Definite integrals without the function

In 1 of 20 recent exams
Paper 1, no calculator
2 to 6 marks

How it is set in the exam

Paper 1 (no calculator), 2–6 marks, often the last part of a question about a graph. Rare (1 of 20 exam versions), but its rules are used inside other questions.

Spot it in the exam

Exam phrases to spot (3)

“given that $\int_1^5 f(x)\,dx = 8$ , find $\int_1^5 (2f(x) + 3)\,dx$ ”
“find $\int_5^1 f(x)\,dx$ ”
“given $f(2) = 5$ and $\int_2^6 f'(x)\,dx = 8$ , find $f(6)$ ”

In disguise (2)

the graph of $f'$ is drawn, with the areas of the regions between it and the $x$ -axis written on it, and one value of $f$ is given: the other values of $f$ come from those areas, with a minus sign for regions below the axis
$\int_a^b g'(x)\,dx$ when $g$ itself is known: no integrating, just $g(b) - g(a)$

Looks like it, but isn't (2)

a formula for $f'$ and a point: Find f from f′ and a point
the graph of $f'$ with no areas on it, and the question is where $f$ has a maximum or is concave up: Graphs of f, f′ and f″

Worked example

Areas on the graph of f′ give values of f

Paper 1 (no calculator)

Question

The graph of $y = f'(x)$ , for $0 \le x \le 7$ , crosses the $x$ -axis at $x = 2$ and $x = 5$ and nowhere else. The table describes the regions between the graph and the $x$ -axis.

$A$
Interval
$0 \le x \le 2$
Position
above the $x$ -axis
Area
$6$
$B$
Interval
$2 \le x \le 5$
Position
below the $x$ -axis
Area
$10$
$C$
Interval
$5 \le x \le 7$
Position
above the $x$ -axis
Area
$3$

Region	Interval	Position	Area
$A$	$0 \le x \le 2$	above the $x$ -axis	$6$
$B$	$2 \le x \le 5$	below the $x$ -axis	$10$
$C$	$5 \le x \le 7$	above the $x$ -axis	$3$

Also, $f(0) = 1$ .
(a) Find $f(2)$ and $f(5)$ . [4]
(b) Find the greatest value and the least value of $f(x)$ for $0 \le x \le 7$ . Justify your answers. [4]
(c) Find $\displaystyle\int_7^0 \left(2f'(x) + 1\right)dx$ . [3]

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Steps

Write what you want in terms of what you know,
with the rules (Definite integrals): constant multiples and sums split; a constant $k$ gives $k(b - a)$ ; reversed limits change the sign; intervals add.
Values of $f$ from $f'$ :
$f(b) = f(a) + \int_a^b f'(x)\,dx$ .
Areas on a graph of $f'$ :
the integral over a region above the axis is $+$ (its area); over a region below, $-$ (its area).
Greatest or least value of $f$ :
read the sign of $f'$ from the graph to find where $f$ turns (Graphs of f, f′ and f″), then compare those values with the ends (Greatest and least values on an interval).

Common mistakes

Adding the areas as if they were the integrals
a region below the axis counts negative.
$\int_a^b k\,dx = k$
it is $k(b - a)$ .
Ignoring reversed limits
$\int_7^0 = -\int_0^7$ .

More mistakes (2)

Splitting a product
$\int f(x)g(x)\,dx$ is not $\int f(x)\,dx \times \int g(x)\,dx$ .
Taking a local maximum as the greatest value
without checking the ends.

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