Approximating Areas with Riemann Sums

AP Calculus · Integration and Accumulation of Change

Approximating Areas with Riemann Sums

Same slices every time — the only choice is where you sample the height, and which way that pushes the error.

key idea$\int_a^b f\,dx\approx\sum_{i=1}^{n} f(x_i^*)\,\Delta x,\quad \Delta x=\frac{b-a}{n}$

The method

Slice $[a,b]$ into $n$ pieces of width $\Delta x=\dfrac{b-a}{n}$ and add the rectangle areas. The methods differ only in where on each slice you sample the height. The CED expects all four representations — graph, table, formula, words.

[Figure: left, right and midpoint Riemann rectangles compared]

Four subintervals on $[0,4]$ for an increasing, concave-up function. Left endpoints under-shoot every rectangle; right endpoints over-shoot; midpoints split the difference.

Method	Height of the $i$-th rectangle	Formula
Left Riemann sum	value at the left edge	$\Delta x\left[f(x_0)+f(x_1)+\dots+f(x_{n-1})\right]$
Right Riemann sum	value at the right edge	$\Delta x\left[f(x_1)+f(x_2)+\dots+f(x_{n})\right]$
Midpoint sum	value at the centre	$\Delta x\left[f(\bar x_1)+\dots+f(\bar x_n)\right]$
Trapezoidal sum	average of the two edges	$\dfrac{\Delta x}{2}\left[f(x_0)+2f(x_1)+\dots+2f(x_{n-1})+f(x_n)\right]$

The left and right sums share every interior term; they differ only in the first and last. That is why their difference is $\Delta x\left[f(b)-f(a)\right]$.

$x$	$0$	$1$	$2$	$3$	$4$
$g(x)$	$1.0$	$1.5$	$3.0$	$5.5$	$9.0$

The same function as the figure, $g(x)=\tfrac12x^{2}+1$, sampled at the subinterval edges. Every example below uses these values.

Worked examples

EX 1

Left and right sums from the table

$n=4$, $\Delta x=1$

Using the table, approximate $\displaystyle\int_{0}^{4} g(x)\,dx$ with a left and then a right Riemann sum, $n=4$.

Find $\Delta x$

$\Delta x=\dfrac{4-0}{4}=1$.

Left sum — drop the last value

Use $g$ at $x=0,1,2,3$: $L=1\left[1.0+1.5+3.0+5.5\right]=11.0$.

Right sum — drop the first value

Use $g$ at $x=1,2,3,4$: $R=1\left[1.5+3.0+5.5+9.0\right]=19.0$.

Check the relationship

$R-L=\Delta x\left[g(4)-g(0)\right]=1(9.0-1.0)=8.0$, which matches $19.0-11.0$. A quick way to catch an arithmetic slip.

EX 2

Which side of the true value?

monotonicity

The exact value is $\displaystyle\int_{0}^{4}\left(\tfrac12x^{2}+1\right)dx=\tfrac{x^{3}}{6}+x\Big|_{0}^{4}=\tfrac{64}{6}+4\approx 14.67$. Explain, without computing it, why $L$ had to be an under-estimate and $R$ an over-estimate.

Establish the direction

$g'(x)=x\gt 0$ on $(0,4)$, so $g$ is increasing throughout.

Reason about the left sum

On each slice the left edge is the lowest point of the function there, so every rectangle sits below the curve. $L$ under-estimates.

Reason about the right sum

The right edge is the highest point on each slice, so every rectangle pokes above the curve. $R$ over-estimates.

Confirm

$11.0\lt 14.67\lt 19.0$. The reasoning is what earns the marks — the arithmetic only confirms it.

State the general rule

For an increasing $f$: left under, right over. For a decreasing $f$: the reverse. Direction, not concavity, decides this one.

EX 3

Trapezoidal sum and the role of concavity

$T=\tfrac{\Delta x}{2}\left[\dots\right]$

Approximate the same integral with a trapezoidal sum, $n=4$, and say whether it over- or under-estimates.

Apply the formula

$T=\dfrac{1}{2}\left[1.0+2(1.5)+2(3.0)+2(5.5)+9.0\right]$.

Compute

$=\dfrac{1}{2}\left[1.0+3.0+6.0+11.0+9.0\right]=\dfrac{30.0}{2}=15.0$.

Note the shortcut

$T=\dfrac{L+R}{2}=\dfrac{11.0+19.0}{2}=15.0$. The trapezoidal sum is always the average of the left and right sums.

Decide the direction of the error

$g''(x)=1\gt 0$, so $g$ is concave up. Each chord lies above the curve, so every trapezoid over-shoots: $T$ over-estimates.

Confirm

$15.0\gt 14.67$. Note the discriminator changed: over/under for trapezoids depends on concavity, not on whether $f$ is increasing.

EX 4

From a verbal description

words → number

Water flows into a tank. The rate is $2$ litres/min at noon, rising to $5$, then $9$, then $14$ litres/min at 3-minute intervals. Estimate the total volume added over the 9 minutes, and say whether your estimate is too big or too small.

Recognise the integral

Total volume is $\displaystyle\int_{0}^{9} r(t)\,dt$ where $r$ is the rate in litres/min and $t$ is minutes.

Set up the sum

Readings at $t=0,3,6,9$ give three subintervals with $\Delta t=3$.

Choose and compute

A left sum uses $t=0,3,6$: $3\left[2+5+9\right]=48$ litres.

Judge the direction

The rate is increasing throughout, so a left sum under-estimates. The true volume exceeds $48$ litres.

State units

$48$ litres — litres/min $\times$ min. Rate problems live or die on the units of the answer.

How wrong can it be?

Direction decides left/right; concavity decides trapezoid

Increasing ⇒ left under, right over. Concave up ⇒ trapezoid over, midpoint under. Mixing the two criteria is the classic error.

$n$ subintervals means $n+1$ data points

A table with five values supports at most four subintervals. Counting rectangles instead of points gives a sum one term too long.

Unequal widths are allowed

When a table's $x$-values are not evenly spaced you cannot factor out a single $\Delta x$ — compute each rectangle's width separately and add.

More slices, less error

Halving $\Delta x$ roughly halves the error of a left or right sum, and quarters it for the trapezoidal sum. Refinement does not change the over/under direction, only the size of the gap.

Course map. Topic 6.2 · LO LIM-5.A — approximate a definite integral using geometric and numeric methods (EK LIM-5.A.1: approximation for functions represented graphically, numerically, analytically or verbally; LIM-5.A.2: left, right, midpoint and trapezoidal sums; LIM-5.A.3: numerical methods and technology).