Which is usually more accurate, the midpoint rule or the trapezoidal rule?

For a curve that is concave up or concave down, the midpoint rectangles under-estimate and over-estimate in the opposite direction to the trapezoidal rule, and the midpoint estimate is usually the more accurate of the two for the same number of strips. Both improve as the number of strips increases.

Which is usually more accurate, the midpoint rule or the trapezoidal rule?

For a concave curve the midpoint and trapezoidal estimates err in opposite directions, and the midpoint estimate is usually the more accurate for the same number of strips. Both improve as the number of strips increases.

Year 12 Methods (Unit 3 & 4) Integration

Estimating the area under a graph

Q: How do you estimate the area under a curve with rectangles?

Divide the interval into n equal strips of width w = (b - a) / n. On each strip build a rectangle whose height is the value of the function at the left endpoint, the right endpoint, or the midpoint. Add the areas: the estimate is w times the sum of the chosen heights.

Q: What is the trapezoidal rule?

The trapezoidal rule joins successive points on the curve with straight chords, making a trapezium on each strip. With n strips of width w the estimate is (w/2)[f(x0) + 2(f(x1) + ... + f(x_{n-1})) + f(xn)].

Q: How do you tell if an estimate is an over- or under-estimate?

For rectangles it depends on increasing or decreasing: on an increasing curve left rectangles under-estimate and right rectangles over-estimate. For the trapezoidal rule it depends on concavity: concave up gives an over-estimate, concave down gives an under-estimate.

Q: Why does using more strips give a better estimate?

Narrower strips follow the curve more closely, so the gap between the strips and the curve shrinks. As the number of strips increases without bound the estimates approach the exact area, which is the definite integral.

Q: How is the estimate linked to the definite integral?

Each estimate is a sum of the form sum f(x_i) times delta x. The definite integral is the limit of these sums as the strip width tends to zero, so the strip estimates approximate the definite integral and the exact area is that limit.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), you estimate the area under a graph by dividing it into equal strips and adding the areas. Each strip can be a left-endpoint, right-endpoint or midpoint rectangle, or a trapezium (the trapezoidal rule). Whether the estimate is an over- or under-estimate follows from the shape of the curve, and using more strips gives a better result — the exact area is the limit of the sums, the definite integral.

To estimate the area between a positive curve \(y=f(x)\) and the \(x\)-axis from \(x=a\) to \(x=b\), split the interval into \(n\) equal strips, each of width \(w=\dfrac{b-a}{n}\), and add up simple shapes that fit each strip.

A rectangle strip takes a single height from the curve: the left-endpoint value, the right-endpoint value, or the midpoint value of the strip. The estimate is then \(w\) times the sum of those heights, a sum of the form \(\sum_i f(x_i)\,\delta x_i\).

A trapezium strip joins the two ends of the strip on the curve with a straight chord; adding the trapezia gives the trapezoidal rule. Because a rectangle or a chord does not follow the curve exactly, an estimate is either an over-estimate or an under-estimate. Taking more strips narrows the gap, and the exact area is the limit of the sums — the definite integral \(\displaystyle\int_a^b f(x)\,dx\).

Key idea. Strip width \(w=\dfrac{b-a}{n}\); a rectangle uses one height (left, right or midpoint), a trapezium uses both end-heights. Add the strips; more strips \(\Rightarrow\) closer to the exact area \(\displaystyle\int_a^b f(x)\,dx\).

Left-endpoint rectangles under an increasing curve — an under-estimate

Trapezoidal strips — each chord joins two points on the curve

Strip width for \(n\) equal strips:

\[w=\frac{b-a}{n}, \qquad x_i=a+i\,w\]

Left- and right-endpoint rectangle sums:

\[A_{\text{left}}=w\big[f(x_0)+f(x_1)+\cdots+f(x_{n-1})\big], \qquad A_{\text{right}}=w\big[f(x_1)+f(x_2)+\cdots+f(x_n)\big]\]

A = w \sum_{i}^{} f (x_{i})

Midpoint rectangle sum (heights at the middle of each strip):

\[A_{\text{mid}}=w\big[f(m_1)+f(m_2)+\cdots+f(m_n)\big], \qquad m_i=\tfrac{1}{2}\left(x_{i-1}+x_i\right)\]

The trapezoidal rule:

\[\int_a^b f(x)\,dx \approx \frac{w}{2}\Big[f(x_0)+2\big(f(x_1)+f(x_2)+\cdots+f(x_{n-1})\big)+f(x_n)\Big]\]

\int_{a}^{b} f (x) d x

The exact area is a limit. As \(n\to\infty\) the strip width \(w\to 0\) and the sums approach \(\displaystyle\int_a^b f(x)\,dx\): the definite integral is defined as this limit of sums \(\sum_i f(x_i)\,\delta x_i\).

