P8.1The fundamental theorem of calculus

Edexcel A level Maths (9MA0) · Pure mathematics › Integration

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Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy4 marks

The curve \(C\) has equation \(y = f(x)\). The point \((-1, -1)\) lies on \(C\).
Given that \(f'(x) = 4x + 3\),

Find \(f(x)\).[4]

Show the answer and mark scheme

Answer: \(f(x) = 2x^{2} + 3x\)

M1 for \(x^{n} \to x^{n + 1}\) for at least one term
A1 for \(2x^{2} + 3x + c\) (may be unsimplified)
M1 for substituting \(x = -1\), \(y = -1\) to find \(c\)
A1 for \(f(x) = 2x^{2} + 3x\)

Worked solution: \(f(x) = \int \left(4x + 3\right) \mathrm{d}x = 2x^{2} + 3x + c\).
At \((-1, -1)\): \(-1 = -1 + c\), so \(c = 0\).
So \(f(x) = 2x^{2} + 3x\).

Question 2Medium6 marks

The curve \(C\) has equation \(y = f(x)\), \(x \gt 0\). The point \((1, 39)\) lies on \(C\).
Given that \(f'(x) = \frac{(x + 4)^{2}}{\sqrt{x}}\),

Find \(f(x)\), giving each term in its simplest form.[6]

Show the answer and mark scheme

Answer: \(f(x) = \frac{2}{5}x^{\frac{5}{2}} + \frac{16}{3}x^{\frac{3}{2}} + 32x^{\frac{1}{2}} + \frac{19}{15}\)

M1 for expanding and dividing by \(\sqrt{x}\) to obtain powers of \(x\): \(x^{\frac{3}{2}} + 8x^{\frac{1}{2}} + 16x^{-\frac{1}{2}}\)
M1 for \(x^{n} \to x^{n + 1}\) for at least one term
A1 for two terms correct
A1 for all terms correct: \(\frac{2}{5}x^{\frac{5}{2}} + \frac{16}{3}x^{\frac{3}{2}} + 32x^{\frac{1}{2}} + c\)
M1 for using \((1, 39)\) to find \(c\)
A1 for \(f(x) = \frac{2}{5}x^{\frac{5}{2}} + \frac{16}{3}x^{\frac{3}{2}} + 32x^{\frac{1}{2}} + \frac{19}{15}\)

Worked solution: \(f'(x) = \frac{x^{2} + 8x + 16}{x^{\frac{1}{2}}} = x^{\frac{3}{2}} + 8x^{\frac{1}{2}} + 16x^{-\frac{1}{2}}\).
\(f(x) = \frac{2}{5}x^{\frac{5}{2}} + \frac{16}{3}x^{\frac{3}{2}} + 32x^{\frac{1}{2}} + c\).
At \(x = 1\): \(39 = \frac{566}{15} + c\), so \(c = \frac{19}{15}\), giving \(f(x) = \frac{2}{5}x^{\frac{5}{2}} + \frac{16}{3}x^{\frac{3}{2}} + 32x^{\frac{1}{2}} + \frac{19}{15}\).

Question 3Hard7 marks

The curve \(C\) has equation \(y = f(x)\), where \(x\) is in radians. The point \((0, -5)\) lies on \(C\).
Given that \(f'(x) = -\mathrm{e}^{-x} - 6\cos 3x - 1\),

(a) Find \(f(x)\).[5]

(b) Find an equation of the tangent to \(C\) at the point where \(x = 0\).[2]

Show the answer and mark scheme

(a) Answer: \(f(x) = \mathrm{e}^{-x} - 2\sin 3x - x - 6\)

M1 for \(\mathrm{e}^{-x} \to \lambda\mathrm{e}^{-x}\)
M1 for \(\cos 3x \to \mu\sin 3x\)
A1 for \(\mathrm{e}^{-x} - 2\sin 3x - x + c\)
M1 for substituting \(x = 0\), \(y = -5\) to find \(c\)
A1 for \(f(x) = \mathrm{e}^{-x} - 2\sin 3x - x - 6\)

Worked solution: \(f(x) = \mathrm{e}^{-x} - 2\sin 3x - x + c\).
At \(x = 0\): \(-5 = 1 + c\), so \(c = -6\), giving \(f(x) = \mathrm{e}^{-x} - 2\sin 3x - x - 6\).

(b) Answer: \(y = -8x - 5\)

M1 for using \(f'(0) = -8\) with the point \((0, -5)\)
A1 for \(y = -8x - 5\)

Worked solution: \(f'(0) = -1 - 6 - 1 = -8\) and the curve passes through \((0, -5)\), so the tangent is \(y = -8x - 5\).

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