)
2
=
4
±
20
2
=
4
±
2
5
2
=
2
±
5
\begin{align*} a &= \frac{4 \pm \sqrt{4^2-4(-1)}}{2} \\ &= \frac{4 \pm \sqrt{20}}{2} \\ &= \frac{4 \pm 2 \sqrt{5}}{2} \\ &= 2 \pm \sqrt{5} \end{align*}
a
=
2
4
±
4
2
−
4
(
−
1
)
=
2
4
±
20
=
2
4
±
2
5
=
2
±
5
From the graph, since
a
>
2
,
{a > 2,}
a
>
2
,
a
=
2
+
5
■
a = 2 + \sqrt{5} \; \blacksquare
a
=
2
+
5
■
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