Given that k k\,k is a positive constant and ∫1k(22x+8)dx=5\displaystyle \int_1^k \left( \frac{2}{2\sqrt{x}} + 8 \right) dx = 5∫1k(2x2+8)dx=5
Show that 8k+2k−15=08k + 2\sqrt{k} - 15 = 08k+2k−15=0
Hence, using algebra, find any values of k k\,k such that ∫1k(22x+8)dx=5\displaystyle \int_1^k \left( \frac{2}{2\sqrt{x}} + 8 \right) dx = 5∫1k(2x2+8)dx=5
411 exam-style questions on Edexcel A Level Maths Integration, covering 13.1 Integrating x^n, 13.2 Indefinite Integrals, 13.3 Finding Functions, 13.4 Definite Integrals, 13.5 Areas under Curves, 13.6 Areas under the x-axis, and 13.7 Areas between curves and lines. Each one has a worked solution and a mark scheme showing where the marks go.