Estimating Areas – OCR FSMQ Additional Maths

1 Rectangular Strips (NM6)

NM6: use rectangular strips to estimate the area between a curve and the $x$-axis. Divide the region into vertical strips, treat each as a rectangle, and add up their areas. Crude, but it always works — and the errors are easy to see.

Two choices of height

Left-hand rectangles: take each strip's height from its left edge
Right-hand rectangles: take each height from its right edge

The bracket this gives you

For an increasing curve, the left-hand estimate is too small and the right-hand estimate is too large.
The true area lies between them.

Worked Example 1 — Both estimates, and a bracket

Estimate $\displaystyle\int_0^4 x^2 \,\mathrm{d}x$ using four rectangles of width $1$, by both methods.

①Values of $x^2$ at $x = 0, 1, 2, 3, 4$ are $0, 1, 4, 9, 16$.

②Left-hand: heights $0, 1, 4, 9$, so area $\approx 1(0+1+4+9) = 14$

③Right-hand: heights $1, 4, 9, 16$, so area $\approx 1(1+4+9+16) = 30$

④So $14 < \text{true area} < 30$.

⑤Exact value: $\Bigg[\dfrac{x^3}{3}\Bigg]_0^4 = \dfrac{64}{3} = 21.33$ ✓ — comfortably inside the bracket.

Averaging the two gives $22$, which is much closer than either. That average is exactly what the trapezium rule computes.

2 The Trapezium Rule (NM7)

Instead of a flat top, give each strip a sloping top joining the curve at both edges. Each strip is then a trapezium, which hugs the curve far more closely than a rectangle.

The trapezium rule — provided in the examination

$\displaystyle\int_a^b y \,\mathrm{d}x \approx \tfrac{1}{2}h\Big[\left(y_0 + y_n\right) + 2\left(y_1 + y_2 + \cdots + y_{n-1}\right)\Big]$
with $n$ strips of width $h = \dfrac{b-a}{n}$, and $n+1$ ordinates $y_0, \ldots, y_n$

Work out $h = \dfrac{b-a}{n}$.
Make a table of the $n+1$ ordinates, from $y_0$ at $x=a$ to $y_n$ at $x=b$.
Add the first and last ordinates.
Add all the middle ordinates and double the total.
Add those two results and multiply by $\tfrac{h}{2}$.

The two commonest errors. Using $n+1$ as the number of strips instead of the number of ordinates; and doubling the end ordinates instead of the middle ones. A clear table prevents both.

Worked Example 2 — Four strips

Use the trapezium rule with four strips to estimate $\displaystyle\int_0^4 x^2 \,\mathrm{d}x$.

①$h = \dfrac{4-0}{4} = 1$, so ordinates at $x = 0, 1, 2, 3, 4$.

$x$	$0$	$1$	$2$	$3$	$4$
$y = x^2$	$0$	$1$	$4$	$9$	$16$
Weight	$\times 1$	$\times 2$	$\times 2$	$\times 2$	$\times 1$

②Ends: $y_0 + y_4 = 0 + 16 = 16$

③Middles: $1 + 4 + 9 = 14$, doubled gives $28$

④$\approx \tfrac12(1)(16 + 28) = \tfrac12(44) = 22$

⑤Exact value $\dfrac{64}{3} = 21.33$, so the estimate is $0.67$ too high — an error of about $3\%$.

The weights row in the table is worth writing out. It makes the "ends once, middles twice" pattern visible and checkable.

Worked Example 3 — More strips, better answer

Repeat with eight strips.

①$h = 0.5$, with ordinates at $0, 0.5, 1, \ldots, 4$.

②Values: $0,\; 0.25,\; 1,\; 2.25,\; 4,\; 6.25,\; 9,\; 12.25,\; 16$

③Ends: $0 + 16 = 16$

④Middles: $0.25 + 1 + 2.25 + 4 + 6.25 + 9 + 12.25 = 35$, doubled gives $70$

⑤$\approx \tfrac12(0.5)(16 + 70) = 0.25 \times 86 = 21.5$

⑥The error has dropped from $0.67$ to $0.17$ — quartered when the number of strips doubled.

The error is roughly proportional to $h^2$. Halving the strip width divides the error by about four, which is why the trapezium rule improves so quickly.

