49. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample:
c. ∫ v du = u·v - ∫ u dv

Verified step by step guidance

1. Recognize that the given statement \( \int v \, du = u \cdot v - \int u \, dv \) resembles the integration by parts formula, which is usually written as \( \int u \, dv = u \cdot v - \int v \, du \).
2. Recall the integration by parts formula: \( \int u \, dv = u \cdot v - \int v \, du \). This formula is derived from the product rule for differentiation.
3. Compare the given statement with the standard formula. Notice that the roles of \( u \) and \( v \) are swapped in the integrals and products.
4. To verify if the statement is true, try substituting \( u \) and \( v \) in the standard formula and see if it matches the given expression or if it leads to a contradiction.
5. Conclude whether the statement is true or false based on the comparison and substitution, and provide a counterexample if it is false by choosing specific functions \( u(x) \) and \( v(x) \) to test the equality.

Key Concepts

Integration by Parts Formula

Integration by parts is a technique derived from the product rule of differentiation. It states that ∫ u dv = u·v - ∫ v du, where u and v are functions of a variable. This formula helps transform complex integrals into simpler ones by choosing appropriate u and dv.

Differentiation and Notation Consistency

In integration by parts, the differentials du and dv correspond to derivatives of u and v respectively. The formula requires careful attention to the order of terms; swapping du and dv changes the meaning and validity of the expression. Correct notation ensures the formula is applied properly.

Counterexamples and Verification

To determine if a statement like ∫ v du = u·v - ∫ u dv is true, one can test it with specific functions u and v. If the equality fails for any example, it is false. This approach helps verify or refute integral identities by direct substitution and evaluation.