Definite Integrals

01 / UNDERSTAND & APPLY

1. A weighted sum becomes an accumulation

Suppose a varying quantity is sampled across an interval. A sample value alone is not the contribution of that piece; multiply it by the piece’s width. Summing value times width gives a weighted accumulation. The definite integral is the stable limit as the pieces become arbitrarily fine, provided that limit exists and does not depend on the allowed sample choices.

For a velocity function, integrating velocity over time gives displacement. Negative velocity contributes negative displacement. For a nonnegative density, integrating density over length gives mass. For a graph, positive heights contribute positively and negative heights negatively. The same mathematical operation represents different quantities because the units of the sampled value and the width differ.

The variable of integration is a dummy variable. The expressions ∫ₐᵇ f(x)dx and ∫ₐᵇ f(t)dt have the same value when the function and bounds are unchanged. In contrast, an external parameter in f(x,a) is held fixed during integration with respect to x. Confusing the dummy variable with an endpoint or parameter leads to errors when differentiating later.

A definite integral is a number once its integrand and bounds are fixed. An indefinite integral denotes a family of antiderivatives. The constant of integration belongs in the latter; it cancels in endpoint subtraction for the former. Writing a trailing + C after evaluating a fixed definite integral misidentifies the object you have calculated.

02 / UNDERSTAND & APPLY

2. The Riemann-sum definition and its conditions

Partition [a,b] with a < b into points a = x₀ < x₁ < ⋯ < xₙ = b. On each subinterval choose a sample ξᵢ and form Σ f(ξᵢ)(xᵢ − xᵢ₋₁). The mesh is the largest subinterval width. If these sums approach the same finite limit as the mesh tends to zero, regardless of sample choices, f is Riemann integrable and the limit is ∫ₐᵇf(x)dx.

Continuous functions on closed finite intervals are Riemann integrable. Bounded functions with only finitely many jump discontinuities are also integrable, so continuity is sufficient rather than necessary. Merely having a formula does not guarantee integrability: an unbounded singularity requires a different, improper-integral treatment. A function that equals one on rationals and zero on irrationals cannot have a sample-independent Riemann limit on a nontrivial interval.

For equal widths Δx = (b − a)/n, a right-endpoint sum uses xᵢ = a + iΔx. A left-endpoint sum uses xᵢ₋₁, and a midpoint sum uses a + (i − 1/2)Δx. For an integrable function, all these standard choices converge to the same integral. They need not be equal for a finite n.

For a monotone increasing function, the left sum lies below the integral and the right sum lies above it. This follows from comparing each sample rectangle with the graph on its own subinterval. If the function is decreasing, the inequalities reverse. The gap for uniform partitions is Δx[f(b) − f(a)] in the increasing case, so it tends to zero for a fixed finite endpoint difference.

03 / UNDERSTAND & APPLY

3. Recognize the width before guessing the function

When given a limit of sums, rewrite each term until a small width and a sample value are visible. The most common pattern is (1/n)Σ f(i/n), corresponding to [0,1]. A factor (b − a)/n suggests an interval of length b − a, while the sample argument a + i(b − a)/n reveals its starting point. Both pieces matter.

For example, n/(n² + i²) = (1/n)/[1 + (i/n)²]. The width is 1/n and the function is 1/(1 + x²). If you overlook the numerator n, the whole sum acquires the wrong scale. A useful check is the size of each summand: n terms of size approximately 1/n can approach a nonzero finite limit.

Index shifts often do not change the limiting integral for continuous functions. Replacing i/n by (i − 1)/n changes right endpoints to left endpoints. Adding or removing finitely many terms of size tending to zero also leaves the limit unchanged. However, a term near a singularity may fail to vanish, so justify this argument with a bound rather than using it mechanically.

Not every sum with n in it is a Riemann sum. Sometimes a factorization reveals a divergent scale or a boundary singularity. Establish continuity or an appropriate integrability argument on the proposed interval before converting. If a denominator vanishes at an endpoint, a standard bounded Riemann-sum argument is not automatically valid.

04 / UNDERSTAND & APPLY

4. The fundamental theorem connects accumulation and slope

If f is continuous and A(x) = ∫ₐˣ f(t)dt, then A′(x) = f(x). A small increase h in the upper limit adds a thin slice whose average height approaches f(x), so [A(x + h) − A(x)]/h approaches f(x). This explains why differentiating an accumulation recovers the local rate.

The evaluation form says that if F is an antiderivative of a continuous f on [a,b], then ∫ₐᵇf(x)dx = F(b) − F(a). Different antiderivatives differ by a constant on an interval, so the constant cancels. This is a powerful computational theorem, not the definition of the definite integral.

The interval condition prevents a serious trap. An antiderivative formula may exist separately on either side of a singularity, but endpoint subtraction across that singularity is invalid. For ∫₋₁¹1/x² dx, using −1/x at the outer endpoints gives a negative number, whereas the nonnegative integrand has a divergent improper integral. The missing hypothesis is continuity or the appropriate valid integral framework across the entire interval.

At a jump discontinuity of f, the accumulation A can still be continuous, but A′ need not equal a nonexistent single limiting height there. The clean fundamental-theorem derivative statement assumes continuity at the point. When a piecewise integrand is involved, evaluate one-sided derivatives rather than quoting the theorem without checking.

05 / UNDERSTAND & APPLY

5. Orientation, splitting and bounds

Reversing the limits reverses the sign: ∫ᵦₐf = −∫ₐᵦf. Equal limits give zero. Adjacent intervals add: ∫ₐᶜf + ∫𝚌ᵇf = ∫ₐᵇf, with oriented-integral conventions. Splitting is particularly useful for piecewise formulas, absolute values, or a symmetry centered at an interior point.

