Exercises — Riemann sums — left, right, midpoint; formal definition of definite integral

What we're doing: just decoding symbols — no computation of areas yet. $\Delta x=\frac{b-a}{n}=\frac{6-2}{8}=\frac{1}{2}.$ Grid points $x_i=a+i\Delta x=2+\frac{i}{2}$: $x_0=2,x_1=2.5,x_2=3,x_3=3.5,x_4=4.$ The third subinterval is $[x_2,x_3]=[3,3.5]$. Its midpoint sample point is $x_3^{\ast }=\frac{x_2+x_3}{2}=\frac{3+3.5}{2}=\mathrm{3.25.}$ Answer: $\Delta x=\frac{1}{2}$; grid starts $2,2.5,3,3.5,4$; third midpoint $x_3^{\ast }=3.25$.

What we're doing: connecting sampling location to over/under behaviour for an increasing $f$. Read the panels by position, not colour.

• Leftmost panel — Left rule. Each rectangle takes its height at the left edge of its strip. On a rising curve the left edge is the lowest point of that strip, so every rectangle sits below the curve → underestimate.
• Rightmost panel — Right rule. Each rectangle takes the right edge, the highest point of a rising strip → rectangles poke above the curve → overestimate.
• Centre panel — Midpoint rule. Height taken at the strip's centre; the sliver poking out above roughly cancels the sliver missing below → closest to the true area. Answer: Leftmost = Left = under; Rightmost = Right = over; Centre = Midpoint = best. (The colours in the figure — burnt orange, plum, teal — merely decorate these three; the position and edge-choice are what identify them.)

Setup: $\Delta x=\frac{2-0}{4}=0.5$, grid $x_i=0.5i$: $0,0.5,1,1.5,2$.

Left samples $x_0,x_1,x_2,x_3=0,0.5,1,1.5$ (drop the last point): $L_4=(f(0)+f(0.5)+f(1)+f(1.5))\Delta x=(0+0.25+1+2.25)(0.5)=3.5\times 0.5=\mathrm{1.75.}$ Right samples $x_1,x_2,x_3,x_4=0.5,1,1.5,2$ (drop the first point): $R_4=(0.25+1+2.25+4)(0.5)=7.5\times 0.5=\mathrm{3.75.}$ Midpoint samples $0.25,0.75,1.25,1.75$: $M_4=(0.0625+0.5625+1.5625+3.0625)(0.5)=5.25\times 0.5=\mathrm{2.625.}$ Check the ordering: $f$ is increasing, so $L_4=1.75<M_4=2.625<R_4=3.75$, and the true area is ${\int }_0^2{x}^{2}dx=\frac{8}{3}\approx 2.667$. Midpoint is closest. ✓

Setup: $\Delta x=\frac{4-0}{4}=1$, grid $0,1,2,3,4$. Right samples $x_1,\dots ,x_4=1,2,3,4$. $R_4=(f(1)+f(2)+f(3)+f(4))(1)=(-1)+(0)+(1)+(2)=2.$ Interpreting the sign: where $f(x)<0$ (for $x<2$) the "height" is negative, so its rectangle contributes a negative area. This is Signed area: area below the axis counts as minus. Here the negative chunk $(-1)$ and positive chunk $(+3)$ net to $+2$. The true integral ${\int }_0^4(x-2)dx=\frac{x^2}{2}-2x{|}_0^4=8-8=0$; our coarse $R_4=2$ overshoots (increasing function ⇒ right overestimates) but the concept — signed contributions — is the point. ✓

The key idea: the "left over/under, right over/under" rule from the parent note only applies to a monotonic function. On $[0,\pi ]$, $\cos x$ rises? No — it falls from $1$ down to $-1$. It is monotonic decreasing on the whole interval.

• Decreasing ⇒ left edge is the tallest point of each strip ⇒ Left overestimates.
• Decreasing ⇒ right edge is the lowest ⇒ Right underestimates.
• Midpoint: best, by cancellation. Confirm exactly: ${\int }_0^{\pi }\cos xdx=\sin x{|}_0^{\pi }=\sin \pi -\sin 0=0-0=0.$ So the true value is $0$; $L_n>0>R_n$ around it. ✓ (This uses the Fundamental Theorem of Calculus as an oracle — legal for checking, not for the from-scratch derivations of L5.)