3 Over-Estimate or Under-Estimate? (NM8)

NM8: recognise whether an estimate would be an over- or under-estimate, and understand how to calculate an improved estimate. The answer depends on curvature, not on whether the curve is rising or falling.

The rule for the trapezium rule

Curve bending upwards (convex, like $y = x^2$) → the chord lies above the curve → over-estimate
Curve bending downwards (concave, like $y = \sqrt{x}$) → the chord lies below → under-estimate

Do not confuse rising with bending. $y = x^2$ is decreasing for $x<0$ and increasing for $x>0$, but it bends upwards throughout, so the trapezium rule over-estimates on either side. It is the second derivative, not the first, that decides.

Worked Example 4 — Justifying the direction of the error

State, with a reason, whether the trapezium rule over- or under-estimates $\displaystyle\int_1^4 x^2 \,\mathrm{d}x$.

①$y = x^2$ gives $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 > 0$, so the curve bends upwards everywhere.

②Each trapezium's sloping top is therefore a chord lying above the curve.

③Every strip includes a sliver of area that is not under the curve, so the total is an over-estimate.

④Confirmed numerically: with three strips the rule gives $21.5$, while the exact value is $\dfrac{63}{3} = 21$.

A sketch is the best justification. Draw the curve with one trapezium on top and the overshoot is obvious — worth a mark on its own.

How to improve an estimate

Use more strips — a smaller $h$ means each chord hugs the curve more closely.
Or average the left-hand and right-hand rectangle estimates.

Worked Example 5 — Bracketing the true value

For $\displaystyle\int_0^2 x^3 \,\mathrm{d}x$, use two strips to find a trapezium-rule estimate, state whether it is over or under, and give a bracket for the true value.

①$h = 1$; ordinates at $x = 0, 1, 2$ are $0, 1, 8$.

②$\approx \tfrac12(1)\Big[(0+8) + 2(1)\Big] = \tfrac12(10) = 5$

③For $x \geqslant 0$, $\dfrac{\mathrm{d}^2}{\mathrm{d}x^2}\left(x^3\right) = 6x \geqslant 0$, so the curve bends upwards and this is an over-estimate.

④Left-hand rectangles give $1(0 + 1) = 1$, an under-estimate.

⑤So $1 < \text{true value} < 5$.

⑥Exact: $\Bigg[\dfrac{x^4}{4}\Bigg]_0^2 = 4$ ✓ — inside the bracket, and just below the trapezium estimate as predicted.

4 Quick Reference

Strip width

$h = \dfrac{b-a}{n}$.

Ordinates

$n+1$ of them for $n$ strips.

The rule

$\tfrac12 h\big[(\text{ends}) + 2(\text{middles})\big]$.

Weights

Ends once, middles twice.

Tabulate

A table of ordinates prevents most errors.

Bends upwards

Over-estimate.

Bends downwards

Under-estimate.

Not rising/falling

Curvature decides, not gradient.

To improve

More strips; error falls like $h^2$.

Bracket

Left and right rectangles trap the true value.

5 Practice Questions

Question 1

How many ordinates are needed for the trapezium rule with $6$ strips, and what is $h$ if the limits are $2$ and $8$?

▶ Show solution

$7$ ordinates, and $h = \dfrac{8-2}{6} = 1$.

Question 2

Use the trapezium rule with two strips to estimate $\displaystyle\int_0^2 x^2 \,\mathrm{d}x$.

▶ Show solution

$h = 1$; ordinates $0, 1, 4$.

$\approx \tfrac12(1)\Big[(0+4) + 2(1)\Big] = \tfrac12(6) = 3$

(Exact: $\tfrac83 = 2.67$.)

Question 3

Use the trapezium rule with four strips to estimate $\displaystyle\int_0^2 x^2 \,\mathrm{d}x$.

▶ Show solution

$h = 0.5$; ordinates at $0, 0.5, 1, 1.5, 2$ are $0, 0.25, 1, 2.25, 4$.

Ends: $0 + 4 = 4$. Middles: $0.25 + 1 + 2.25 = 3.5$, doubled $= 7$.

$\approx \tfrac12(0.5)(4 + 7) = 0.25 \times 11 = 2.75$

Closer to $2.67$ than the two-strip answer of $3$.

Question 4

State, with a reason, whether your answer to Question 3 is an over- or under-estimate.

▶ Show solution

$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 > 0$, so $y = x^2$ bends upwards.

