Corrected exercises: numerical integration, 201-SN3-RE (201-NYB)

First reason: the function has NO elementary antiderivative. This is the case for $e^{-x^{2}}$ (the normal distribution in statistics), $\frac{\sin x}{x}$ (optics) and $\sqrt{1+x^{3}}$: no finite combination of the usual functions integrates them, and this is a theorem, not an admission of ignorance. Second reason: there is NO formula for $f$ at all, only measured values. A speed reading every two seconds, a river flow recorded every hour, a pressure curve coming out of a sensor: the Fundamental Theorem requires an antiderivative, hence a formula, and there is none. In both cases numerical integration is not an approximation of convenience, it is the only route available.

$h=\frac{b-a}{n}=\frac{2-1}{4}=0.25$. Points: $x_{0}=1$, $x_{1}=1.25$, $x_{2}=1.5$, $x_{3}=1.75$, $x_{4}=2$. Values of $f(x)=\frac{1}{x}$: $1$; $0.8$; $0.6666667$; $0.5714286$; $0.5$. Left rectangles (leave out the LAST value): $L_{4}=0.25\left[1+0.8+0.6666667+0.5714286\right]=0.25\times 3.0380952=0.7595238$. The gap with $\ln 2$ is $0.0663766$: a large overestimate, as expected, since $\frac{1}{x}$ is decreasing and each rectangle is built on its higher edge.

Midpoints of the four subintervals: $1.125$; $1.375$; $1.625$; $1.875$. Values: $0.8888889$; $0.7272727$; $0.6153846$; $0.5333333$. So $M_{4}=0.25\times 2.7648795=0.6912199$. Error: $\left|0.6912199-0.6931472\right|=0.0019273$, against $0.0663766$ for $L_{4}$: the MIDPOINT rule is about 34 times better. This could be predicted: on each subinterval, the midpoint rectangle crosses the curve and partly makes up on one side for what it cuts off on the other, whereas the left rectangle always errs in the same direction. An error that alternates cancels out; an error that accumulates never does.

These sums ARE Riemann sums, except that we do not take the limit. By definition, $\int_{a}^{b}f(x)dx=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_{i}^{*})\Delta x$ for any choice of sample points $x_{i}^{*}$ in the subintervals: taking the left endpoint gives $L_{n}$, the right endpoint $R_{n}$, the centre $M_{n}$. The chapter on the definite integral used these sums to DEFINE the integral and prove the Fundamental Theorem, by pushing them to infinity; numerical integration goes the other way and uses them to COMPUTE, stopping at a finite $n$. The whole question then becomes: what error do we make by stopping, and that is the subject of Part B.

On the subinterval $\left[x_{i-1},x_{i}\right]$, replace the curve with the segment joining $\left(x_{i-1},f(x_{i-1})\right)$ to $\left(x_{i},f(x_{i})\right)$. The resulting figure is a trapezoid with “bases” $f(x_{i-1})$ and $f(x_{i})$ and height $h$, so its area is $\frac{h}{2}\left[f(x_{i-1})+f(x_{i})\right]$. Summing over the $n$ subintervals: $T_{n}=\frac{h}{2}\sum_{i=1}^{n}\left[f(x_{i-1})+f(x_{i})\right]$. The coefficients $1, 2,\ldots,2, 1$ then explain themselves: each INTERIOR point $x_{1},\ldots,x_{n-1}$ belongs to TWO consecutive trapezoids (it is the right end of one and the left end of the next), so it is counted twice; the two END points $x_{0}$ and $x_{n}$ belong to a single trapezoid and are counted once. Each roof joins the curve at both ends of its subinterval, so the rule is exact for every linear function, where a left or right rectangle is exact only for constants: this is why the error falls like $\frac{1}{n^{2}}$ instead of $\frac{1}{n}$ (exercise 6).

