Math · Calculus I · Worked example

Left, right and midpoint sums for x²

Estimate $\int_{0}^{1} x^{2} d x$ with $n = 2$ rectangles, using left, right and midpoint heights. Compare each estimate with the exact value $1 / 3$ .

\int_{0}^{1} x^{2} d x, n = 2

Choose the partition

With $n = 2$ pieces, the width is $Δ x = (1 - 0) / 2 = 1 / 2$ . The endpoints are $x_{0} = 0$ , $x_{1} = 1 / 2$ and $x_{2} = 1$ .

Δ x = \frac{1 - 0}{2} = \frac{1}{2}

Left sum

The left ends are $0$ and $1 / 2$ , with heights $0^{2} = 0$ and ${(1 / 2)}^{2} = 1 / 4$ . Add the heights and multiply by the width $1 / 2$ .

L_{2} = \frac{1}{2} (0 + \frac{1}{4}) = \frac{1}{8}

Right sum

The right ends are $1 / 2$ and $1$ , with heights $1 / 4$ and $1$ . The right sum shares the height $1 / 4$ with the left sum and swaps $0$ for $1$ .

R_{2} = \frac{1}{2} (\frac{1}{4} + 1) = \frac{5}{8}

Midpoint sum

The midpoints of $[0, 1 / 2]$ and $[1 / 2, 1]$ are $1 / 4$ and $3 / 4$ . Their heights are $1 / 16$ and $9 / 16$ . Add them and multiply by the width.

\frac{1}{2} (\frac{1}{16} + \frac{9}{16}) = \frac{5}{16}

Compare with the exact value

The exact value is $1 / 3 \approx 0.333$ , which the Fundamental Theorem of Calculus finds from the antiderivative $x^{3} / 3$ . Because $x^{2}$ rises on $[0, 1]$ , the left sum $0.125$ is too low and the right sum $0.625$ too high. The midpoint sum $0.3125$ misses by only $1 / 48 \approx 0.021$ .

\int_{0}^{1} x^{2} d x = \frac{1^{3}}{3} - \frac{0^{3}}{3} = \frac{1}{3}

Result

$L_{2} = 1 / 8 = 0.125$ , $R_{2} = 5 / 8 = 0.625$ and $M_{2} = 5 / 16 = 0.3125$ , against the exact $1 / 3 \approx 0.333$ .

Your turn

1. One like the example

Estimate $\int_{1}^{3} x^{2} d x$ with $n = 2$ rectangles, using left, right and midpoint heights. The exact value is $26 / 3$ .

Show the answer and explanation

$L_{2} = 5$ , $R_{2} = 13$ and $M_{2} = 8.5$ , against the exact $26 / 3 \approx 8.667$ .

$Δ x = (3 - 1) / 2 = 1$ , with endpoints $1$ , $2$ and $3$ . The left heights are $1$ and $4$ , and the right heights $4$ and $9$ . The midpoint heights are ${1.5}^{2} = 2.25$ and ${2.5}^{2} = 6.25$ . $x^{2}$ rises on $[1, 3]$ , so the left sum is low and the right sum high. The midpoint sum misses by only $1 / 6$ .

\begin{array}{l} Δ x = \frac{3 - 1}{2} = 1 \\ 1 (1 + 4) = 5 \\ 1 (4 + 9) = 13 \\ 1 (2.25 + 6.25) = 8.5 \\ \int_{1}^{3} x^{2} d x = \frac{3^{3}}{3} - \frac{1^{3}}{3} = \frac{26}{3} \end{array}

2. A variation

Estimate $\int_{0}^{2} (4 - x^{2}) d x$ with $n = 4$ , using left and right sums. This function falls. The exact value is $16 / 3$ .

Show the answer and explanation

$L_{4} = 6.25$ and $R_{4} = 4.25$ , against the exact $16 / 3 \approx 5.333$ .

$Δ x = 0.5$ , and the points $0$ , $0.5$ , $1$ , $1.5$ and $2$ have heights $4$ , $3.75$ , $3$ , $1.75$ and $0$ . The left sum uses the first four heights and the right sum the last four. Because $4 - x^{2}$ falls on $[0, 2]$ , the left sum is too big and the right sum too small, the reverse of a rising function.

\begin{array}{l} Δ x = \frac{2 - 0}{4} = 0.5 \\ 0.5 (4 + 3.75 + 3 + 1.75) = 6.25 \\ 0.5 (3.75 + 3 + 1.75 + 0) = 4.25 \\ \int_{0}^{2} (4 - x^{2}) d x = 8 - \frac{8}{3} = \frac{16}{3} \end{array}

3. Somewhere new

Estimate $\int_{0}^{3} (x - 2) d x$ with three midpoint rectangles. What does the sign of the answer say?

Show the answer and explanation

$- 1.5$ , which is exact. More of the region lies below the x-axis than above it.

$Δ x = 1$ , and the midpoints $0.5$ , $1.5$ and $2.5$ have heights $- 1.5$ , $- 0.5$ and $0.5$ , which add to $- 1.5$ . The line is below the axis from $0$ to $2$ , a triangle of area $2$ , and above it from $2$ to $3$ , a triangle of area $0.5$ . The signed area is $0.5 - 2 = - 1.5$ . For a straight line, each midpoint rectangle gains exactly the small triangle it loses, so the estimate is exact.

\begin{array}{l} Δ x = \frac{3 - 0}{3} = 1 \\ 1 (- 1.5 - 0.5 + 0.5) = - 1.5 \\ \int_{0}^{3} (x - 2) d x = \frac{9}{2} - 6 = - \frac{3}{2} \end{array}

Keep exploring

In Riemann Sums, raise $n$ from $2$ to $4$ and then $8$ . The midpoint sum becomes $0.328125$ and then $0.33203125$ : each doubling of $n$ makes the error four times smaller.

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Choose the partition

Left sum

Right sum

Midpoint sum

Compare with the exact value

Result

Your turn

1. One like the example

L2=5L2​=5, R2=13R2​=13 and M2=8.5M2​=8.5, against the exact 26⁢/3≈8.66726/3≈8.667.

2. A variation

L4=6.25L4​=6.25 and R4=4.25R4​=4.25, against the exact 16⁢/3≈5.33316/3≈5.333.

3. Somewhere new

−1.5−1.5, which is exact. More of the region lies below the x-axis than above it.

Keep exploring

Study it, then try it

$L_{2} = 5$ , $R_{2} = 13$ and $M_{2} = 8.5$ , against the exact $26 / 3 \approx 8.667$ .

$L_{4} = 6.25$ and $R_{4} = 4.25$ , against the exact $16 / 3 \approx 5.333$ .

$- 1.5$ , which is exact. More of the region lies below the x-axis than above it.