Riemann sums and area bounds

“One use of the integral calculus is to enable us to ascertain the values of areas bounded by curves.”

— Silvanus P. Thompson

Explore a curved area

We know the area of a rectangle: width times height. How could rectangles help us find the area beneath a curved graph? Start with the region under \(f(x)=x^2\) from \(0\) to \(2\) and explore it below.

The lower rectangles fit inside the curved region; the upper rectangles cover it with some extra area. Their totals put the area between two numbers. Narrower strips suggest a way to bring those numbers together. The calculus question is what happens as every strip becomes narrow.

We can also choose other heights from the curve within each strip. Those choices give estimates between the lower and upper sums. We will now give these observations precise names and notation.

For now, take \(f\) to be continuous and nonnegative on a closed interval \([a,b]\), with \(a<b\). The region lies between the graph, the horizontal axis, and the vertical lines \(x=a\) and \(x=b\).

Naming the pieces

There are \(n\) strips and \(n+1\) partition points. The widths need not be equal. If they are equal, then

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

Riemann sums

A Riemann sum depends on both the partition and the sample points. Its rectangles may cross the curve, and the resulting estimate may be above or below the area.

Lower and upper sums

Every sampled height satisfies \(\ell_i\le f(x_i^*)\le u_i\). Multiplying by the positive strip widths and adding gives, for every Riemann sum \(S\) on the same partition,

\[\underline{S}(f,P)\le S\le\overline{S}(f,P).\]

The lower rectangles lie inside the curved region, and the upper rectangles cover it. Thus the area \(A\) satisfies

\[\underline{S}(f,P)\le A\le\overline{S}(f,P).\]

For equal-width partitions with \(n\) strips, we abbreviate these sums as \(\underline{S}_n\) and \(\overline{S}_n\). Their difference measures the uncertainty left by these bounds. If the lower and upper sums agree, their common value determines the area exactly.

Find the minimum and maximum on each whole strip, wherever they occur.

Beyond continuous functions (optional)

For a bounded function that is not continuous, a strip’s function values need not have an attained minimum or maximum. The general definitions use the infimum, the greatest lower bound of those values, and the supremum, their least upper bound, in place of \(\ell_i\) and \(u_i\).

These still define lower and upper Darboux sums and bound every sampled Riemann sum on the same partition. If an extreme bound is not attained, no sample point supplies that height, so the corresponding Darboux sum need not itself be a sampled Riemann sum. Continuity avoids this complication in the present lesson.

Example 1

Bounding a parabolic region

Return to the initial setup in the opening investigation: \(f(x)=x^2\) from \(0\) to \(2\) with four equal strips. Each width is \(2/4=1/2\). We can now calculate the two bounds shown there.

Since \(x^2\) increases on this interval, each strip’s minimum occurs at its left end and its maximum at its right end. The minimum heights are \(0,\ 1/4,\ 1,\ 9/4\), so

\[\begin{aligned} \underline{S}_4&=\frac12\left(0+\frac14+1+\frac94\right)\\ &=\frac74=1.75. \end{aligned}\]

The maximum heights are \(1/4,\ 1,\ 9/4,\ 4\), giving

\[\begin{aligned} \overline{S}_4&=\frac12\left(\frac14+1+\frac94+4\right)\\ &=\frac{15}{4}=3.75. \end{aligned}\]

Therefore

\[1.75\le A\le3.75.\]

The calculation is exact for the rectangles. The gap of \(2\) records how much room these bounds leave for the curved region’s area.

Example 2

A maximum inside the strip

For \(f(x)=4x(2-x)\) on \([0,2]\), use one strip of width \(2\). The function is nonnegative and has minimum \(0\) at the ends. Writing

\[f(x)=4-4(x-1)^2\]

shows that its maximum is \(4\) at the interior point \(x=1\). Thus

\[\begin{aligned} \underline{S}_1&=2\cdot0=0,\\ \overline{S}_1&=2\cdot4=8. \end{aligned}\]

Choose Rise and fall with one rectangle in the opening investigation. The lower rectangle has zero height; the upper rectangle reaches the peak. These bounds give \(0\le A\le8\).

