The curve C C\,C has equation
y=184(2x−k),x≠k2 y = \frac{18}{4(2x - k)}, \quad x \not= \frac{k}{2} y=4(2x−k)18,x=2kwhere k k\,k is a positive constant and k≠2k \neq 2k=2.
Find dydx\displaystyle \frac{dy}{dx}dxdy giving your answer in simplest form in terms of kkk.
The point P P\,P with x x\,x coordinate 1 lies on CCC. Given that the gradient of the curve at P P\,P is -9, find the two possible values of kkk.
Given also that k<2k < 2k<2, find the equation of the normal to C C\,C at PPP, writing your answer in the form ax+by+c=0ax + by + c = 0ax+by+c=0, where a,b a, b\,a,b and c c\,c are integers to be found.
Show, using algebraic integration, that
∫13184(2x−k) dx=λln(5) \int_{1}^{3} \frac{18}{4(2x - k)} \, dx = \lambda \ln(5) ∫134(2x−k)18dx=λln(5)where λ \lambda\,λ is a constant to be found.
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.