Why can the constant of integration be omitted from the antiderivative when evaluating a definite integral?

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Understand the concept of a definite integral: A definite integral calculates the net area under a curve between two specific bounds, say from x = a to x = b. It results in a numerical value rather than a general function.

Recall the Fundamental Theorem of Calculus: The definite integral of a function f(x) from a to b is computed using an antiderivative F(x), such that \( \int_a^b f(x) \, dx = F(b) - F(a) \).

Recognize the role of the constant of integration: When finding an antiderivative, a constant of integration (C) is added because indefinite integrals represent a family of functions. However, for definite integrals, this constant cancels out during subtraction.

Perform the subtraction step: When evaluating \( F(b) - F(a) \), the constant of integration (C) appears in both F(b) and F(a). Since \( C - C = 0 \), the constant does not affect the final result of the definite integral.

Conclude why the constant can be omitted: Because the constant of integration cancels out during the subtraction process, it is unnecessary to include it when evaluating a definite integral.

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주요 개념

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Definite Integral

A definite integral calculates the net area under a curve between two specified limits. It is represented as ∫[a, b] f(x) dx, where 'a' and 'b' are the bounds of integration. The result is a numerical value that reflects the accumulation of quantities, such as area, over the interval [a, b].

Antiderivative

An antiderivative of a function f(x) is a function F(x) such that F'(x) = f(x). When finding an antiderivative, we often include a constant of integration (C) because the derivative of a constant is zero, meaning multiple functions can yield the same derivative. However, this constant does not affect the evaluation of definite integrals.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links differentiation and integration, stating that if F is an antiderivative of f on an interval [a, b], then ∫[a, b] f(x) dx = F(b) - F(a). This theorem shows that when evaluating a definite integral, the constant of integration cancels out, as it is present in both F(b) and F(a), making it unnecessary for the final result.