Math Repository
about
topic
al
year
ly
Yearly
2021
P2 Q2
Topical
Areas & Volumes
21 P2 Q2
2021 H2 Mathematics Paper 2 Question 2
Definite Integrals: Areas and Volumes
Answers
(a)
h
=
4
5
{h=\frac{4}{5}}
h
=
5
4
(b)
4
5
∑
n
=
1
5
(
f
(
1
+
4
5
n
)
)
{{\displaystyle \frac{4}{5}} {\displaystyle \sum_{n=1}^5} \left( f\left( 1 + \frac{4}{5}n \right) \right) }
5
4
n
=
1
∑
5
(
f
(
1
+
5
4
n
)
)
(c)
Lower bound
=
701
125
≈
5.61
{= \frac{701}{125} \approx 5.61}
=
125
701
≈
5.61
Upper bound
=
821
125
≈
6.57
{= \frac{821}{125} \approx 6.57}
=
125
821
≈
6.57
(d)
Full solutions
(a)
h
=
4
5
■
h=\frac{4}{5} \; \blacksquare
h
=
5
4
■
The area of the rectangles,
∑
n
=
0
4
(
f
(
1
+
n
h
)
)
h
{\displaystyle \sum_{n=0}^4 \left( f(1+nh) \right)h}
n
=
0
∑
4
(
f
(
1
+
nh
)
)
h
is smaller than the area under graph,
A
.
{A.}
A
.
(b)
4
5
∑
n
=
1
5
(
f
(
1
+
4
5
n
)
)
{\displaystyle \frac{4}{5}} {\displaystyle \sum_{n=1}^5} \left( f\left( 1 + \frac{4}{5}n \right) \right)
5
4
n
=
1
∑
5
(
f
(
1
+
5
4
n
)
)
(c)
Lower bound
=
4
5
∑
n
=
0
4
(
f
(
1
+
4
5
n
)
)
=
4
5
(
f
(
1
)
+
f
(
9
5
)
+
f
(
13
5
)
+
f
(
17
5
)
+
f
(
21
5
)
)
=
4
5
(
21
20
+
581
500
+
669
500
+
789
500
+
941
500
)
=
701
125
■
\begin{align*} & \textrm{Lower bound} \\ & = {\displaystyle \frac{4}{5}} {\displaystyle \sum_{n=0}^4} \left( f\left( 1 + \frac{4}{5}n \right) \right) \\ & = \frac{4}{5} \left( f(1) + f\left( \frac{9}{5} \right) + f\left( \frac{13}{5} \right) + f\left( \frac{17}{5} \right) + f\left( \frac{21}{5} \right) \right) \\ & = \frac{4}{5} \left( \frac{21}{20} + \frac{581}{500} + \frac{669}{500} + \frac{789}{500} + \frac{941}{500} \right) \\ & = \frac{701}{125} \; \blacksquare \end{align*}
Lower bound
=
5
4
n
=
0
∑
4
(
f
(
1
+
5
4
n
)
)
=
5
4
(
f
(
1
)
+
f
(
5
9
)
+
f
(
5
13
)
+
f
(
5
17
)
+
f
(
5
21
)
)
=
5
4
(
20
21
+
500
581
+
500
669
+
500
789
+
500
941
)
=
125
701
■
Upper bound
=
4
5
∑
n
=
1
5
(
f
(
1
+
4
5
n
)
)
=
4
5
(
f
(
9
5
)
+
f
(
13
5
)
+
f
(
17
5
)
+
f
(
21
5
)
+
f
(
5
)
)
=
4
5
(
581
500
+
669
500
+
789
500
+
941
500
+
9
4
)
=
821
125
■
\begin{align*} & \textrm{Upper bound} \\ & = {\displaystyle \frac{4}{5}} {\displaystyle \sum_{n=1}^5} \left( f\left( 1 + \frac{4}{5}n \right) \right) \\ & = \frac{4}{5} \left( f\left( \frac{9}{5} \right) + f\left( \frac{13}{5} \right) + f\left( \frac{17}{5} \right) + f\left( \frac{21}{5} \right) + f\left( 5 \right) \right) \\ & = \frac{4}{5} \left( \frac{581}{500} + \frac{669}{500} + \frac{789}{500} + \frac{941}{500} + \frac{9}{4} \right) \\ & = \frac{821}{125} \; \blacksquare \end{align*}
Upper bound
=
5
4
n
=
1
∑
5
(
f
(
1
+
5
4
n
)
)
=
5
4
(
f
(
5
9
)
+
f
(
5
13
)
+
f
(
5
17
)
+
f
(
5
21
)
+
f
(
5
)
)
=
5
4
(
500
581
+
500
669
+
500
789
+
500
941
+
4
9
)
=
125
821
■
(d)
Back to top ▲