Definite Integrals and Areas

Evaluate definite integrals exactly and use them to find areas under a curve, below the x-axis, between a line and a curve, and between two curves.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 39 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

From integral to area

A farmer's field has one curved edge. There is no simple area formula for that shape. Integration gives the exact area.

So far your integrals had no limits and ended with + c. A definite integral has two limits, a bottom one and a top one, and its answer is a number.

Formula

\int_{a}^{b} f (x) d x = [F (x)]_{a}^{b} = F (b) - F (a)

$b$ — the upper limit. Put it in first.
$a$ — the lower limit. Subtract its value.
$F (x)$ — the integral of $f (x)$ , written without + c.

Why no + c? It would appear in both values and cancel: $(F (b) + c) - (F (a) + c) = F (b) - F (a)$ .

Worked example

1
Find $\int_{1}^{3} (3 x^{2} - 2 x) d x$ .
2
$[x^{3} - x^{2}]_{1}^{3}$ — Integrate. Write the answer in square brackets with the limits on the right.
3
$= (3^{3} - 3^{2}) - (1^{3} - 1^{2})$ — Top limit in, minus bottom limit in. Keep each in its own bracket.
4
$= 18 - 0 = 18$ — 27 − 9 = 18 and 1 − 1 = 0.

Rule

Learn this. No board prints it. Mark schemes give a method mark for showing the limits put in, so always write the substitution line.

Key idea

Over 2,200 years ago, Archimedes found the exact area of a region cut from a parabola. He filled it with smaller and smaller triangles and added them up, which is the same idea as an integral.

Key idea

Integrate, then top minus bottom. A definite integral is a number, so there is no + c.

Exact answers

On the non-calculator paper (0606 and 4037 Paper 1), the limits are often numbers like $\frac{π}{6}$ or $ln 3$ . Use exact values, and leave answers with surds, π or ln when they do not simplify.

Worked example

1
Find the exact value of $\int_{0}^{\frac{π}{6}} 4 cos 2 x d x$ .
2
$[2 sin 2 x]_{0}^{\frac{π}{6}}$ — Cosine gives sine. Divide 4 by 2.
3
$= 2 sin \frac{π}{3} - 2 sin 0$ — Put in the limits. $2 \times \frac{π}{6} = \frac{π}{3}$ .
4
$= 2 \times \frac{3}{2} - 0 = 3$ — Exact value: $sin \frac{π}{3} = \frac{3}{2}$ .

Worked example

1
Find the exact value of $\int_{0}^{l n 3} 2 e^{2 x} d x$ .
2
$[e^{2 x}]_{0}^{l n 3}$ — Keep the power, divide 2 by 2.
3
$= e^{2 l n 3} - e^{0}$ — Put in the limits.
4
$= e^{l n 9} - 1 = 9 - 1 = 8$ — $2 ln 3 = ln 9$ , and $e^{l n 9} = 9$ . Any number to the power 0 is 1.

Example

0606 and 4037: $\int_{1}^{3} \frac{4}{2 x - 1} d x = [2 ln (2 x - 1)]_{1}^{3}$ $= 2 ln 5 - 2 ln 1 = 2 ln 5$ , since $ln 1 = 0$ .

Common mistake

Thinking the lower limit 0 always gives 0. $cos 0 = 1$ and $e^{0} = 1$ , so those terms do not vanish.

Common mistake

Dropping brackets when you subtract. $- (1 - 2)$ is $+ 1$ . Write each value in its own bracket first.

Key idea

Exact values in, exact answer out. Simplify ln and e terms with the log laws.

Area under a curve

Think of the region under a curve as many thin strips standing on the x-axis. Each strip has height $y$ and a tiny width. Adding them all up is exactly what the integral does.

Formula

Area = \int_{a}^{b} y d x

$y$ — the equation of the curve. It must be above the x-axis between the limits.
$a$ — the x-value where the region starts.
$b$ — the x-value where the region ends.

Steps

1
1. Find the limits. If they are not given, solve $y = 0$ to see where the curve meets the x-axis.
2. Integrate $y$ .
3. Put in the limits: top minus bottom.

