Edexcel A level Maths (9MA0) · Pure mathematics › Integration
Practise Definite integrals and areas under curves. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
The diagram shows part of the curve \(C\) with equation \(y = x^{3} - 5x^{2} + 6x\). The curve crosses the \(x\)-axis at the origin and at the points where \(x = 2\) and \(x = 3\). The region \(R_{1}\) is bounded by \(C\) and the \(x\)-axis for \(0 \leqslant x \leqslant 2\). The region \(R_{2}\) is bounded by \(C\) and the \(x\)-axis for \(2 \leqslant x \leqslant 3\). In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
[object Object]
(a) Find the exact area of \(R_{1}\).[4]
(b) Find the exact total area of the shaded regions.[4]
Show the answer and mark scheme
(a)Answer: \(\frac{8}{3}\)
M1 for \(x^{n} \to x^{n + 1}\) for at least one term
A1 for \(\frac{1}{4}x^{4} - \frac{5}{3}x^{3} + 3x^{2}\)
M1 for evaluating \(\int_{2}^{3} y \,\mathrm{d}x\) with their integrated expression
A1 for \(-\frac{5}{12}\)
M1 for using the modulus of this value (the region lies below the \(x\)-axis) and adding to \(R_{1}\)
A1 for \(\frac{37}{12}\)
Worked solution: \(\int_{2}^{3} y \,\mathrm{d}x = \left[\frac{1}{4}x^{4} - \frac{5}{3}x^{3} + 3x^{2}\right]_{2}^{3} = \left(\frac{9}{4}\right) - \left(\frac{8}{3}\right) = -\frac{5}{12}\), which is negative because \(R_{2}\) lies below the \(x\)-axis. So \(R_{2} = \frac{5}{12}\) and the total area is \(\frac{8}{3} + \frac{5}{12} = \frac{37}{12}\).