With the values already computed: $T_{4}=\frac{0.25}{2}\left[1+2(0.8)+2(0.6666667)+2(0.5714286)+0.5\right]=0.125\left[1+1.6+1.3333333+1.1428571+0.5\right]=0.125\times 5.5761905=0.6970238$. Error: $\left|0.6970238-0.6931472\right|=0.0038766$. Compare with the $0.0663766$ of the left rectangles: here the trapezoidal rule is 17 times more accurate, for exactly the same number of function evaluations. It is the best return on effort in the chapter.

$R_{4}$ is obtained by leaving out the FIRST value: $R_{4}=0.25\left[0.8+0.6666667+0.5714286+0.5\right]=0.25\times 2.5380952=0.6345238$. Average: $\frac{L_{4}+R_{4}}{2}=\frac{0.7595238+0.6345238}{2}=\frac{1.3940476}{2}=0.6970238=T_{4}$ ✓. The relation can be read straight off the formula: adding $L_{n}$ and $R_{n}$ counts each interior point twice and the endpoints once, which is exactly the pattern $1, 2,\ldots,2, 1$; all that is left is to divide by 2. In practice, this means the trapezoidal rule costs NOTHING more than the rectangles: it simply averages the two errors, which have opposite signs.

$T_{4}=0.6970238>\ln 2$: the rule OVERESTIMATES. Explanation by concavity: $f(x)=\frac{1}{x}$ has second derivative $f''(x)=\frac{2}{x^{3}}>0$ on $[1, 2]$, so the function is CONVEX (concave up). Now the chord joining two points of a convex curve lies ABOVE the curve: each trapezoid therefore holds a little more than the true area, and the sum overestimates. General rule: $f$ convex gives an overestimate with the trapezoids, $f$ concave an underestimate. The MIDPOINT rule errs in the opposite direction (it underestimated here, $M_{4}<\ln 2$ ✓), which is no coincidence and suggests that a weighted average of the two should be excellent: that is exactly what Simpson's rule is, since one can show that $S_{2n}=\frac{2M_{n}+T_{n}}{3}$.

Principle: instead of approximating $f$ with line segments, approximate it with ARCS OF PARABOLAS. A parabola $y=ax^{2}+bx+c$ has three coefficients, so THREE points are needed to determine it: the subintervals are grouped IN PAIRS and a parabola is passed through $\left(x_{0},f(x_{0})\right)$, $\left(x_{1},f(x_{1})\right)$, $\left(x_{2},f(x_{2})\right)$, then through the next three, and so on. The condition follows at once: since the subintervals must be paired, their number $n$ must be EVEN. This also explains the alternating coefficients $1, 4, 2, 4,\ldots,4, 1$: the points with an ODD index are the middle points of each parabola (coefficient 4), the interior points with an EVEN index are shared by two neighbouring parabolas (coefficient 2), and the two endpoints belong to only one parabola (coefficient 1). Three points per arc instead of two, plus the free extra degree of part d): the error then falls like $\frac{1}{n^{4}}$ (exercise 6).

$S_{4}=\frac{0.25}{3}\left[f(1)+4f(1.25)+2f(1.5)+4f(1.75)+f(2)\right]=\frac{0.25}{3}\left[1+3.2+1.3333333+2.2857143+0.5\right]=0.0833333\times 8.3190476=0.6932540$. Error: $\left|0.6932540-0.6931472\right|=0.0001068$. Compared with $T_{4}$, whose error was $0.0038766$: Simpson is about 36 times more accurate, using exactly the SAME five function values. Only the coefficients change. This is why Simpson's rule is the default method whenever there is a choice: for the same computational cost, it gains one or two orders of magnitude.

With $n=5$, the subintervals cannot be grouped in pairs: a single one would be left at the end, on which no parabola can be built for lack of a third point. The formula simply does not apply, and applying it mechanically would produce an inconsistent pattern of coefficients, hence a wrong result without any warning. This is the most frequent pitfall of the chapter on exams. The student must switch to $n=4$ or $n=6$ (one of the two neighbouring even integers), or else use the trapezoidal rule, which imposes no parity condition. Reflex to build: before writing anything for Simpson, check that $n$ is even.