The chords therefore lie above the curve, making it an over-estimate.

Confirmed: $2.75 > 2.67$.

Question 5

Use left-hand rectangles of width $1$ to estimate $\displaystyle\int_1^4 x^2 \,\mathrm{d}x$.

▶ Show solution

Heights at $x = 1, 2, 3$ are $1, 4, 9$.

Area $\approx 1(1 + 4 + 9) = 14$

(Exact: $21$. The curve is increasing, so left-hand rectangles under-estimate badly.)

Question 6

Repeat Question 5 with right-hand rectangles, and give a bracket for the true value.

▶ Show solution

Heights at $x = 2, 3, 4$ are $4, 9, 16$.

Area $\approx 1(4+9+16) = 29$

So $14 < \text{true area} < 29$. (The exact value $21$ lies inside.)

Question 7

Use the trapezium rule with three strips to estimate $\displaystyle\int_1^4 x^2 \,\mathrm{d}x$, and compare with Questions 5 and 6.

▶ Show solution

$h = 1$; ordinates $1, 4, 9, 16$.

Ends: $1 + 16 = 17$. Middles: $4 + 9 = 13$, doubled $= 26$.

$\approx \tfrac12(1)(17 + 26) = 21.5$

Far better than either rectangle estimate — and it is exactly their average, $\tfrac{14+29}{2} = 21.5$.

Question 8

Explain why the trapezium rule gives the exact answer for $\displaystyle\int_0^5 (2x+1)\,\mathrm{d}x$.

▶ Show solution

$y = 2x+1$ is a straight line, so it has zero curvature.

Each trapezium's sloping top lies exactly on the line, with nothing over- or under-counted.

The trapezia fit the region perfectly, so the rule is exact — for any number of strips.

(Check with one strip: $\tfrac12(5)(1 + 11) = 30$, and $\int_0^5(2x+1)\mathrm{d}x = [x^2+x]_0^5 = 30$ ✓)

Question 9

A student uses the trapezium rule with four strips and gets $8.4$, then with eight strips and gets $8.1$. Estimate the true value, with a reason.

▶ Show solution

The estimates are decreasing, and the error falls by roughly a factor of four each time the strips double.

If the error with four strips is $E$, the error with eight is about $\tfrac{E}{4}$. The difference between the estimates is $0.3$, which is about $E - \tfrac{E}{4} = \tfrac{3E}{4}$.

So $E \approx 0.4$, and the true value is roughly $8.4 - 0.4 = 8.0$.

Equivalently, the eight-strip error is about $0.1$, giving $8.1 - 0.1 = 8.0$ — the same answer.

Question 10

The speed of a car, in m s⁻¹, is recorded every $2$ seconds:

$t$ (s)	$0$	$2$	$4$	$6$	$8$	$10$
$v$ (m s⁻¹)	$0$	$6$	$11$	$15$	$18$	$20$

(a) Explain what the area under the speed–time graph represents. (b) Use the trapezium rule with all the data to estimate it. (c) State, with a reason, whether this is likely to be an over- or under-estimate. (d) Estimate the car's acceleration at $t = 4$.

▶ Show solution

(a) The area under a speed–time graph is the distance travelled, in metres — because distance is speed multiplied by time, and the area accumulates exactly that product.

(b) There are $6$ ordinates, so $5$ strips, with $h = 2$.

Ends: $y_0 + y_5 = 0 + 20 = 20$

Middles: $6 + 11 + 15 + 18 = 50$, doubled $= 100$

Distance $\approx \tfrac12(2)(20 + 100) = 1 \times 120 = \mathbf{120}$ m

(c) Look at how the speed is rising: the increases are $6, 5, 4, 3, 2$ — getting smaller each time.

So the graph is rising ever more gently: it bends downwards (concave). The chords therefore lie below the true curve, and the estimate is an under-estimate.

(d) Acceleration is the gradient of the speed–time graph. Use the straddling chord from $t=2$ to $t=6$:

$a \approx \dfrac{15 - 6}{6 - 2} = \dfrac{9}{4} = \mathbf{2.25}$ m s⁻²

(The one-sided chords give $\tfrac{11-6}{2} = 2.5$ and $\tfrac{15-11}{2} = 2$; the straddling value $2.25$ is their average and the better estimate.)

Estimating Areas (NM6–NM8) · OCR FSMQ Additional Maths · Created with MathJax