Mathematics III (Engineering)
Assessment 4 Answers
Year 2026
0027 68 628 1800
,
, QUESTION 1
𝑓(𝑡) = 𝑡 2 − 𝑡, (−𝜋 < 𝑡 < 𝜋 ), 𝑓(𝑡) = 𝑡 + 2𝜋
1 𝜋
𝑎0 = ∫ 𝑓(𝑡)𝑑𝑡
𝜋 −𝜋
1 𝜋 2
= ∫ (𝑡 − 𝑡)𝑑𝑡
𝜋 −𝜋
𝜋
1 𝑡3 𝑡2
= [ − ]
𝜋 3 2 −𝜋
1 𝜋3 𝜋2 (−𝜋)3 (−𝜋)2
= [( − ) − ( − )]
𝜋 3 2 3 2
1 𝜋3 𝜋2 𝜋3 𝜋2
= [ − + + ]
𝜋 3 2 3 2
1 2𝜋 3
= .
𝜋 3
2𝜋 2
𝑎0 =
3
……………………………………………………………………………………………………………
Assessment 4 Answers
Year 2026
0027 68 628 1800
,
, QUESTION 1
𝑓(𝑡) = 𝑡 2 − 𝑡, (−𝜋 < 𝑡 < 𝜋 ), 𝑓(𝑡) = 𝑡 + 2𝜋
1 𝜋
𝑎0 = ∫ 𝑓(𝑡)𝑑𝑡
𝜋 −𝜋
1 𝜋 2
= ∫ (𝑡 − 𝑡)𝑑𝑡
𝜋 −𝜋
𝜋
1 𝑡3 𝑡2
= [ − ]
𝜋 3 2 −𝜋
1 𝜋3 𝜋2 (−𝜋)3 (−𝜋)2
= [( − ) − ( − )]
𝜋 3 2 3 2
1 𝜋3 𝜋2 𝜋3 𝜋2
= [ − + + ]
𝜋 3 2 3 2
1 2𝜋 3
= .
𝜋 3
2𝜋 2
𝑎0 =
3
……………………………………………………………………………………………………………