Worked example

1
Find the area enclosed by the curve $y = 6 x - x^{2} - 5$ and the x-axis.
2
$6 x - x^{2} - 5 = 0 \Rightarrow (x - 1) (x - 5) = 0$ — Find where the curve meets the x-axis. These are the limits.
3
$\int_{1}^{5} (6 x - x^{2} - 5) d x = [3 x^{2} - \frac{x ^{3}}{3} - 5 x]_{1}^{5}$ — Integrate each term.
4
$= \frac{25}{3} - (- \frac{7}{3})$ — At 5: 75 − 125/3 − 25 = 25/3. At 1: 3 − 1/3 − 5 = −7/3.
5
$= \frac{32}{3}$ — The area, in square units.

Worked example

1
The area under $y = 2 x + 1$ from $x = 0$ to $x = k$ is 12. Find $k$ .
2
$[x^{2} + x]_{0}^{k} = k^{2} + k$ — Integrate and put in the limits. The answer has $k$ in it.
3
$k^{2} + k = 12 \Rightarrow (k + 4) (k - 3) = 0$ — Set it equal to the area and solve.
4
$k = 3$ — $k$ must be bigger than 0, so reject −4.

Key idea

No limits given? Find where the curve crosses the x-axis. Those are your limits.

Below the x-axis

Where the curve is below the x-axis, $y$ is negative. So the integral there comes out negative. An area can never be negative, so take its size.

Formula

Area = \int_{a}^{b} y d x when y < 0

$y$ — the curve, below the x-axis between $a$ and $b$ . The bars mean: ignore the minus sign.

Worked example

1
Find the total area between $y = x^{2} - 4 x + 3$ and the x-axis, from $x = 0$ to $x = 3$ .
2
$x^{2} - 4 x + 3 = (x - 1) (x - 3)$ — The curve crosses the x-axis at 1 and 3. From 0 to 1 it is above; from 1 to 3 it is below.
3
$\int_{0}^{1} y d x = [\frac{x ^{3}}{3} - 2 x^{2} + 3 x]_{0}^{1} = \frac{4}{3}$ — The part above the axis.
4
$\int_{1}^{3} y d x = 0 - \frac{4}{3} = - \frac{4}{3}$ — The part below the axis: negative, as expected.
5
$Area = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$ — Add the sizes of the two parts.

If you integrate straight from 0 to 3 you get $\frac{4}{3} - \frac{4}{3} = 0$ . The two parts cancel, and the answer is wrong.

Common mistake

Integrating across a point where the curve crosses the x-axis. Always split there first.

Key idea

A sketch saves marks. See where the curve crosses, split there, and add the sizes of the parts.

Between a line and a curve

Now the region is trapped between two graphs. Each thin strip goes from the lower graph up to the upper graph, so its height is top minus bottom.

Formula

Area = \int_{a}^{b} (y_{top} - y_{bottom}) d x

$y_{top}$ — the graph that is higher between the limits.
$y_{bottom}$ — the graph that is lower.
$a, b$ — the x-values where the two graphs meet. Find them by setting the equations equal.

Worked example

1
Find the area between the curve $y = 4 x - x^{2}$ and the line $y = x$ .
2
$4 x - x^{2} = x \Rightarrow x (3 - x) = 0$ — Set the equations equal. They meet at $x = 0$ and $x = 3$ .
3
$\int_{0}^{3} (4 x - x^{2} - x) d x = \int_{0}^{3} (3 x - x^{2}) d x$ — At $x = 1$ the curve is at 3 and the line at 1, so the curve is on top.
4
$= [\frac{3 x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{3} = \frac{27}{2} - 9 = \frac{9}{2}$ — Integrate and put in the limits.

Another way, often used in mark schemes: area under the curve minus the shape under the line. Here $\int_{0}^{3} (4 x - x^{2}) d x = 9$ , and the triangle under the line is $\frac{1}{2} \times 3 \times 3 = \frac{9}{2}$ . So the area is $9 - \frac{9}{2} = \frac{9}{2}$ again.

