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2018
P2 Q1
Topical
Integration
18 P2 Q1
2018 H2 Mathematics Paper 2 Question 1
Integration Techniques
Answers
(i)
3
(
2
9
x
+
4
)
3
2
+
45
{3 \left( \frac{2}{9} x + 4 \right)^{\frac{3}{2}} + 45}
3
(
9
2
x
+
4
)
2
3
+
45
(ii)
(
54
,
237
)
{(54, 237)}
(
54
,
237
)
Full solutions
(i)
d
y
d
x
=
(
1
3
y
−
15
)
1
3
(
1
3
y
−
15
)
−
1
3
d
y
d
x
=
1
\begin{gather*} \frac{\mathrm{d}y}{\mathrm{d}x} = \left( \frac{1}{3} y - 15 \right)^{\frac{1}{3}} \\ \left( \frac{1}{3} y - 15 \right)^{- \frac{1}{3}} \frac{\mathrm{d}y}{\mathrm{d}x} = 1 \end{gather*}
d
x
d
y
=
(
3
1
y
−
15
)
3
1
(
3
1
y
−
15
)
−
3
1
d
x
d
y
=
1
∫
(
1
3
y
−
15
)
−
1
3
d
y
=
∫
1
d
x
(
1
3
y
−
15
)
2
3
2
3
(
1
3
)
=
x
+
C
9
2
(
1
3
y
−
15
)
2
3
=
x
+
C
′
(
1
3
y
−
15
)
2
3
=
2
9
x
+
C
\begin{gather*} \int \left( \frac{1}{3} y - 15 \right)^{- \frac{1}{3}} \; \mathrm{d}y = \int 1 \; \mathrm{d}x \\ \frac{\left( \frac{1}{3} y - 15 \right)^{\frac{2}{3}}}{\frac{2}{3}\left( \frac{1}{3} \right)} = x + C \\ \frac{9}{2} \left(\frac{1}{3} y - 15\right)^{\frac{2}{3}} = x + C' \\ \left( \frac{1}{3} y - 15 \right)^{\frac{2}{3}} = \frac{2}{9}x + C \end{gather*}
∫
(
3
1
y
−
15
)
−
3
1
d
y
=
∫
1
d
x
3
2
(
3
1
)
(
3
1
y
−
15
)
3
2
=
x
+
C
2
9
(
3
1
y
−
15
)
3
2
=
x
+
C
′
(
3
1
y
−
15
)
3
2
=
9
2
x
+
C
When
x
=
0
,
y
=
69
{x=0, y=69}
x
=
0
,
y
=
69
0
+
C
=
(
1
3
(
69
)
−
15
)
2
3
C
=
4
\begin{align*} 0 + C &= \left( \frac{1}{3}\left( 69 \right) - 15 \right)^{\frac{2}{3}} \\ C &= 4 \end{align*}
0
+
C
C
=
(
3
1
(
69
)
−
15
)
3
2
=
4
1
3
y
−
15
=
(
2
9
x
+
4
)
3
2
1
3
y
=
(
2
9
x
+
4
)
3
2
+
15
f
(
x
)
=
3
(
2
9
x
+
4
)
3
2
+
45
■
\begin{gather*} \frac{1}{3} y - 15 = \left( \frac{2}{9} x + 4 \right)^{\frac{3}{2}} \\ \frac{1}{3} y = \left( \frac{2}{9} x + 4 \right)^{\frac{3}{2}} + 15 \\ f(x) = 3 \left( \frac{2}{9} x + 4 \right)^{\frac{3}{2}} + 45 \; \blacksquare \end{gather*}
3
1
y
−
15
=
(
9
2
x
+
4
)
2
3
3
1
y
=
(
9
2
x
+
4
)
2
3
+
15
f
(
x
)
=
3
(
9
2
x
+
4
)
2
3
+
45
■
(ii)
When the gradient is
4
,
{4,}
4
,
(
1
3
y
−
15
)
1
3
=
4
1
3
y
−
15
=
4
3
y
=
237
\begin{gather*} \left( \frac{1}{3} y - 15 \right)^{\frac{1}{3}} = 4 \\ \frac{1}{3} y - 15 = 4^3 \\ y = 237 \end{gather*}
(
3
1
y
−
15
)
3
1
=
4
3
1
y
−
15
=
4
3
y
=
237
2
9
x
+
4
=
(
1
3
(
237
)
−
15
)
2
3
2
9
x
=
16
−
4
x
=
54
\begin{align*} \frac{2}{9} x + 4 &= \left( \frac{1}{3} (237) - 15 \right)^{\frac{2}{3}} \\ \frac{2}{9}x &= 16 - 4 \\ x &= 54 \end{align*}
9
2
x
+
4
9
2
x
x
=
(
3
1
(
237
)
−
15
)
3
2
=
16
−
4
=
54
Coordinates of the point on the curve where the gradient is
4
:
{4:}
4
:
(
54
,
237
)
■
(54, 237) \; \blacksquare
(
54
,
237
)
■
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