How to estimate the area under a graph

Choose the number of strips \(n\) and find the width \(w=\dfrac{b-a}{n}\). The strip boundaries are \(x_0=a,\ x_1,\ \ldots,\ x_n=b\).
Read off the heights. For left rectangles use \(f\) at \(x_0,\ldots,x_{n-1}\); for right rectangles use \(x_1,\ldots,x_n\); for midpoint rectangles use the middle of each strip; for trapezia use both end-heights.
Add the strips. A rectangle sum is \(w\) times the sum of the heights; the trapezoidal rule is \(\dfrac{w}{2}\big[\text{ends}+2(\text{interior})\big]\).
Decide over or under. For rectangles, use increasing/decreasing; for the trapezoidal (and midpoint) rule, use concavity. State whether the estimate is above or below the true area.

Over- or under-estimate. On an increasing curve, left rectangles under-estimate and right rectangles over-estimate. A concave-up curve makes the trapezoidal rule over-estimate; a concave-down curve makes it under-estimate.

Example 1 — Left-endpoint rectangles

Estimate the area under \(y=x^2\) from \(x=0\) to \(x=4\) using \(4\) left-endpoint rectangles of width \(1\).

Solution

Heights at the left endpoints \(x=0,1,2,3\).

\(A\)	\(\approx\)	\(1\big[f(0)+f(1)+f(2)+f(3)\big]\)
	\(=\)	\(1\big[0+1+4+9\big]=14\)

Estimate \(=14\) square units (an under-estimate, since \(y=x^2\) is increasing).

A = 14

Example 2 — Right rectangles, over/under

For the same curve \(y=x^2\) on \([0,4]\) with \(4\) strips, find the right-endpoint estimate and say whether it is an over- or under-estimate.

Solution

Heights at the right endpoints \(x=1,2,3,4\).

\(A\)	\(\approx\)	\(1\big[f(1)+f(2)+f(3)+f(4)\big]\)
	\(=\)	\(1\big[1+4+9+16\big]=30\)

The curve is increasing, so the right rectangles rise above it: an over-estimate (the true area lies between \(14\) and \(30\)).

A = 30

Example 3 — Trapezoidal rule

Use the trapezoidal rule with \(4\) strips of width \(1\) to estimate \(\displaystyle\int_0^4 x^2\,dx\). Is it an over- or under-estimate?

Solution

Ends \(f(0),f(4)\); twice the interior \(f(1),f(2),f(3)\).

\(A\)	\(\approx\)	\(\dfrac{1}{2}\big[0+2(1+4+9)+16\big]\)
	\(=\)	\(\dfrac{1}{2}\big[0+28+16\big]=22\)

The curve is concave up, so each trapezium lies above it: an over-estimate (exact \(=\tfrac{64}{3}\approx 21.33\)).

A = 22

Example 4 — More strips → the integral

Estimate the area under \(y=x^2\) from \(x=0\) to \(x=2\) with right rectangles, using \(2\) strips then \(4\) strips. What do the estimates approach?

Solution

\(n=2\ (w=1)\)	\(:\)	\(1[f(1)+f(2)]=5\)
\(n=4\ (w=0.5)\)	\(:\)	\(0.5[0.25+1+2.25+4]=3.75\)

As \(n\to\infty\) the estimates fall towards the exact area \(\displaystyle\int_0^2 x^2\,dx=\tfrac{8}{3}\approx 2.67\).

\int_{0}^{2} x^{2} d x = \frac{8}{3}

Common pitfalls

Remember to multiply by the strip width. A rectangle estimate is \(w\) times the sum of the heights, not just the sum of the heights. When \(w\neq 1\) forgetting the \(w\) is the most common slip.

Use the correct endpoints. Left rectangles use \(x_0,\ldots,x_{n-1}\); right rectangles use \(x_1,\ldots,x_n\). Mixing them (or reading the wrong side of a strip) changes the answer.

Over/under depends on the shape. For rectangles it is increasing vs decreasing; for the trapezoidal rule it is concave up vs concave down. Do not assume every estimate is too big.

Frequently asked questions

How do you estimate the area under a curve with rectangles?

Split \([a,b]\) into \(n\) strips of width \(w=\dfrac{b-a}{n}\). On each strip use one height (left, right or midpoint) and add: the estimate is \(w\) times the sum of the heights.

What is the trapezoidal rule?

Join successive points on the curve with chords, making a trapezium per strip: \(\dfrac{w}{2}\big[f(x_0)+2(f(x_1)+\cdots+f(x_{n-1}))+f(x_n)\big]\).

How do you tell if an estimate is an over- or under-estimate?

Rectangles: on an increasing curve, left under-estimates and right over-estimates. Trapezoidal: concave up over-estimates, concave down under-estimates.

Why does using more strips give a better estimate?

Narrower strips hug the curve more closely, so the error shrinks. As \(n\to\infty\) the estimates approach the exact area, the definite integral.

How is the estimate linked to the definite integral?

Each estimate is a sum \(\sum_i f(x_i)\,\delta x_i\). The definite integral \(\displaystyle\int_a^b f(x)\,dx\) is the limit of these sums as \(w\to0\).

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Finding the exact area: the definite integral