Use the substitution u=1+tu = 1 + \sqrt{t}u=1+t to show that the integral
∫12t1+t dt \int \frac{12\sqrt{t}}{1+\sqrt{t}} \, dt ∫1+t12tdtcan be written in the form
∫(24u−48+24u) du \int \left( 24u - 48 + \frac{24}{u} \right) \, du ∫(24u−48+u24)duThe mass of a fungal colony, mmm grams, grows at a rate modelled by the equation
dmdt=12t1+t \frac{dm}{dt} = \frac{12\sqrt{t}}{1+\sqrt{t}} dtdm=1+t12twhere ttt is the number of days since the colony was first observed, for 1≤t≤91 \le t \le 91≤t≤9.
Determine the total increase in the mass of the colony from the end of day 1 to the end of day 9. Show each stage of your working and give your answer to one decimal place.
363 exam-style questions on CCEA A Level Maths 3.6 Integration (A-level only), covering 3.6.1 Integration (A-level only), 3.6.2 Integration (A-level only), 3.6.3 Integration (A-level only), 3.6.4 Integration (A-level only), 3.6.5 Integration (A-level only), 3.6.6 Integration (A-level only), and 3.6.7 Integration (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.