Linearity allows constants to move outside an integral and sums to split. It does not turn products into products of integrals, nor quotients into quotients of integrals. The integral is a weighted accumulation, and multiplying two totals is a different operation from accumulating pointwise products.

For a ≤ b and m ≤ f(x) ≤ M, integrating the inequalities gives m(b − a) ≤ ∫ₐᵇf ≤ M(b − a). A positive integrand on an interval of positive length should give a positive integral, assuming it is positive on a nontrivial subinterval. These bounds can detect missing scale factors in a Riemann sum or an incorrect sign in endpoint subtraction.

The average value of a continuous function on [a,b], with a < b, is f_avg = [1/(b − a)]∫ₐᵇf. Continuity ensures that f attains this average at least once in the interval. The average is a value of the function, while the corresponding c satisfying f(c) = f_avg is a location; report whichever the question requests.

06 / UNDERSTAND & APPLY

6. Reflection identities are substitutions in disguise

The substitution x = a + b − t reverses an interval and gives ∫ₐᵇf(x)dx = ∫ₐᵇf(a + b − x)dx. Adding the two forms can simplify a difficult-looking numerator or denominator. This is an identity for an integrable function; the reflected integrand must be interpreted on the same valid interval.

On [−a,a], an odd integrand has zero integral and an even integrand has twice its integral over [0,a]. These are special reflection cases. The word integrand includes every factor. Multiplying an odd numerator by an odd factor makes an even product, and shifting the denominator may destroy symmetry entirely.

For ratios such as f(x)/[f(x) + f(a + b − x)], the reflected ratio adds to 1 whenever the denominator is nonzero and the integrals exist. The integral is then half the interval length. If the denominator vanishes, inspect the singularities before applying the shortcut. A formal identity away from a pole does not by itself prove the improper integral exists.

Periodicity is another interval tool. For an integrable periodic function of period T, an integral over any full period has the same value, and an integer number of full periods multiplies that value. A partial final period must be handled separately. Periodicity does not imply that a full-period integral is zero; that additionally depends on the function’s mean.

07 / UNDERSTAND & APPLY

7. Change the bounds when you change the variable

For a definite integral, substitution changes both the differential and the endpoints. If u = g(x), compute the u-values corresponding to the original x-limits. Then finish entirely in u. Alternatively, find an antiderivative, return to x, and use the original x-limits. Mixing a u-antiderivative with x-endpoints is the invalid combination.

The substitution theorem can be justified by the chain rule and endpoint evaluation when the integrand is continuous and g is differentiable on the interval. In familiar one-to-one changes of variable, it also has a direct strip-width interpretation. A decreasing substitution reverses the transformed limits; the sign from du and the limit reversal must be handled consistently.

Integration by parts has a definite form: ∫ₐᵇu v′ dx = [uv]ₐᵇ − ∫ₐᵇu′v dx. The boundary term is part of the calculation. Dropping it is especially tempting when the remaining integral looks familiar. Choose methods for the structure of the integrand; a definite integral does not always need an elementary antiderivative if symmetry or geometry already gives its value.

08 / UNDERSTAND & APPLY

8. Differentiate the endpoints with the chain rule

If G(x) = ∫ₐ^g(x)f(t)dt and f is continuous on the relevant range, write G = A∘g with A(u) = ∫ₐᵘf(t)dt. The chain rule gives G′(x) = f(g(x))g′(x). The derivative is not simply f(x), because the endpoint may move at a different rate from x.

With two moving endpoints, split at a fixed reference value: ∫ₕ₍ₓ₎^g(x)f(t)dt = A(g(x)) − A(h(x)). Differentiation gives f(g(x))g′(x) − f(h(x))h′(x). The lower endpoint contributes with a minus sign. This formula remains valid if the endpoints cross, because oriented integrals already handle their order.

If the integrand itself depends on x as a parameter, the endpoint-only formula is incomplete. Additional differentiation under the integral sign requires appropriate conditions and contributes an integral of the parameter derivative. In the present moving-boundary formula, f depends only on the dummy variable t. Check that distinction before using it.

An accumulation function can be studied without evaluating its integral. Its derivative is the original integrand, so the sign of f determines where accumulation increases or decreases. Its second derivative, where it exists, is f′. This is often the intended route when the integrand has no elementary antiderivative.

09 / UNDERSTAND & APPLY

9. Original practice: limit-of-sums recognition

For each sum, identify the width, sampled function and interval before evaluating. These details are the actual conversion proof.

10 / UNDERSTAND & APPLY

10. Original practice: evaluate, bound and differentiate

Use the structural property that best fits each problem. In every application of a theorem, verify the domain and continuity or integrability needed for that theorem.

11 / UNDERSTAND & APPLY

11. The reliable order of operations

For an unfamiliar definite integral, inspect its domain first. Next look for symmetry, a useful split, a substitution or a standard antiderivative. Keep the orientation of bounds explicit. After evaluating, compare with sign and size bounds. A supposedly negative integral of a nonnegative continuous function on an increasing interval signals an error before any detailed rechecking.

For a limit of sums, factor out the width and identify the sample argument. State the interval and justify integrability. If the sum’s scale looks inconsistent, estimate a typical term times the number of terms. For a moving-limit derivative, identify the dummy variable, each endpoint function and its chain-rule factor.

Remember what each theorem actually says. Riemann integrability is a sample-independent limiting property. The fundamental theorem turns valid accumulation into differentiation or endpoint subtraction. Reflection and parity are changes of variable. None of these permits cancellation across a divergent singularity or omission of an endpoint derivative.

The most effective check is a different viewpoint: compare an antiderivative calculation with a finite-sum identity, a reflection result with a bound, or a derivative of an accumulation with a small added strip. Understanding these connections makes the formulas easier to remember and much harder to misuse.