Unequal widths carry unequal weights

Example 3

The middle strip is twice as wide

Partition \([0,2]\) at \(0,\ 1/2,\ 3/2,\ 2\) for \(f(x)=x^2\). The widths are \(1/2,\ 1,\ 1/2\). Since the function increases, the maximum heights are \(1/4,\ 9/4,\ 4\).

Three upper rectangles for x squared. The middle strip, from one half to three halves, is twice as wide as either outer strip. The heights are one quarter, nine quarters and four. — The widths are \(\tfrac12,1,\tfrac12\). Each rectangle uses the maximum height on its own strip.

The upper sum is

\[\begin{aligned} \overline{S}(f,P)&=\frac14\cdot\frac12+\frac94\cdot1\\ &\quad+4\cdot\frac12\\ &=\frac{35}{8}=4.375. \end{aligned}\]

The minimum heights are \(0,\ 1/4,\ 9/4\), so the lower sum is

\[\begin{aligned} \underline{S}(f,P)&=0\cdot\frac12+\frac14\cdot1\\ &\quad+\frac94\cdot\frac12\\ &=\frac{11}{8}=1.375. \end{aligned}\]

These bounds give \(1.375\le A\le4.375\). Each height must be multiplied by its own strip’s width. Replacing all widths by \(2/3\) would describe different rectangles.

For a general Riemann sum on this same partition, choose the sample points \(1/4,\ 5/4,\ 7/4\). Each belongs to its own strip. Their heights are \(1/16,\ 25/16,\ 49/16\), giving

\[\begin{aligned} S&=\frac1{16}\cdot\frac12+\frac{25}{16}\cdot1\\ &\quad+\frac{49}{16}\cdot\frac12\\ &=\frac{25}{8}=3.125. \end{aligned}\]

As guaranteed, \(11/8\le25/8\le35/8\). Different sample points can change \(S\), but it remains between the same lower and upper sums. These bounds alone do not tell us whether this particular estimate is above or below the area.

Quiz

Enable JavaScript for interactive answer checking.Divide \([2,5]\) into six equal subintervals. What are the width and the third division point \(x_3\)?

\(\Delta x=\tfrac12,\ x_3=\tfrac72\)

\(\Delta x=\tfrac56,\ x_3=\tfrac52\)

\(\Delta x=\tfrac12,\ x_3=\tfrac32\)

Explanation

The width is \((5-2)/6=1/2\), and division points start at \(a=2\): \(x_3=2+3/2=7/2\).A positive function is strictly decreasing on a strip. Where are its minimum and maximum?

Minimum at the left end; maximum at the right end

Minimum at the right end; maximum at the left end

Both at the centre of the strip

Explanation

The heights decrease as the input increases. The smallest height is therefore at the right end and the largest at the left end. These supply the lower and upper rectangles, respectively.Two strips have widths \(1\) and \(3\), with sampled heights \(4\) and \(2\). What is their Riemann sum?

\(6\)

\(12\)

\(10\)

Explanation

Each sampled height needs its own strip width: \(4\cdot1+2\cdot3=10\). Adding heights alone loses the widths; using one average width gives incorrect weights.For \(f(x)=(x-1)^2\) on \([0,2]\), use a single strip. What are the lower and upper sums?

Both are \(2\)

Lower \(0\), upper \(2\)

Lower \(0\), upper \(1\)

Explanation

The minimum is \(0\) at the interior point \(x=1\), and the maximum is \(1\) at the ends. Multiplying by the width \(2\) gives the sums \(0\) and \(2\).For a continuous, nonnegative function, a lower sum is \(5.2\) and an upper sum is \(5.6\). What is guaranteed about the area estimate \(5.4\)?

Its absolute error is at most \(0.2\)

Its absolute error is exactly \(0.2\)

Its error is zero

Explanation

The area lies in \([5.2,5.6]\), so its distance from \(5.4\) is at most \(0.2\). The bounds do not determine the actual error.A partition has lower sum \(5\) and upper sum \(7\). What must hold for every Riemann sum \(S\) on that same partition?

\(5\le S\le7\)

It equals the area

It cannot exceed the area

Explanation

Each sampled height lies between the minimum and maximum on its strip. Multiplying by the positive widths and adding gives \(5\le S\le7\). This does not determine whether \(S\) is above or below the area.