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

True — on each strip the left edge is the lowest point of a rising curve, so every rectangle sits under the curve. This fails the instant the function stops rising, so "increasing" is essential.

True — on a line the triangle poking out above the midpoint height exactly matches the triangle missing below it, so the errors cancel on every strip regardless of how many strips you use.

Careful — this alone is not sufficient; the definition demands the same limit for every sample choice, not just left and right. In practice for continuous $f$ it is enough, but as a logical statement it is false (a function could agree on edges yet diverge for cleverer samples).

False — heights alone are not area. You must multiply each height by the strip width: area is $\sum f(x_i^{\ast })\Delta x_i$. Dropping $\Delta x$ gives a number with the wrong units entirely.

False in general — for a non-uniform partition you could add many tiny strips while leaving one fat strip untouched, so $\Vert P\Vert$ never shrinks. Only for uniform partitions does $n\to \infty$ force $\Vert P\Vert \to 0$.

True — it computes signed area, so any part of $f$ below the $x$-axis contributes negatively. See Signed area; the integral is area only when $f\ge 0$ on $[a,b]$.

True — this is a core theorem (see Continuity and Integrability). But the converse is false: a function can be integrable while having jumps, so continuity is sufficient, not necessary.

False — that average is the Trapezoidal rule, not the midpoint rule (see Trapezoidal Rule). They are genuinely different approximations, and midpoint usually beats the trapezoid despite using one sample per strip.

False — for a uniform partition $R_n-L_n=\Delta x(f(b)-f(a))$, so the sums are equal whenever $f(a)=f(b)$, which many non-constant functions satisfy (e.g. a symmetric bump).

True — for a monotone function left over/under-estimates and right does the opposite, so the truth lies between them; midpoint, being close to the truth, also sits between. (For non-monotone $f$ this ordering can break.)

The left edge of strip $i$ is $x_{i-1}=a+(i-1)\Delta x$, not $x_i$. Using $x_i$ silently computes the right sum instead — a classic off-by-one.

You may pick easy samples only after knowing $f$ is integrable, because integrability guarantees all choices give the same limit. Choosing samples to prove integrability itself is circular — you must know the limit is choice-independent first.

$x^2$ is increasing on $[0,1]$, so the right sum overestimates — every $R_n>1/3$, approaching from above. Right sums undercount only for decreasing functions.

It is not Riemann integrable at all: sampling rationals gives sum $1$, sampling irrationals gives $0$, so no single limit exists. "Almost everywhere" is a Lebesgue idea, not a Riemann one.

More samples does not mean more accurate. Midpoint's single centred sample benefits from symmetric error cancellation; its error is typically about half the trapezoid's and of opposite sign. Count of samples ≠ quality.

It does not vanish — it becomes $dx$. The symbol $dx$ in $\int fdx$ is the surviving trace of the width $\Delta x$ after the limit, which is why forgetting it is such a serious error.

Continuity is sufficient but not necessary. A function with finitely many jumps (e.g. a step function) is still integrable, because the misbehaving strips shrink to zero total width as $\Vert P\Vert \to 0$.

Because if left, right, and any other sampling all agreed, the answer reflects a genuine property of $f$ rather than an artefact of our arbitrary choice. A choice-dependent "answer" would be meaningless as the area.

The average could stay small while one stubborn fat strip remains, spoiling the approximation there. Forcing the maximum to zero guarantees every strip vanishes, so the staircase hugs the curve everywhere.

It stands for Sum — the integral is the limit of the Riemann sum $\sum f(x_i^{\ast })\Delta x$, so Leibniz turned the summation into a stretched S and the discrete $\Delta x$ into the infinitesimal $dx$.

The centred sample lets the strip's overshoot above the curve on one side cancel its undershoot on the other. Left/right samples sit at an extreme edge, so their errors accumulate in one direction with nothing to cancel them.

Any finite $n$ leaves a nonzero gap between the staircase and the curve, so it is an approximation, not the exact area. Only the limit $n\to \infty$ removes the last sliver of error; that exact value is what "definite integral" means.

On a line the midpoint height equals the average height across the strip, so the rectangle area equals the true (trapezoidal) area exactly. Left and right heights are not the average, so they over/undershoot by the slope times half the width.

It is $0$ — the interval has zero width, so every $\Delta x_i=0$ and every rectangle has zero area. There is literally no region to measure.

The integral is $0$: as strips thin, the probability any sample lands exactly on the spike vanishes, and even a strip containing it contributes height $\times$ (a width shrinking to $0$). A single point has no area.

All three are identical and exactly equal to $c(b-a)$ for every $n$, since the height is $c$ no matter where you sample. A flat function has no rise for any rule to miss.

Strips where $f<0$ contribute negative rectangle areas, so the sum (and the limit) computes signed area — positive parts minus negative parts, not total geometric area. See Area under a curve vs Signed area.

No — the fat strip dominates the error and $\Vert P\Vert$ is large, so the approximation is poor there. Accuracy is controlled by the widest strip, not by how many strips you have.

It is $0$. Since $L_n-R_n=\Delta x(f(a)-f(b))$ and $\Delta x\to 0$ with $f(a),f(b)$ fixed and finite, the whole difference collapses — both sums converge to the same integral.