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Question 89

Given that k k\,k is a positive constant and ∫1k(72x+10)dx=28\displaystyle \int_1^k \left( \frac{7}{2\sqrt{x}} + 10 \right) dx = 28∫1k​(2x​7​+10)dx=28

a.

Show that 10k+7k−45=010k + 7\sqrt{k} - 45 = 010k+7k​−45=0

[4]
b.

Hence, using algebra, find any values of k k\,k such that ∫1k(72x+10)dx=28\displaystyle \int_1^k \left( \frac{7}{2\sqrt{x}} + 10 \right) dx = 28∫1k​(2x​7​+10)dx=28

[4]
Markscheme

Integration Questions

  1. A Level
  2. /Maths
  3. /Integration

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.

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