Recover a function and evaluate a definite integral

Try

Both x² and x²+4 have derivative 2x. If you are told only the derivative, can you choose between them? If you are additionally told F(1)=5, which function remains? Draw the two vertically shifted curves and compare their tangent slopes at matching inputs.

Predict

Predict whether the arbitrary constant changes the definite integral from 0 to 2. Evaluate the difference between the primitive’s endpoint values before deciding. The same constant appears twice, so it cancels; that does not mean every antiderivative has constant zero.

Experiment

For f(x)=3x²−2x, compare two primitives.

Primitive	Value at 0	Value at 2	Endpoint difference
F(x)=x³−x²	0	4	4
G(x)=x³−x²+7	7	11	4

Differentiate both expressions to verify the integrand. Then use the initial condition F(1)=5 for the separate derivative problem F′=2x: the family x²+C gives C=4 and F(3)=13. This initial-value calculation selects a function rather than merely calculating a definite accumulation.

Understand

An antiderivative of f on an interval is a function F with F′=f there. If two antiderivatives exist on the same interval, their difference has derivative zero and is constant by the mean value theorem. On disconnected domains, separate constants may occur on separate intervals; do not silently link both sides of a singularity.

The second fundamental theorem says that for continuous f on [a,b] and an antiderivative F, ∫ₐᵇf(x)dx=F(b)−F(a). The notation [F(x)]ₐᵇ means this upper-minus-lower difference. To justify it, the accumulation A(x)=∫ₐˣf has A′=f by the first theorem. Thus F−A is constant. Compare that identity at the two endpoints to obtain the evaluation formula.

Basic antiderivatives reverse familiar derivatives. For exponent n≠−1, x^(n+1)/(n+1) differentiates to x^n on a valid real interval. The exception 1/x has primitive ln|x| for x≠0. Exponential e^x integrates to itself; cosine integrates to sine; sine integrates to negative cosine, using radians. Verification by differentiation is an independent way to check a proposed symbolic answer.

An indefinite integral denotes a family, not a number. A definite integral is a signed numerical quantity once the function and bounds are specified. An initial-value condition chooses one member of a family. Keeping these three objects distinct prevents misplaced constants and forgotten boundary terms.

The formula cannot be applied across a point where the integrand is singular without further analysis. For 1/x across zero, ln|x| exists on either side but the hypotheses of this ordinary continuous-interval version fail. Improper integrals will introduce separate one-sided limits; writing one endpoint subtraction across the gap is not a repair.

Apply

For the practice calculations, write the primitive, verify it, then evaluate the endpoint difference. For an initial condition, solve for C before evaluating another input. For the power-rule exception, explain why division by zero is invalid and show the logarithmic derivative instead. In the theorem proof, identify exactly where continuity and the interval assumption enter.

Continue

Next reverse the chain rule using substitution. A valid transformation changes the integrand, the differential and—when evaluating a definite integral—the bounds consistently.