Exact value: $\int_{0}^{2}x^{3}dx=\left[\frac{x^{4}}{4}\right]_{0}^{2}=\frac{16}{4}=4$. By Simpson with $n=2$, so $h=1$ and the points $0$, $1$, $2$: $S_{2}=\frac{1}{3}\left[0^{3}+4(1)^{3}+2^{3}\right]=\frac{1}{3}\left[0+4+8\right]=\frac{12}{3}=4$ ✓, exactly. The result reveals that Simpson's rule is EXACT for every polynomial of degree at most 3, although it was only built to be exact on parabolas (degree 2). This free extra degree comes from a symmetry cancellation: on each pair of subintervals, the error made on the cubic term on one side exactly cancels the error on the other side. It is precisely this exactness at degree 3 that explains the error bound in $f^{(4)}$ of exercise 6: the method only starts to err from degree 4 on, so the error is governed by the fourth derivative.

Trapezoidal error: $T_{4}-\ln 2=0.6970238-0.6931472=+0.0038766$, an OVERESTIMATE. Midpoint error: $M_{4}-\ln 2=0.6912199-0.6931472=-0.0019273$, an UNDERESTIMATE. The two errors have OPPOSITE signs, which already guarantees a bracket: $M_{4}<\ln 2<T_{4}$.

The ratio is $\frac{+0.0038766}{-0.0019273}\approx-2.01$, very close to $-2$. General relation: $E_{T}\approx-2E_{M}$; in other words, the midpoint error is HALF the trapezoidal error, with the opposite sign. Explanation by concavity: the function $\frac{1}{x}$ is convex, so the chord lies above the curve (the trapezoids overestimate) while the tangent line at the midpoint lies below it (the midpoint rule underestimates, because the midpoint rectangle has the same area as the trapezoid built on the tangent line). The factor 2 comes from the error formulas themselves: $E_{T}=-\frac{f''(c)(b-a)^{3}}{12n^{2}}$ against $E_{M}=+\frac{f''(c)(b-a)^{3}}{24n^{2}}$ (these formulas count the error as exact value minus estimate, the reverse of the convention used in part a, which does not change the ratio), and $24=2\times 12$. The midpoint rule is therefore TWICE as accurate as the trapezoids for the same number of evaluations, which makes it surprisingly competitive despite its apparent simplicity.

Computation: $\frac{2(0.6912199)+0.6970238}{3}=\frac{1.3824398+0.6970238}{3}=\frac{2.0794636}{3}=0.6931545$. And $S_{8}=0.6931545$ as well: the equality is EXACT, to every decimal place, not approximate. This number is Simpson's estimate with eight subintervals, whose error is only $7.3\times 10^{-6}$.

Proof. Let $h=\frac{b-a}{n}$. Simpson's rule with $2n$ subintervals uses the $n+1$ points of the coarse partition AND the $n$ midpoints. Grouping its coefficients $1, 4, 2, 4,\ldots,4, 1$ (with step $\frac{h}{2}$), the points of the coarse partition receive the even-position coefficients and the midpoints the coefficients 4. A direct computation gives $S_{2n}=\frac{h}{6}\left[f(x_{0})+f(x_{n})+2\sum_{\text{interior}}f(x_{i})+4\sum_{\text{midpoints}}f(m_{i})\right]$, which we recognize as $\frac{1}{3}T_{n}+\frac{2}{3}M_{n}$, that is $\frac{T_{n}+2M_{n}}{3}$ ✓. PRACTICAL CONSEQUENCE: if $T_{n}$ and $M_{n}$ have already been computed, Simpson with $2n$ subintervals comes FOR FREE, by a simple weighted average, without a single extra function evaluation. This is the principle of extrapolation: combine two mediocre estimates whose errors have opposite signs and a known ratio to build a much better one. Romberg integration generalizes this idea in a cascade.

The formula is $L=\int_{a}^{b}\sqrt{1+\left[f'(x)\right]^{2}}\,dx$. Here $f(x)=x^{2}$ gives $f'(x)=2x$, so $L=\int_{0}^{1}\sqrt{1+4x^{2}}\,dx$.

With $h=0.25$ and $g(x)=\sqrt{1+4x^{2}}$: $g(0)=1$; $g(0.25)=\sqrt{1.25}\approx 1.118034$; $g(0.5)=\sqrt{2}\approx 1.414214$; $g(0.75)=\sqrt{3.25}\approx 1.802776$; $g(1)=\sqrt{5}\approx 2.236068$. So $S_{4}=\frac{0.25}{3}\left[1+4(1.118034)+2(1.414214)+4(1.802776)+2.236068\right]=\frac{0.25}{3}\left(17.747734\right)\approx 1.478978$.

Exact value: $\frac{\sqrt{5}}{2}+\frac{\ln\left(2+\sqrt{5}\right)}{4}=1.118034+0.360909=1.478943$. Simpson's error is therefore $\left|1.478978-1.478943\right|\approx 3.5\times 10^{-5}$, a relative accuracy of $0.0024$ percent with only five function evaluations. Going to $n=8$, the error drops to $1.1\times 10^{-6}$, as the order 4 of the method predicts.

Because the formula imposes a SQUARE ROOT of a polynomial: $\sqrt{1+\left[f'(x)\right]^{2}}$. Antiderivatives of such expressions are elementary only in very special cases. As soon as $f$ is a little more complicated than a parabola, you land on so-called elliptic integrals, which are PROVEN to have no elementary antiderivative: this is the case for the perimeter of an ellipse, the length of a sine curve, or a general catenary. The parabola of part a) is a happy exception, which is also why it serves as the example in every textbook. GEOMETRIC BRACKET: the length of the arc is necessarily greater than that of the CHORD joining $(0, 0)$ to $(1, 1)$, that is $\sqrt{2}\approx 1.414$, since the segment is the shortest path; and it is less than the sum of the two sides of the enclosing rectangle, $1+1=2$. So $1.414<L<2$, and the value $1.479$ found fits well ✓. This kind of rough bracket costs ten seconds and immediately catches an error of a factor or of a sign.

Bounds: $\left|E_{T}\right|\leq\frac{K(b-a)^{3}}{12n^{2}}$ with $K\geq\left|f''\right|$ on $[a,b]$, and $\left|E_{S}\right|\leq\frac{M(b-a)^{5}}{180n^{4}}$ with $M\geq\left|f^{(4)}\right|$. For $f(x)=x^{-1}$: $f'=-x^{-2}$, $f''=2x^{-3}$, $f'''=-6x^{-4}$, $f^{(4)}=24x^{-5}$. On $[1, 2]$, these two functions decrease in absolute value, so their maximum is reached at $x=1$: $K=\left|f''(1)\right|=2$ and $M=\left|f^{(4)}(1)\right|=24$. Always look for the maximum of the ABSOLUTE value over the interval, not the value at some arbitrary point.

Trapezoids, $n=4$, $b-a=1$: $\left|E_{T}\right|\leq\frac{2\times 1}{12\times 16}=\frac{2}{192}=0.0104167$. Actual error observed: $0.0038766\leq 0.0104167$ ✓. Simpson, $n=4$: $\left|E_{S}\right|\leq\frac{24\times 1}{180\times 256}=\frac{24}{46080}=0.0005208$. Actual error: $0.0001068\leq 0.0005208$ ✓. Both bounds hold, and they are pessimistic by a factor of 3 to 5, which is normal and desirable: a bound must guarantee, not predict.

With $n=8$ (so $h=0.125$): $T_{8}=0.6941219$, with error $0.0009747$; and $S_{8}=0.6931545$, with error $0.0000073$. Reduction factors: for the trapezoids, $\frac{0.0038766}{0.0009747}\approx 3.98$, that is $4$; for Simpson, $\frac{0.0001068}{0.0000073}\approx 14.5$, that is about $16$. These factors can be read directly from the formulas: the trapezoidal error goes like $\frac{1}{n^{2}}$, so doubling $n$ divides it by $2^{2}=4$; Simpson's error goes like $\frac{1}{n^{4}}$, so it is divided by $2^{4}=16$. The trapezoidal rule is said to be of ORDER 2 and Simpson's rule of ORDER 4. Decisive practical consequence: to gain one more significant figure, you must roughly triple $n$ with the trapezoids, but only multiply it by $1.8$ with Simpson.

They are bounds because an unknown quantity is replaced by its worst case. The exact theorems state that there EXISTS a point $c\in[a,b]$ such that $E_{T}=-\frac{f''(c)(b-a)^{3}}{12n^{2}}$, and likewise for Simpson with $f^{(4)}(c)$. The problem is that the theorem guarantees that this $c$ exists without locating it: we do not know where it is. Without it, we bound $\left|f''(c)\right|$ by the maximum $K$ over the whole interval, which turns a useless equality into a usable inequality. To get the exact error you would therefore need $c$; but if you knew it, you would also know the exact value of the integral, and the approximation would serve no purpose. The reasoning is the same as for the Mean Value Theorem of differential calculus, and for the Taylor remainder: an inaccessible piece of information is traded for a computable guarantee.

We require $\left|E_{T}\right|\leq\frac{K(b-a)^{3}}{12n^{2}}<10^{-4}$ with $K=2$ and $b-a=1$, that is $\frac{2}{12n^{2}}<10^{-4}$, or $\frac{1}{6n^{2}}<10^{-4}$, so $n^{2}>\frac{10^{4}}{6}=1666.67$, so $n>40.82$. Since $n$ is an integer, we need $n\geq 41$: take $n=41$. Check of the bound: $\frac{2}{12\times 1681}=9.91\times 10^{-5}<10^{-4}$ ✓.

We require $\frac{M(b-a)^{5}}{180n^{4}}<10^{-4}$ with $M=24$, that is $\frac{24}{180n^{4}}<10^{-4}$, so $n^{4}>\frac{24\times 10^{4}}{180}=1333.33$, so $n>6.04$. The next integer would be 7, but $n$ must be EVEN for Simpson: so take $n=8$. Check: $\frac{24}{180\times 4096}=3.26\times 10^{-5}<10^{-4}$ ✓. Note in passing that $n=6$ would not be enough, since $\frac{24}{180\times 1296}=1.03\times 10^{-4}$, very slightly above the threshold: rounding up to the next even integer is not an unnecessary precaution, here it is the difference between guaranteeing and not guaranteeing.

Trapezoids: 41 subintervals, so 42 evaluations of $f$. Simpson: 8 subintervals, so 9 evaluations. Simpson needs about FIVE TIMES fewer computations for the same guaranteed accuracy. The gap widens as the requirement gets stricter: for an error less than $10^{-8}$, the trapezoids would need $n=4083$ while Simpson would settle for $n=62$, a ratio of 66 instead of 5. This is the concrete meaning of the difference between order 2 and order 4, and it is the reason why no serious software uses the trapezoids when Simpson is available.

It is a mistake because the condition obtained is $n>40.82$, a STRICT inequality pointing up: rounding to 40 gives an $n$ that does NOT satisfy the condition, and the required accuracy is then no longer guaranteed. Direct check: with $n=40$, the bound is $\frac{2}{12\times 1600}=1.042\times 10^{-4}>10^{-4}$, it does exceed the threshold. You must therefore round UP to the next integer, here 41, and not to the nearest integer. The general rule of the chapter: in a computation of $n$ that guarantees an accuracy, ALWAYS round up, then move up to the next even integer if the method is Simpson's. Taking a larger $n$ than necessary only costs computing time; taking one that is too small invalidates the guarantee, which is the only thing the exercise asked you to produce.

With $h=0.25$ and $f(x)=e^{-x^{2}}$: $f(0)=1$; $f(0.25)=e^{-0.0625}=0.9394131$; $f(0.5)=e^{-0.25}=0.7788008$; $f(0.75)=e^{-0.5625}=0.5697829$; $f(1)=e^{-1}=0.3678794$. So $S_{4}=\frac{0.25}{3}\left[1+4(0.9394131)+2(0.7788008)+4(0.5697829)+0.3678794\right]=\frac{0.25}{3}\times 8.9622646=0.7468554$. The true value is $0.7468241$: the estimate is correct to within $3\times 10^{-5}$. The Fundamental Theorem is useless because $e^{-x^{2}}$ has NO elementary antiderivative: it is not that we cannot find it, it is proven that none exists among the usual functions. Yet this integral is one of the most important in science, since it gives the probabilities of the normal distribution; the normal tables of your statistics courses were built exactly this way, by numerical integration.

There are 7 readings, so $n=6$ subintervals of width $h=2$ s (and $n$ is even, so Simpson applies). Trapezoids: $T_{6}=\frac{2}{2}\left[0+2(5+12+18+21+22)+22.5\right]=1\times\left[0+2(78)+22.5\right]=178.5$ m. Simpson: $S_{6}=\frac{2}{3}\left[0+4(5)+2(12)+4(18)+2(21)+4(22)+22.5\right]=\frac{2}{3}\left[0+20+24+72+42+88+22.5\right]=\frac{2}{3}\times 268.5=179.0$ m. The two estimates of the distance travelled in 12 seconds are therefore $178.5$ m and $179.0$ m, in agreement to within $0.3$ percent. The principle used is that distance is the integral of speed, $d=\int_{0}^{12}v(t)\,dt$.

NO, not with the formulas of this chapter. Both error bounds involve $\max\left|f''\right|$ or $\max\left|f^{(4)}\right|$, and here we have no FORMULA for $v(t)$: only seven numbers. So we can neither differentiate the function, nor even claim that it is four times differentiable, nor know what it does BETWEEN two readings (the vehicle may have braked hard at $t=3$ s without the data showing it). This is an important limitation, often passed over in silence: numerical integration of experimental data gives an estimate with no mathematical guarantee. The only way to quantify the uncertainty would be statistical or physical, relying on the precision of the sensor and on an assumption about how smooth the motion is, not on the theorems of this chapter. In practice, the gap between the results of two different methods (here $0.5$ m) serves as an empirical indicator, without being a proof.

The order of the questions. 1) Is an elementary ANTIDERIVATIVE within reach, possibly after substitution, integration by parts or partial fractions? If so, the Fundamental Theorem gives the EXACT value, and it is always the best answer. 2) If not, is there a FORMULA for $f$? If so, two options: the Taylor SERIES expansion integrated term by term, which gives a value as accurate as you want with control of the remainder (the method seen in the chapter on power series), or NUMERICAL INTEGRATION, with Simpson by default and a computable error bound. The choice between the two is a matter of convenience: the series is excellent near the centre of the expansion, the numerical method does not care. 3) Do we only have DATA? Then only numerical integration applies, with no mathematical error bound, and Simpson is preferred if the number of subintervals is even, the trapezoids otherwise. The course follows exactly this order, from the most exact to the most general: this chapter is the last resort, and that is precisely why it is the most used outside the classroom.

Yes, perfectly. The trapezoidal bound goes like $\frac{1}{n^{2}}$, so multiplying $n$ by 10 should divide the error by $10^{2}=100$. The observed values confirm it exactly: $6.25\times 10^{-4}$, then $6.25\times 10^{-6}$, then $6.25\times 10^{-8}$, a factor of 100 each time. The coefficient $6.25\times 10^{-2}$ stays constant, which means that the error is approximately $\frac{0.0625}{n^{2}}$. One can also check consistency with the theoretical bound $\frac{K(b-a)^{3}}{12n^{2}}=\frac{2}{12n^{2}}\approx\frac{0.167}{n^{2}}$: the actual error is about 2.7 times smaller than the bound, which is normal, since a bound is made to guarantee and not to predict.