Worked example

1
0606 and 4037: find the exact area between the curve $y = \frac{7}{2 x - 1}$ and the line $y = 9 - 2 x$ .
2
$\frac{7}{2 x - 1} = 9 - 2 x \Rightarrow 4 x^{2} - 20 x + 16 = 0$ — Multiply both sides by $2 x - 1$ and tidy.
3
$4 (x - 1) (x - 4) = 0, x = 1, 4$ — The limits. The line is on top between them.
4
$Trapezium = \frac{1}{2} (7 + 1) \times 3 = 12$ — Under the line: at $x = 1$ , $y = 7$ ; at $x = 4$ , $y = 1$ .
5
$\int_{1}^{4} \frac{7}{2 x - 1} d x = [\frac{7}{2} ln (2 x - 1)]_{1}^{4} = \frac{7}{2} ln 7$ — Under the curve. $ln 1 = 0$ .
6
$Area = 12 - \frac{7}{2} ln 7$ — Line shape minus curve area. Leave it exact.

Common mistake

Subtracting the wrong way round and getting a negative area. Test one x-value between the limits to see which graph is on top.

Key idea

Top minus bottom works everywhere, even below the x-axis, because the difference is always positive.

Two curves, and the y-axis

Two curves work exactly like a line and a curve: find where they meet, then integrate top minus bottom.

Worked example

1
Find the area enclosed by $y = x^{2}$ and $y = 8 - x^{2}$ .
2
$x^{2} = 8 - x^{2} \Rightarrow x^{2} = 4, x = \pm 2$ — Set them equal to find the limits.
3
$\int_{- 2}^{2} (8 - x^{2} - x^{2}) d x = \int_{- 2}^{2} (8 - 2 x^{2}) d x$ — At $x = 0$ : 8 is above 0, so $8 - x^{2}$ is on top.
4
$= [8 x - \frac{2 x ^{3}}{3}]_{- 2}^{2} = \frac{32}{3} - (- \frac{32}{3}) = \frac{64}{3}$ — Careful with the negative limit: put in −2 inside brackets.

4PM1 also lists the area between a curve and the y-axis. Turn everything round: make $x$ the subject and integrate with respect to $y$ .

Formula

Area = \int_{c}^{d} x d y

$x$ — the curve written as $x$ in terms of $y$ .
$c$ — the lower y-value.
$d$ — the upper y-value.

Worked example

1
4PM1: find the area between $y = x^{2}$ ( $x \geq 0$ ), the y-axis, and the lines $y = 1$ and $y = 4$ .
2
$x = y = y^{\frac{1}{2}}$ — Make $x$ the subject.
3
$\int_{1}^{4} y^{\frac{1}{2}} d y = [\frac{2}{3} y^{\frac{3}{2}}]_{1}^{4}$ — Integrate with respect to $y$ . The limits are y-values.
4
$= \frac{2}{3} (8 - 1) = \frac{14}{3}$ — $4^{\frac{3}{2}} = 8$ and $1^{\frac{3}{2}} = 1$ .

Common mistake

Using x-limits in an integral with $d y$ . The limits must match the letter you integrate with.

Key idea

Same method every time: find where the graphs meet, decide which is on top, integrate the difference.

Which method? Read the question

Read the question and look at the sketch. What you see tells you what to do.

You see	You do
$\int_{a}^{b}$ with numbers	Integrate, then top limit minus bottom limit. No + c
“Exact value”	Use exact values of sin, cos, e and ln. No decimals
Region between a curve and the x-axis	Solve $y = 0$ for the limits, then integrate $y$
Part of the curve below the x-axis	Split at the crossing point and add the sizes
A line and a curve	Find where they meet; integrate top − bottom, or curve area minus triangle or trapezium
Two curves	Find where they meet; integrate top − bottom
Area next to the y-axis (4PM1)	Make $x$ the subject; integrate $x d y$ with y-limits
The area is given, find $k$	Integrate with $k$ as a limit, set equal to the area, solve

Common mistake

Rounding too early. Keep exact values to the end, and give decimals only when the question asks.

Key idea

Sketch first. The picture tells you the limits, which graph is on top, and where to split.

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