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2019
P1 Q2
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Areas & Volumes
19 P1 Q2
2019 H2 Mathematics Paper 1 Question 2
Definite Integrals: Areas and Volumes
Answers
(i)
a
=
0.682
{a=0.682}
a
=
0.682
(ii)
b
=
1.892
{b=1.892}
b
=
1.892
Full solutions
(i)
Using a GC,
a
=
0.68233
=
0.682
(3 dp)
■
\begin{align*} a &= 0.68233 \\ &= 0.682 \textrm{ (3 dp)} \; \blacksquare \end{align*}
a
=
0.68233
=
0.682
(3 dp)
■
(ii)
∫
0.68233
b
(
x
3
+
x
−
1
)
d
x
=
−
2
∫
−
1
0
(
x
3
+
x
−
1
)
d
x
[
1
4
x
4
+
1
2
x
2
−
x
]
0.68233
b
=
3.5
1
4
b
4
+
1
2
b
2
−
b
−
(
−
0.39535
)
=
3.5
1
4
b
4
+
1
2
b
2
−
b
−
3.1046
=
0
\begin{gather*} \int_{0.68233}^b (x^3 + x - 1) \; \mathrm{d}x = -2\int_{-1}^0 (x^3 + x - 1) \; \mathrm{d}x \\ \left[ \frac{1}{4} x^4 + \frac{1}{2} x^2 - x \right]_{0.68233}^b = 3.5 \\ \frac{1}{4} b^4 + \frac{1}{2} b^2 - b - (-0.39535) = 3.5 \\ \frac{1}{4} b^4 + \frac{1}{2} b^2 - b -3.1046 = 0 \end{gather*}
∫
0.68233
b
(
x
3
+
x
−
1
)
d
x
=
−
2
∫
−
1
0
(
x
3
+
x
−
1
)
d
x
[
4
1
x
4
+
2
1
x
2
−
x
]
0.68233
b
=
3.5
4
1
b
4
+
2
1
b
2
−
b
−
(
−
0.39535
)
=
3.5
4
1
b
4
+
2
1
b
2
−
b
−
3.1046
=
0
Using a GC,
b
=
1.892
(3 dp)
■
b=1.892 \textrm{ (3 dp)} \; \blacksquare
b
=
1.892
(3 dp)
■
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