Quantitative Modelling 1
DSC1520
Department of Decision Sciences
Assignment 02 for 2021 (compulsory)
Due Date: 25 June 2021
Unique Number: 387523
Instructions
• Work through Chapters 7 and 8 in the study guide before attempting this assignment.
• Answer all the questions.
• Submit your answers electronically through myUnisa.
Question 1
Evaluate
1 1
∫ 𝑥 3 (1 + + ) 𝑑𝑥
𝑥 2 √𝑥 2
3 𝑥3 𝑥3
[1] 𝑥 + + +𝑐
𝑥2 √𝑥 2
[2] 3𝑥 4 + 2𝑥 3 + 𝑥 2 + 𝑐
𝑥4 𝑥3 𝑥2
[3] + + +𝑐
4 3 2
3 𝑥2
[4] 𝑥 + +𝑐
2
Answer:
Refer to Chapter 7.2.1 The power rule for integration, in your study guide.
, 1 1
∫ 𝑥 3 (1 + + ) 𝑑𝑥
𝑥 2 √𝑥 2
1 1
∫ 𝑥 3 (1 + + ) 𝑑𝑥
𝑥2 𝑥
∫ 𝑥 3 + 𝑥 + 𝑥 2 𝑑𝑥
𝑥4 𝑥2 𝑥3
= + + +𝐶
4 3 3
𝑥4 𝑥3 𝑥2
= + + +𝐶
4 3 2
Question 2
Evaluate the definite integral
1
∫0 18𝑒 3𝑥+1 𝑑𝑥 (round to an integer)
[1] 322
[2] 311
[3] 932
[4] 965
Answer:
Refer to chapter 7.3 The definite integral and the area under a curve, in your study guide.
1
∫ 18𝑒 3𝑥+1 𝑑𝑥
0
Since 18 is constant with respect to 𝑥, move 18 out of the integral.
1
18 ∫ 𝑒 3𝑥+1 𝑑𝑥
0
1
Let 𝑢 = 3𝑥 + 1. Then 𝑑𝑢 = 3𝑑𝑥 , so 3 𝑑𝑢 = 𝑑𝑥. Rewrite using 𝑢 and 𝑑𝑢.
𝑑𝑢
Let 𝑢 = 3𝑥 + 1. Find 𝑑𝑥 .
Rewrite:
DSC1520
Department of Decision Sciences
Assignment 02 for 2021 (compulsory)
Due Date: 25 June 2021
Unique Number: 387523
Instructions
• Work through Chapters 7 and 8 in the study guide before attempting this assignment.
• Answer all the questions.
• Submit your answers electronically through myUnisa.
Question 1
Evaluate
1 1
∫ 𝑥 3 (1 + + ) 𝑑𝑥
𝑥 2 √𝑥 2
3 𝑥3 𝑥3
[1] 𝑥 + + +𝑐
𝑥2 √𝑥 2
[2] 3𝑥 4 + 2𝑥 3 + 𝑥 2 + 𝑐
𝑥4 𝑥3 𝑥2
[3] + + +𝑐
4 3 2
3 𝑥2
[4] 𝑥 + +𝑐
2
Answer:
Refer to Chapter 7.2.1 The power rule for integration, in your study guide.
, 1 1
∫ 𝑥 3 (1 + + ) 𝑑𝑥
𝑥 2 √𝑥 2
1 1
∫ 𝑥 3 (1 + + ) 𝑑𝑥
𝑥2 𝑥
∫ 𝑥 3 + 𝑥 + 𝑥 2 𝑑𝑥
𝑥4 𝑥2 𝑥3
= + + +𝐶
4 3 3
𝑥4 𝑥3 𝑥2
= + + +𝐶
4 3 2
Question 2
Evaluate the definite integral
1
∫0 18𝑒 3𝑥+1 𝑑𝑥 (round to an integer)
[1] 322
[2] 311
[3] 932
[4] 965
Answer:
Refer to chapter 7.3 The definite integral and the area under a curve, in your study guide.
1
∫ 18𝑒 3𝑥+1 𝑑𝑥
0
Since 18 is constant with respect to 𝑥, move 18 out of the integral.
1
18 ∫ 𝑒 3𝑥+1 𝑑𝑥
0
1
Let 𝑢 = 3𝑥 + 1. Then 𝑑𝑢 = 3𝑑𝑥 , so 3 𝑑𝑢 = 𝑑𝑥. Rewrite using 𝑢 and 𝑑𝑢.
𝑑𝑢
Let 𝑢 = 3𝑥 + 1. Find 𝑑𝑥 .
Rewrite:
