Integrating with Two Conditions

From f″(x) to f′(x): integrate with c₁ and fix it from a given gradient

In the exam for Edexcel International AS Maths · 6 real past-paper questions below

From f″(x) back to f′(x)

You can already find a curve from its gradient and one point: integrate, then use the point to fix $c$ . AS questions go one level higher. They give you $f^{''} (x)$ , so you integrate twice, and each time you need a fact to fix a constant.

Formula

f^{''} (x) \to f^{'} (x) \to f (x)

$f^{''} (x)$ — what the question gives you: the gradient of the gradient.
$f^{'} (x)$ — integrate once and add $k$ . A gradient fact fixes $k$ .
$f (x)$ — integrate again and add $d$ . A point on the curve fixes $d$ .

First trip: $f^{''} (x)$ to $f^{'} (x)$ . The fact you need is a gradient, such as “ $f^{'} (x) = 15$ at P”. It goes into $f^{'} (x)$ , because that is the gradient.

Worked example

1
Given $f^{''} (x) = \frac{3}{x} + 2$ , $x > 0$ , and $f^{'} (x) = 15$ at the point P where $x = 4$ , find $f^{'} (x)$ .
2
$f^{''} (x) = 3 x^{- \frac{1}{2}} + 2$ — Rewrite the root as a power first.
3
$f^{'} (x) = \frac{3 x ^{\frac{1}{2}}}{\frac{1}{2}} + 2 x + k$ — Up one, divide by the new power. Add $k$ .
4
$f^{'} (x) = 6 x + 2 x + k$ — Dividing by one half doubles: 3 becomes 6.
5
$15 = 64 + 2 (4) + k$ — Put $x = 4$ and $f^{'} (x) = 15$ into $f^{'} (x)$ .
6
$15 = 12 + 8 + k, k = - 5$ — Solve for $k$ .
7
$f^{'} (x) = 6 x + 2 x - 5$ — Write $f^{'} (x)$ in full. You need it for the next trip.

Common mistake

Putting the gradient into $f^{''} (x)$ . The value 15 is a gradient, so it belongs to $f^{'} (x)$ . Integrate first, then substitute.

Key idea

One integration, one new constant, one fact to fix it.

Gradient from a tangent or normal

The gradient at P is not always given as a number. Often the question gives the tangent or the normal at P instead. Read the gradient off the line.

Rule

Tangent at P with gradient $m$ : the curve's gradient at P is $m$ . Normal at P with gradient $m$ : the curve's gradient at P is $- \frac{1}{m}$ , because the normal is at right angles to the tangent.

Memory trick

Normal means “flip and change sign”. A normal gradient of $\frac{3}{4}$ flips to $\frac{4}{3}$ , then changes sign to $- \frac{4}{3}$ .

Worked example

1
The normal to a curve at P, where $x = 1$ , has equation $x + 2 y = 7$ . Find $f^{'} (1)$ and the y coordinate of P.
2
$y = - \frac{1}{2} x + \frac{7}{2}$ — Make $y$ the subject to see the gradient.
3
$m_{normal} = - \frac{1}{2}$ — The number in front of $x$ .
4
$f^{'} (1) = - 1 \div (- \frac{1}{2}) = 2$ — Flip and change sign: the tangent gradient is 2.
5
$y = - \frac{1}{2} (1) + \frac{7}{2} = 3$ — P is on the normal too, so the line gives P's y coordinate. P is (1, 3).

An earlier part may go the other way: you know the gradient at P and must write the line. Use $y - y_{1} = m (x - x_{1})$ .

Worked example

1
At P(3, 5), $f^{'} (x) = 4$ . Find the normal at P in the form $y = m x + c$ .
2
$m_{normal} = - \frac{1}{4}$ — Flip 4 and change sign.
3
$y - 5 = - \frac{1}{4} (x - 3)$ — Line through P with the normal gradient.
4
$y = - \frac{1}{4} x + \frac{23}{4}$ — Expand: $5 + \frac{3}{4} = \frac{23}{4}$ .

Common mistake

Using the normal's gradient as the curve's gradient. The curve's gradient is the tangent's gradient. Flip and change sign first.

Key idea

A tangent or normal line gives you two facts at once: the gradient at P, and the y coordinate of P.

Integrate again for f(x)

Second trip: $f^{'} (x)$ to $f (x)$ . Now the fact you need is a point on the curve. It goes into $f (x)$ .

Worked example

1
Carry on: $f^{'} (x) = 6 x + 2 x - 5$ , and P(4, 10) lies on the curve. Find $f (x)$ .
2
$f^{'} (x) = 6 x^{\frac{1}{2}} + 2 x - 5$ — Root as a power again.
3
$f (x) = \frac{6 x ^{\frac{3}{2}}}{\frac{3}{2}} + x^{2} - 5 x + d$ — Integrate each term. Use a new letter, $d$ .
4
$f (x) = 4 x^{\frac{3}{2}} + x^{2} - 5 x + d$ — 6 divided by 3/2 is 4.
5
$10 = 4 (4)^{\frac{3}{2}} + 4^{2} - 5 (4) + d$ — Put $x = 4$ and $y = 10$ into $f (x)$ .
6
$10 = 32 + 16 - 20 + d, d = - 18$ — $4^{\frac{3}{2}} = (4)^{3} = 8$ , so the first term is 32.
7
$f (x) = 4 x^{\frac{3}{2}} + x^{2} - 5 x - 18$ — The final answer, all on one line.

Common mistake

Calling both constants $c$ and then mixing them up. The mark scheme allows the same letter, but a new letter keeps you safe. And $k$ is a number now: $- 5$ integrates to $- 5 x$ .

Common mistake

Forgetting the first constant. Without $k$ your $f^{'} (x)$ is wrong, so $f (x)$ is wrong too, and the marks that depend on it go.

Key idea

Physics does this every day. Integrate acceleration to get speed, then integrate speed to get position. The starting speed and the starting position are the two facts that fix the two constants.

Key idea

Same point P, two different jobs: its gradient fixes $k$ , its coordinates fix $d$ .

Unknown letters a and b

Another AS type gives $f^{'} (x)$ with unknown constants, such as $a$ and $b$ . Each fact in the question gives you one equation. Two unknowns need two equations, which you solve together.

Steps

1
1. A gradient fact: put it into $f^{'} (x)$ .
2. Integrate, with $+ c$ . Keep $a$ and $b$ as letters.
3. A point fact, or the y-intercept: put it into $f (x)$ .
4. Solve the equations together. Write $f (x)$ with numbers.

Worked example

1
$f^{'} (x) = 3 x^{2} + a x + b$ . The curve $y = f (x)$ crosses the y-axis at 4, passes through P(2, 10), and has gradient 7 at P. Find $f (x)$ .
2
$12 + 2 a + b = 7 \Rightarrow 2 a + b = - 5$ — Gradient 7 at $x = 2$ : put it into $f^{'} (x)$ .
3
$f (x) = x^{3} + \frac{a}{2} x^{2} + b x + c$ — Integrate. $a$ and $b$ are just numbers, so they wait in front.
4
$c = 4$ — The y-intercept is $f (0)$ . Every x term is 0, so $c = 4$ .
5
$8 + 2 a + 2 b + 4 = 10 \Rightarrow a + b = - 1$ — Put P(2, 10) into $f (x)$ .
6
$a = - 4, b = 3$ — Subtract the equations: $a = - 4$ . Then $b = 3$ .
7
$f (x) = x^{3} - 2 x^{2} + 3 x + 4$ — Check: $f^{'} (2) = 12 - 8 + 3 = 7$ . Correct.

Sometimes an earlier part says “show that $2 a + b = - 5$ ”. Put the fact in, then show at least one line before the given answer. The answer is printed, so the marks are for the working.

Common mistake

Putting the point into $f^{'} (x)$ . A point is on the curve, so it goes into $f (x)$ . Only a gradient goes into $f^{'} (x)$ .

Key idea

Count the unknowns, then count the facts. You need one fact for each unknown.

Lay it out for the marks

These questions are worth 6 to 8 marks, and about half are method marks. Many are dependent (dM1): you only get them if the method mark before them was earned. So one missing line can cost three marks.

Line you write	Mark
Integrate $f^{''} (x)$ , one power right	M1
Correct $f^{'} (x)$ with $+ k$	A1
Put the gradient fact in, find $k$	dM1 A1
Integrate again, one power right	dM1
Correct $f (x)$ with $+ d$	A1
Put the point in, find $d$	dM1
Final $f (x)$ , all on one line	A1

Rule

Label each line: $f^{'} (x) = \dots$ and $f (x) = \dots$ . Show the substitution line with the numbers in. Write the final answer on one line, with powers worked out. If the question says “you must show all stages of your working”, an answer from a calculator alone scores nothing.

Read the question to choose the route.

You see	You do
$f^{''} (x)$ and a gradient at P	Integrate with $+ k$ , use the gradient, then integrate again
A tangent at P	Its gradient is the curve's gradient at P; it also gives P's y
A normal at P	Curve's gradient at P is $- \frac{1}{m}$ ; it also gives P's y
$f^{'} (x)$ with $a$ and $b$	One equation per fact, solve together, then write $f (x)$
“Show that …”	Every line shown before the printed answer

Common mistake

Stopping at $f^{'} (x)$ . When the question asks for $f (x)$ , you must integrate twice.

Key idea

Two integrations, two constants, two facts. Gradient into $f^{'} (x)$ , point into $f (x)$ .

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