Four economists, Amara, Biruk, Chen, and Dalia, are attempting to determine the indefinite integral representing a total utility function:
∫1v dv \int \frac{1}{v} \, dv ∫v1dvfor v>0v > 0v>0. Each economist proposes a different general form for the result:
Amara: ∫1v dv=lnv\int \frac{1}{v} \, dv = \ln v∫v1dv=lnv
Biruk: ∫1v dv=Alnv\int \frac{1}{v} \, dv = A \ln v∫v1dv=Alnv
Chen: ∫1v dv=ln(kv)\int \frac{1}{v} \, dv = \ln(kv)∫v1dv=ln(kv)
Dalia: ∫1v dv=lnv+C\int \frac{1}{v} \, dv = \ln v + C∫v1dv=lnv+C
(i) Explain why Amara's result is incomplete.
(ii) Explain why Biruk's answer is incorrect as a general anti-derivative of 1v\frac{1}{v}v1.
Using the properties of logarithms, demonstrate why Chen and Dalia's forms can be considered equivalent for certain values of the constants kkk and CCC.
Practise AQA A Level Maths 1.11 H: Integration with exam-style questions for A Level Maths. 436 questions covering 1.11.1 Fundamental Theorem of Calculus, 1.11.2 Integrating standard functions, 1.11.3 Definite integrals and areas, 1.11.4 Integration as the limit of a sum (A-level only), 1.11.5 Integration by substitution and by parts (A-level only), 1.11.6 Integration using partial fractions (A-level only), 1.11.7 Differential equations with separable variables (A-level only), and 1.11.8 Interpreting solutions of differential equations (A-level only), matched to the AQA A Level Maths (7357) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.