What we're doing: subtract two sums term by term and watch a telescope collapse. $L_n={\sum }_{i=1}^{n}f(x_{i-1})\Delta x=\Delta x(f(x_0)+f(x_1)+\cdots +f(x_{n-1})),$ $R_n={\sum }_{i=1}^{n}f(x_i)\Delta x=\Delta x(f(x_1)+f(x_2)+\cdots +f(x_n)).$ Why line these up: every interior term $f(x_1),\dots ,f(x_{n-1})$ appears in both lists, so it cancels in the difference. Only the two ends survive: $R_n-L_n=\Delta x(f(x_n)-f(x_0))=(f(b)-f(a))\Delta x.$ What it means: $\Delta x=\frac{b-a}{n}\to 0$ as $n\to \infty$, and $f(b)-f(a)$ is a fixed number. So $R_n-L_n\to 0$: left and right sums squeeze together toward the same limit — a first glimpse of why the integral is well-defined (see Continuity and Integrability). ✓

Strategy: reverse-engineer the pieces $f$, $a$, $b$, $\Delta x$ from the pattern $\sum f(x_i)\Delta x$.

• The lone $\frac{1}{n}$ is the width $\Delta x$. Since $\Delta x=\frac{b-a}{n}$, this means $b-a=1$.
• The sample point is $\frac{i}{n}$. With $a=0$, $x_i=a+i\Delta x=\frac{i}{n}$ ✓ — a right sum on $[0,1]$.
• The height is $\sqrt{1+x}$ evaluated at $x=\frac{i}{n}$, so $f(x)=\sqrt{1+x}$. Therefore ${\lim }_{n\to \infty }{\sum }_{i=1}^n\frac{1}{n}\sqrt{1+\frac{i}{n}}={\int }_0^1\sqrt{1+x}dx.$ Evaluate (antiderivative of $(1+x{)}^{1/2}$ is $\frac{2}{3}(1+x{)}^{3/2}$): ${\int }_0^1\sqrt{1+x}dx=\frac{2}{3}(1+x{)}^{3/2}{|}_0^1=\frac{2}{3}(2^{3/2}-1)=\frac{2}{3}(2\sqrt{2}-1)\approx \mathrm{1.2190.}$ Answer: $\frac{2}{3}(2\sqrt{2}-1)\approx 1.219$. This is exactly the trick Area under a curve problems use in reverse. ✓

(a) The Trapezoidal Rule averages left and right: $T_4=\frac{1}{2}(1.75+3.75)=\frac{1}{2}(5.5)=\mathrm{2.75.}$ (b) Errors (signed, true − estimate): $\text{error}(T_4)=2.6667-2.75=-0.0833,\text{error}(M_4)=2.6667-2.625=+\mathrm{0.0417.}$ Sizes: $|-0.0833|=0.0833$ vs $0.0417$. So midpoint's error is about half the trapezoid's, and opposite in sign — the classic $2:1$ relationship. Answer: $T_4=2.75$; midpoint wins by roughly a factor $2$. ✓ (Push $n$ higher and both shrink; see Simpson's Rule which cleverly combines $M$ and $T$ to kill even more error.)

Step 1 — set up the grid. $a=1,b=3$, so $\Delta x=\frac{3-1}{n}=\frac{2}{n}$, and $x_i=a+i\Delta x=1+\frac{2i}{n}.$ Why: right sums evaluate at these grid points.

Step 2 — write the sum. With $f(x)=2x+1$, $f(x_i)=2(1+\frac{2i}{n})+1=3+\frac{4i}{n}.$ $R_n={\sum }_{i=1}^n(3+\frac{4i}{n})\frac{2}{n}=\frac{2}{n}{\sum }_{i=1}^{n}3+\frac{2}{n}\cdot \frac{4}{n}{\sum }_{i=1}^{n}i.$ Why we may split like this: a sum of two things added together is the sum of each piece separately (addition can be regrouped), and any factor that does not depend on the counter $i$ — here $\frac{2}{n}$ and $\frac{4}{n}$ — is the same in every term, so it can be pulled outside the $\sum$ as a common multiplier. What's left inside is a pure $\sum i$ we have a formula for.

Step 3 — apply summation formulas. ${\sum }_{i=1}^{n}3=3n$ (adding the constant $3$ a total of $n$ times) and ${\sum }_{i=1}^{n}i=\frac{n(n+1)}{2}$: $R_n=\frac{2}{n}(3n)+\frac{8}{n^2}\cdot \frac{n(n+1)}{2}=6+\frac{4(n+1)}{n}=6+4+\frac{4}{n}=10+\frac{4}{n}.$ Why: substituting the closed forms turns the sum into ordinary algebra we can simplify.

Step 4 — take the limit. As $n\to \infty$, the tail $\frac{4}{n}\to 0$ (a fixed top over an exploding bottom shrinks to nothing): ${\int }_1^3(2x+1)dx={\lim }_{n\to \infty }(10+\frac{4}{n})=10.$ Cross-check with the antiderivative: $x^2+x{|}_1^3=(9+3)-(1+1)=12-2=10.$ ✓ Watch the endpoints: the answer is the difference $12-2=10$, not the upper value $12$ alone — the Riemann limit already subtracts the lower endpoint for you. ✓

Step 1 — grid. $\Delta x=\frac{1}{n}$, $x_i=\frac{i}{n}$. Step 2 — sum. $R_n={\sum }_{i=1}^n(\frac{i}{n}{)}^3\frac{1}{n}=\frac{1}{n^4}{\sum }_{i=1}^n{i}^3.$ Why $\frac{1}{n^4}$ comes out front: $\frac{1}{n^3}$ from cubing $\frac{1}{n}$, times $\frac{1}{n}$ from $\Delta x$, and none of it depends on the counter $i$, so it pulls outside the $\sum$ leaving a pure $\sum i^3$. Step 3 — plug $\sum i^3={\left[\frac{n(n+1)}{2}\right]}^2=\frac{n^2(n+1{)}^2}{4}$: $R_n=\frac{1}{n^4}\cdot \frac{n^2(n+1{)}^2}{4}=\frac{(n+1{)}^2}{4n^2}=\frac{n^2+2n+1}{4n^2}.$ Step 4 — limit. Divide top and bottom by $n^2$ (the fastest-growing power, so the surviving pieces become constants): $\frac{1+2/n+1/n^2}{4}\to \frac{1}{4}$. ${\int }_0^1{x}^{3}dx=\frac{1}{4}.✓$

(a) Zero-width interval — argued through the mesh. We are not allowed to set $\Delta x\equiv 0$ and call it a partition; the definition works with genuine partitions whose mesh $\Vert P\Vert ={\max }_i\Delta x_i$ is then driven to $0$. So take any honest partition of the interval $[a,a]$. But $[a,a]$ is a single point: its total length is $b-a=a-a=0$, and the widths of its subintervals are non-negative numbers summing to $0$, hence every $\Delta x_i=0$. Therefore, for any partition (its mesh is already $0$) and any choice of sample points $x_i^{\ast }$, $S_n={\sum }_{i=1}^{n}f(x_i^{\ast })\Delta x_i={\sum }_{i=1}^{n}f(x_i^{\ast })\cdot 0=0.$ Every admissible Riemann sum equals $0$, so the limit as $\Vert P\Vert \to 0$ is trivially $0$: ${\int }_a^{a}f(x)dx=0.$ The point: we did not assume "$\Delta x=0$ is a partition." We looked at legitimate partitions of a degenerate interval and found the widths are forced to $0$ by the interval having zero length. ✓

(b) Dirichlet function. Let $D(x)=\{\begin{array}{ll}1,& x\text{ rational}\\ 0,& x\text{ irrational.}\end{array}$ It is bounded ($0\le D\le 1$). Fix any partition of $[0,1]$ into $n$ strips of width $\Delta x=\frac{1}{n}$. In each subinterval, no matter how tiny, there is both a rational and an irrational number (both kinds are dense — see Limits of sequences). The definition demands the same limit for every legal choice of sample points $x_i^{\ast }$. Test two legal choices:

• All $x_i^{\ast }$ rational: every height $D(x_i^{\ast })=1$, so $S_n={\sum }_{i=1}^{n}1\cdot \Delta x=n\cdot \frac{1-0}{n}=1\text{for every }n\Rightarrow {\lim }_{\Vert P\Vert \to 0}{S}_n=1.$
• All $x_i^{\ast }$ irrational: every height $D(x_i^{\ast })=0$, so $S_n={\sum }_{i=1}^{n}0\cdot \Delta x=0\text{for every }n\Rightarrow {\lim }_{\Vert P\Vert \to 0}{S}_n=0.$ Two legal sampling choices give limits $1$ and $0$. Since $1\ne 0$, no single number $I$ can be the limit for all choices, so $D$ fails the integrability requirement — it is not Riemann integrable. This is exactly why the parent's phrase "for every choice of sample points" is the definition's teeth, not decoration (contrast with Continuity and Integrability, where continuity guarantees one common limit). ✓