The rate of increase of a substance's temperature, T T\,T in degrees Celsius, over time t t\,t minutes is modeled by the equation
dTdt=52t+3,t≥0 \frac{dT}{dt} = \frac{5}{2t + 3}, \quad t \ge 0 dtdT=2t+35,t≥0Calculate the exact change in temperature between t=1t = 1t=1 and t=6t = 6t=6 minutes, giving your answer in its simplest form.
g(x)=2x3−11x2−8x+87(x−4)2g(x) = \dfrac{2x^3 - 11x^2 - 8x + 87}{(x - 4)^2}g(x)=(x−4)22x3−11x2−8x+87 for x>4x > 4x>4.
Given that g(x)=Ax+B+C(x−4)2g(x) = Ax + B + \dfrac{C}{(x - 4)^2}g(x)=Ax+B+(x−4)2C where AAA, B B\,B and C C\,C are constants to be determined, find
∫g(x) dx \int g(x) \, dx ∫g(x)dx438 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.