Edexcel A level Maths (9MA0) · Pure mathematics › Integration
Practise Integrating standard functions. unlimited generated questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
In this question \(x \gt 0\).
Find \(\int \left(12x^{3} - 6x^{2} - 2x + \frac{4}{x^{3}}\right) \,\mathrm{d}x\), giving each term in its simplest form.[4]
Show the answer and mark scheme
Answer: \(3x^{4} - 2x^{3} - x^{2} - 2x^{-2} + c\)
M1 for \(x^{n} \to x^{n + 1}\) for at least one term
A1 for two terms correct (may be unsimplified)
A1 for three terms correct
A1 for a fully correct, simplified answer including \(+ c\)
Worked solution: Writing each term as a power of \(x\): \(12x^{3} - 6x^{2} - 2x + 4x^{-3}\). \(\int \left(12x^{3} - 6x^{2} - 2x + 4x^{-3}\right) \,\mathrm{d}x = 3x^{4} - 2x^{3} - x^{2} - 2x^{-2} + c\)
Question 2Medium7 marks
\(f(x) = \frac{-2x^{3} - 4x - 6}{x^{3}}, \quad x \gt 0\) In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
(a) Write \(f(x)\) in the form \(A + Bx^{-2} + Cx^{-3}\), where \(A\), \(B\) and \(C\) are constants to be found.[2]
(b) Hence find \(\int f(x) \,\mathrm{d}x\).[3]
(c) Hence find the exact value of \(\int_{1}^{2} f(x) \,\mathrm{d}x\).[2]
Show the answer and mark scheme
(a)Answer: \(A = -2\), \(B = -4\), \(C = -6\)
M1 for expanding the numerator and dividing each term by the denominator
A1 for \(A = -2\), \(B = -4\), \(C = -6\)
Worked solution: Expanding and dividing term by term: \(f(x) = -2 - 4x^{-2} - 6x^{-3}\), so \(A = -2\), \(B = -4\) and \(C = -6\).
(b)Answer: \(-2x + 4x^{-1} + 3x^{-2} + c\)
M1 for \(x^{n} \to x^{n + 1}\) for at least one term
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Find the exact value of \(\int_{0}^{4} \left(\frac{6}{x + 1} + 6(x + 1)\right) \,\mathrm{d}x\), giving your answer in the form \(a + b\ln 5\), where \(a\) and \(b\) are rational numbers.[5]
Show the answer and mark scheme
Answer: \(72 + 6\ln 5\)
M1 for \(\frac{6}{x + 1} \to \lambda\ln(x + 1)\)
M1 for \((x + 1) \to \mu(x + 1)^{2}\)
A1 for \(6\ln(x + 1) + 3(x + 1)^{2}\)
dM1 for substituting both limits and subtracting
A1 for \(72 + 6\ln 5\)
Worked solution: \(\left[6\ln(x + 1) + 3(x + 1)^{2}\right]_{0}^{4}\). At the upper limit \(x + 1 = 5\); at the lower limit it equals 1, and \(\ln 1 = 0\). So the value is \(6\ln 5 + 3\left(25 - 1\right) = 72 + 6\ln 5\).