Find f from f′ and a point

In 9 of 20 recent exams
1% of the marks
Paper 1 or 2
4 to 6 marks

How it is set in the exam

Paper 1 (no calculator) and Paper 2 (calculator), 4–6 marks: a short Section A question, or the last part of a longer one. Common (9 of 20 exam versions, 3 of them on Paper 2 (calculator)). The algebra is the same on both papers; on Paper 2 (calculator) the constant may be a decimal.

Spot it in the exam

Exam phrases to spot (3)

“ $f'(x) = \ldots$ and $f(0) = 4$ ; find $f(x)$ ”
“the graph of $f$ passes through the point $(1, 3)$ ”
“given that $\frac{dy}{dx} = \ldots$ and $y = 3$ when $x = 1$ , find $y$ in terms of $x$ ”

In disguise (2)

"the gradient of the curve at any point $(x, y)$ is …": that is $\frac{dy}{dx}$ as a formula in $x$
the point is named by a feature instead of coordinates, such as an intercept on an axis or a stationary point

Variants (3)

$f''(x)$ with two conditions: integrate twice, one constant each time, and one condition for each constant
a velocity and a starting position: $s(t) = \int v(t)\,dt$ , with $s(0)$ fixing the constant. In a motion question the type is Displacement, position and distance travelled, and this is its first step
$f'(x)$ has an unknown constant too: a gradient fact fixes it first (e.g. $f'(1) = 0$ ), then integrate and use the point

Looks like it, but isn't (4)

$f$ is given with unknown constants, and facts about its gradient fix them: Find unknown constants (differentiate, don't integrate:
no point is given: Integrate: choose the method, with $+c$ in the answer
$f'(x)$ and one value of $f$ are given, and the tangent is asked for: Tangent or normal at a given point, with no integrating
one value of $f$ and a number for $\int_a^b f'(x)\,dx$ , with no formula: Definite integrals without the function

Worked example

Linear insides, and a decimal constant

Paper 2 (calculator)

Question

The function $f$ is such that $f'(x) = 6\cos 3x + 2e^{0.5x}$ , and $f(1) = 8$ .
(a) Find $f(x)$ . [5]
(b) Find $f(2)$ . [2]

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Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

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Steps

Integrate
$f'(x)$ term by term (Integrate: choose the method), and write $+c$ straight away.
Substitute the point
into that expression, $c$ included. The method mark for this step needs the $+c$ to be there.
Solve for $c$ ,
then write the answer as a function, $f(x) = \ldots$ (or $y = \ldots$ ), with $c$ replaced by its value. On keep a decimal $c$ at full accuracy (keep full-accuracy values).
$f''$ given:
integrate once ( $+c_1$ ) and use the condition on $f'$ (a gradient, a stationary point); integrate again ( $+c_2$ ) and use the condition on $f$ (a point).
Check:
differentiate your $f$ , and substitute the point.

Common mistakes

Substituting the point before writing $+c$ ,
or never writing it: the method mark needs the $c$ in the expression.
Differentiating instead of integrating
markschemes give nothing for it.
Putting the point into $f'$ instead of $f$
in Example 2.3, $f(1) = -1$ belongs to $f$ , and $f'(1) = 0$ to $f'$ .

More mistakes (3)

$e^0 = 0$ or $\cos 0 = 0$
when the point is at $x = 0$ : both are $1$ .
Stopping at the value of $c$
the answer is the function, $f(x) = \ldots$
Rounding $c$ too early
on : keep the stored value for later parts.

Easy to confuse with

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