Left, Right and Midpoint Riemann Sums

Strips, and a height for each

Take the region under the curve $y=\frac{x^2}{4}$y = x²/4, above the x-axis, from x = 0 to x = 4. Cut the interval into 4 strips of equal width. Each strip is (4 − 0) ÷ 4 = 1 wide. In general, an interval from a to b cut into n strips gives each strip the width $\left(b-a\right)/n$(b − a) / n, written $\Delta x$Δx.

Over each strip stand a rectangle with a flat top. A flat top has one height, but the curve has a different height at every point of the strip, so the sum has to say where that height is read. A left sum reads it at the left edge of each strip.

The left edges are at x = 0, 1, 2 and 3, where $\frac{x^2}{4}$x²/4 is 0, 0.25, 1 and 2.25. Each rectangle is 1 wide, so the left sum is 0 × 1 + 0.25 × 1 + 1 × 1 + 2.25 × 1 = 3.5.

The left rectangles under $y=\frac{x^2}{4}$y = x²/4 from x = 0 to x = 4. Each top meets the curve at its left corner, and the curve climbs above it across the strip. The first rectangle has height 0, and the four add to 3.5.

The right sum

A right sum reads each height at the right edge of its strip: x = 1, 2, 3 and 4, where the heights are 0.25, 1, 2.25 and 4. The right sum is 0.25 + 1 + 2.25 + 4 = 7.5.

The right sum uses the same heights as the left sum, moved along by one strip. The first left height, 0 at x = 0, drops out, and the last right height, 4 at x = 4, comes in. So the right sum is bigger by (4 − 0) × 1 = 4, and 7.5 − 3.5 = 4.

The right rectangles under the same curve. Each top meets the curve at its right corner, so each rectangle stands above the curve everywhere else in its strip. The four add to 7.5.

The midpoint sum

A midpoint sum reads each height at the middle of its strip: x = 0.5, 1.5, 2.5 and 3.5, where the heights are 0.0625, 0.5625, 1.5625 and 3.0625. The midpoint sum is 0.0625 + 0.5625 + 1.5625 + 3.0625 = 5.25.

On a rising curve the middle of each strip is higher than its left edge and lower than its right edge, so the midpoint sum lands between the left sum 3.5 and the right sum 7.5. Each midpoint top crosses the curve halfway along its strip: over the left half it stands above the curve, over the right half below, and the two pieces almost cancel.

The midpoint rectangles. Each top crosses the curve at the middle of its strip, part over and part under. The four add to 5.25.

Against the exact area

Integrating $\frac{x^2}{4}$x²/4 gives $\frac{x^3}{12}$x³/12, so the exact area is $\frac{4^3}{12}=\frac{64}{12}=\frac{16}{3}$4³/12 = 64/12 = 16/3, about 5.333. The left sum 3.5 is about 1.833 short, the right sum 7.5 is about 2.167 over, and the midpoint sum 5.25 is only $\frac{1}{12}$1/12, about 0.083, short.

Cut the same interval into 8 strips, each 0.5 wide. The left sum becomes 4.375, the right sum 6.375 and the midpoint sum 5.3125. All three have moved toward $\frac{16}{3}$16/3. The left and right sums are now 2 apart, which is (4 − 0) × 0.5, and the midpoint sum is $\frac{1}{48}$1/48 short, a quarter of its error with 4 strips.

Four left rectangles under $y=x^2$y = x² from x = 0 to x = 2. They add to 1.75, against the exact area $\frac{8}{3}\approx 2.667$8/3 ≈ 2.667. Switch to right heights for 3.75, or to midpoints for 2.625. Then drag n up to 24: the left and right sums close in from either side, and the midpoint error falls to 0.001.

Under or over

On a rising curve, the left edge of each strip is its lowest point. A left rectangle touches the curve there, and the curve climbs above its top across the rest of the strip, so every left rectangle falls short and the left sum underestimates. The right edge is the highest point of each strip, so every right rectangle pokes above the curve and the right sum overestimates. The exact area lies between them: for $y=\frac{x^2}{4}$y = x²/4, $3.5<\frac{16}{3}<7.5$3.5 < 16/3 < 7.5.

A falling curve reverses this, because now the left edge is the tall side of each strip. Take $y=4-\frac{x^2}{4}$y = 4 − x²/4 from 0 to 4 in 4 strips. The heights at x = 0, 1, 2, 3 and 4 are 4, 3.75, 3, 1.75 and 0. The left sum is 4 + 3.75 + 3 + 1.75 = 12.5, and the right sum is 3.75 + 3 + 1.75 + 0 = 8.5. The exact area is $4\times 4-\frac{16}{3}=\frac{32}{3}$4 × 4 − 16/3 = 32/3, about 10.667, so here the left sum is over and the right sum is under.

If a curve rises over part of the interval and falls over the rest, some left rectangles are short and others are tall, and neither sum is sure to be over or under.

Left rectangles under the falling curve $y=4-\frac{x^2}{4}$y = 4 − x²/4. Each top meets the curve at its left corner, where the strip is tallest, and the curve drops below it. The four add to 12.5, more than the exact area $\frac{32}{3}$32/3.

The usual mistakes

Leaving out the width. With strips 1 wide it makes no difference, but with 8 strips 0.5 wide the left heights add to 8.75, and the left sum is 8.75 × 0.5 = 4.375.

Reading the wrong edge. The left sum of $y=\frac{x^2}{4}$y = x²/4 reads x = 0, 1, 2 and 3 and gives 3.5; reading x = 1, 2, 3 and 4 gives 7.5, which is the right sum.

Taking the midpoint sum as the average of the left and right sums. Those average to (3.5 + 7.5) ÷ 2 = 5.5, which is the trapezium rule. The midpoint sum is 5.25.

Calling the left sum an underestimate on every curve. It is under only where the curve rises; on $y=4-\frac{x^2}{4}$y = 4 − x²/4 it is 12.5, over the exact $\frac{32}{3}$32/3.

An hourly meter

In the application below, a works’ electrical load is read every hour, and the energy used is the area under the load curve. The table gives every height the three sums need, and the rising load says which sum is low and which is high.