No, the law stops being followed. Going from $n=10^{6}$ to $n=10^{7}$, the error should be divided by 100, so drop from $6.25\times 10^{-14}$ to about $6.25\times 10^{-16}$. It actually drops to $6.7\times 10^{-16}$, which is still close, but any further progress becomes impossible: at $n=10^{8}$ the computed error is zero, not because the result is exact, but because the resolution of the machine has been reached. What is happening is that the TRUNCATION error (the one in the formulas, in $\frac{1}{n^{2}}$) has come down to the level of the ROUND-OFF error, which comes from storing numbers in floating point and from accumulating millions of additions.

A real number is stored in a computer with about 16 significant figures, a relative precision of $2.2\times 10^{-16}$. Since the value sought, $\ln 2\approx 0.693$, is of order 1, no method can deliver a result more accurate than a few times $10^{-16}$ in absolute value: it is an impassable FLOOR, independent of the formula used. And at $n=10^{7}$ we are already there. Going further would multiply the computing time by 10, 100, 1000, without ever gaining a single correct digit. Worse, the sum of $10^{8}$ terms itself accumulates round-off, so that beyond a certain point the accuracy can actually GET WORSE. The error formula, for its part, naively promises infinite accuracy: it describes exact mathematics, not machine arithmetic.

In practice, choose the smallest $n$ that reaches the accuracy ACTUALLY needed, never the largest possible. A physical measurement is rarely known to better than four digits, so aiming for $10^{-6}$ is already generous, and $n$ of the order of a few hundred is more than enough with the trapezoids. The comparison with Simpson is telling: to reach $10^{-8}$, the trapezoids need $n\approx 2500$ while Simpson settles for $n=44$, its actual error being about $\frac{0.03}{n^{4}}$, nearly 60 times fewer evaluations. And since Simpson is of order 4, it hits the machine floor with a few thousand points where the trapezoids need tens of millions. The practical rule is therefore twofold: choose the method of highest order available, then the smallest $n$ that meets the need, and remember that below about $10^{-15}$, no refinement of the method helps any more.

The trapezoidal rule requires evaluating the function at BOTH endpoints, but $f(0)=\frac{1}{\sqrt{0}}$ does not exist: the function tends to $+\infty$ at 0. The computation stops on a division by zero. This is a structural flaw of the so-called CLOSED rules, those that use the endpoints of the interval, and it affects Simpson's rule as well.

Successive errors: $2-1.8089=0.1911$; $2-1.9395=0.0605$; $2-1.9809=0.0191$; $2-1.9940=0.0060$. Ratios: $\frac{0.1911}{0.0605}\approx 3.16$; $\frac{0.0605}{0.0191}\approx 3.16$; $\frac{0.0191}{0.0060}\approx 3.16$. The error is therefore divided by only about $3.16$, that is $\sqrt{10}$, each time $n$ is multiplied by 10.

The expected order 2 would mean a division by $10^{2}=100$ with each tenfold increase; we observe $\sqrt{10}\approx 3.16$, which corresponds to convergence like $n^{-1/2}$ only. The method is therefore DRAMATICALLY slower than promised: to gain a single decimal digit, $n$ must be multiplied by 100, and reaching $10^{-6}$ would require $n\approx 10^{12}$, which is out of reach. The violated assumption concerns the DERIVATIVE: the bound $\left|E_{M}\right|\leq\frac{K(b-a)^{3}}{24n^{2}}$ assumes there is a $K$ bounding $\left|f''\right|$ over the whole interval. But $f(x)=x^{-1/2}$ gives $f''(x)=\frac{3}{4}x^{-5/2}$, which tends to $+\infty$ as $x\to 0^{+}$: no finite $K$ exists, the bound is EMPTY, and it says strictly nothing. The method still converges, but nothing guarantees how fast any more, and the observation shows that it is much more slowly.