A research probe is submerged in a fluid where the vertical force, F F\,F kilonewtons, exerted on its hull is modeled by the function
F(d)=54d2+4d−26,d>0 F(d) = \frac{54}{d^2} + 4d - 26, \quad d > 0 F(d)=d254+4d−26,d>0where d d\,d is the depth in metres below the surface.
Using calculus,
determine the range of depths for which the vertical force F(d)F(d)F(d) is increasing.
show that ∫39(54d2+4d−26)dd=0\displaystyle \int_{3}^{9} \left( \frac{54}{d^2} + 4d - 26 \right) dd = 0∫39(d254+4d−26)dd=0.
The points A(3,−8)A(3, -8)A(3,−8) and B(6,−0.5)B(6, -0.5)B(6,−0.5) lie on the curve F(d)F(d)F(d).
Given that ∫36(54d2+4d−26)dd=−15\displaystyle \int_{3}^{6} \left( \frac{54}{d^2} + 4d - 26 \right) dd = -15∫36(d254+4d−26)dd=−15.
(i) state the value of ∫69(54d2+4d−26)dd\displaystyle \int_{6}^{9} \left( \frac{54}{d^2} + 4d - 26 \right) dd∫69(d254+4d−26)dd.
(ii) find the value of the constant k k\,k such that ∫36(54d2+4d+k)dd=0\displaystyle \int_{3}^{6} \left( \frac{54}{d^2} + 4d + k \right) dd = 0∫36(d254+4d+k)dd=0.
864 exam-style questions on Edexcel A Level Maths Integration, covering 11.1 Integrating Standard Functions, 11.2 Integrating f(ax + b), 11.3 Using Trigonometric Identities, 11.4 Reverse Chain Rule, 11.5 Integration by Substitution, 11.6 Integration by Parts, 11.7 Partial Fractions, 11.8 Finding Areas, 11.9 The Trapezium Rule, 11.10 Solving Differential Equations, and 11.11 Modelling with Differential Equations. Each one has a worked solution and a mark scheme showing where the marks go.