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.
438 exam-style questions on OCR A Level Maths 1.8 Integration, covering 1.8.1 Fundamental theorem of calculus (A-level only), 1.8.2 Integrating x^n, 1.8.3 Integrating standard functions (A-level only), 1.8.4 Evaluating definite integrals, 1.8.5 Area between a curve and the x-axis, 1.8.6 Area between two curves, 1.8.7 Integration as the limit of a sum (A-level only), 1.8.8 Integration by substitution (A-level only), 1.8.9 Integration by parts (A-level only), 1.8.10 Use of partial fractions in integration (A-level only), 1.8.11 Differential equations with separable variables (A-level only), 1.8.12 Interpreting the solution of a differential equation (A-level only), and 1.8 Integration. Each one has a worked solution and a mark scheme showing where the marks go.