A-level · Integration

Definite integration and area under a curveExam-style questions

Finding the area of a region with a curved edge. Thin strips and the limit of a sum; why an antiderivative measures area (the fundamental theorem); evaluating with square brackets, top limit minus bottom limit, with no + C; why a region below the x-axis counts as negative, so an integral and an area can differ; splitting and reversing the limits; and finding a limit that is a letter.

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5 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 2 Level 3 Level 4 Level 5

1
(a) Ravi and Ellen were each asked to find $\displaystyle\int (3x - 2)^2\,dx$ .

Ravi wrote: $\displaystyle\int (3x - 2)^2\,dx = \tfrac19(3x - 2)^3 + c$

Ellen wrote: $\displaystyle\int (3x - 2)^2\,dx = \int (9x^2 - 12x + 4)\,dx = 3x^3 - 6x^2 + 4x + c$

Ravi multiplies out his answer to get $3x^3 - 6x^2 + 4x - \tfrac89 + c$ . He says that Ellen's answer must be wrong because it does not contain the term $-\tfrac89$ .

Explain whether you agree with Ravi.

(1 mark)
(b) In this part of the question $k$ is a constant greater than $1$ .

(i) Find $\displaystyle\int_1^k \frac{12}{(2x + 1)^2}\,dx$ . Express your answer as a single fraction in terms of $k$ .

(ii) The region bounded by the curve $y = \dfrac{12}{(2x + 1)^2}$ , the $x$ -axis and the lines $x = 1$ and $x = k$ has area $\dfrac43$ .

Determine the value of $k$ .

(5 marks)
(c) In this question you must show detailed reasoning.

Find, in its simplest form, the exact value of $\displaystyle\int_0^{\frac{\pi}{4}} \frac{\sec^2 x}{1 + 3\tan x}\,dx$ .

(4 marks)
Level 4 · 10 marksTry it