Unit notes

A definite integral represents signed accumulation; an antiderivative is a function whose derivative is the integrand. The symbol dx identifies the variable of integration. Constants of integration matter for indefinite integrals but cancel when evaluating a definite integral using endpoint values.

A Riemann sum multiplies a rate or density value by a small interval width and adds the contributions. The definite integral is the limit as a suitable partition becomes fine. Units multiply: liters per minute integrated over minutes gives liters.

Area below the horizontal axis contributes negatively to a signed integral. Geometric area uses magnitudes and may require splitting at zeros. Reversing integration limits reverses the sign; adding adjacent intervals adds their integrals. A definite integral is a number, not a family with +C.

Left, right, and midpoint sums use different representative values in each subinterval. A trapezoidal sum averages endpoint heights and multiplies by width. With unequal subintervals, calculate each width individually rather than using one average width.

For an increasing function, a left sum underestimates and a right sum overestimates the signed integral on a partition. Concavity governs standard midpoint and trapezoidal error direction: concave-up curves lie below chords, so trapezoids overestimate. Keep monotonicity and concavity arguments separate.

If f is continuous, the derivative of an accumulation function F(x) = ∫ from a to x of f(t)dt is f(x). A variable upper bound g(x) requires the chain rule: f(g(x))g′(x). The integration variable inside is a dummy variable and need not be x.

If an antiderivative H is known, the definite integral from a to b is H(b) − H(a). This does not require adding an arbitrary constant because it cancels. For an indefinite integral, include +C to represent the family of antiderivatives.

The power antiderivative divides by the new exponent after increasing it by one, except for exponent −1, whose antiderivative is ln|x| on intervals avoiding zero. Trigonometric and exponential antiderivatives should be checked by differentiating.

Algebra can reveal a simpler integrand: divide polynomials, split terms, or complete a square when appropriate. An antiderivative may exist even when no elementary formula is convenient. Calculator evaluation of a definite integral does not supply a symbolic antiderivative.

Choose u to represent an inner expression whose derivative appears as a factor. Replace both the expression and differential consistently. For a definite integral, either transform the bounds into u-values or return fully to x before using the original bounds.

A missing constant factor is a frequent error. If du = 3 dx, then dx = du3. A valid substitution changes the entire integral into one variable; leaving unrelated x terms behind means more algebra or a different substitution is needed.

Partition an interval into subintervals, multiply a representative function value by each width, and add. A Riemann sum approximates accumulation. Equal widths are convenient but not required; a table can require different widths. Left, right, and midpoint sums differ in sample location. A trapezoidal sum averages the two endpoint heights on each subinterval.

For an increasing function, a left sum underestimates and a right sum overestimates the integral when comparing the usual signed rectangles over an interval. Concavity determines the common midpoint and trapezoidal error directions for smooth functions. State the relevant behavior instead of assuming every trapezoidal approximation is too large. If the integrand is a rate in liters per minute and widths are in minutes, each product and the total are in liters.

If F′=f on an interval, the definite integral of f from a to b equals F(b)−F(a). This computes an accumulation using an antiderivative. In the other direction, if G(x)=∫ from a to x of f(t)dt and f is continuous, then G′(x)=f(x). The dummy variable t is not the same role as the moving upper limit x.

For a moving upper limit g(x), the chain rule gives d/dx[∫ from a to g(x) of f(t)dt]=f(g(x))g′(x). If both bounds move, subtract the lower-bound contribution. An accumulation function can be increasing even when it is negative; its derivative depends on the current integrand, while its value depends on previous signed area.

Look for a function and its derivative within an integrand. Let u be the inner expression and replace both the expression and the differential consistently. If u=x²+1, then du=2x dx, so x dx=du2. A missing constant factor changes the answer. Not every composition can be integrated by a simple substitution.

For a definite integral, either change the bounds to u-values and stay in u, or return to x before using the original bounds. Never combine u-expressions with x-bounds. Check an indefinite answer by differentiating it. For logarithmic antiderivatives, absolute values preserve the appropriate real-valued expression on intervals that avoid zero.

If F is an antiderivative of a suitably continuous f, then ∫ from a to b of f(x) dx equals F(b)−F(a). The definite integral is a number, while an indefinite integral describes a family of antiderivatives with a constant. Keep these roles separate.

For A(x)=∫ from a to x of f(t) dt, A′(x)=f(x) under the usual continuity condition. With upper limit g(x), apply the chain rule: A′(x)=f(g(x))g′(x). A variable lower limit contributes a negative term. The dummy integration variable is not an extra independent quantity to differentiate.

An accumulation function can increase while remaining negative. Its derivative depends on the current integrand value, while its height depends on prior accumulated area and any initial value. This mirrors the distinction between position and velocity.

Simplify sums and powers before integrating. Substitution reverses a chain rule: choose u so that du matches a factor in the integrand, including constants. If bounds are changed to u-values, finish in u consistently; if returning to x, use the original bounds.

The antiderivative of 1x is ln|x| on appropriate intervals, not a power-rule result with division by zero. A general power rule excludes exponent −1. Differentiate a proposed antiderivative to verify it; this is often the fastest error check.

A definite integral can be evaluated by geometry, properties, an antiderivative, or numerical approximation. Use the information given rather than forcing an unavailable formula. Reversing bounds changes the sign; splitting intervals adds contributions; neither operation turns signed area automatically into total geometric area.

For ∫2x cos(x²)dx, choose u=x² and du=2x dx. The full differential factor is already present, leaving ∫cos u du. Returning to x gives sin(x²)+C. The point is not merely replacing x² with a letter; it is matching how that inner quantity changes.

For a definite integral, map both endpoints through u. If x runs from 0 to 2, then u runs from 0 to 4. Evaluating sin u at x-bounds would mix coordinate systems. Alternatively, substitute back to x before using the original endpoints. Both approaches agree when used consistently.

A substitution may need a constant correction: if du=3dx but only dx appears, replace dx by du3. Differentiate the final answer to catch missing constants. For a proposed substitution that leaves unresolved x and u factors, reconsider whether the algebra is complete.

For F(x)=C+∫_aˣf(t)dt, the value C fixes the starting height. The integrand determines whether F rises or falls, while the accumulated signed area determines how far it has moved. A zero of f can produce an extremum of F only if the sign changes appropriately.

The second derivative F″=f′, where it exists, connects concavity of F to the slope of the integrand graph. A positive but decreasing f makes F increase and become concave down. This combines two separate observations from the same rate graph.

If the upper endpoint is a function, chain-rule effects can reverse or scale the rate of accumulation. For ∫₀^(1−x)f(t)dt, the derivative is −f(1−x). The moving endpoint travels in the negative direction as x increases, explaining the minus sign geometrically.