Problem 65 - Applied Calculus

Question

Choose the correct formula: a. \(\int u d v=u v+\int v d u \quad\) or b. \(\int u d v=u v-\int v d u\)

Step-by-Step Solution

Answer

Formula b: \(\int u d v=u v-\int v d u\).

Step 1: Understand Integration by Parts Formula

Integration by parts is derived from the product rule of differentiation and is used to integrate the product of two functions. It requires us to identify a function to differentiate (\(u\)) and another to integrate (dv\(dv\)). The correct formula is important for simplifying the integration process.

Step 2: Recall the Integration by Parts Formula

The formula for integration by parts is given by:\[\int u \, dv = uv - \int v \, du\]

Key Concepts

Integration Techniques

Understanding integration techniques is crucial for tackling more complex integrals. Integration is the process of finding the integral of a function, which is the inverse operation of differentiation. However, when dealing with products of functions, standard integration methods may not be sufficient.
In these cases, specific techniques such as integration by parts become invaluable. Integration by parts is particularly useful when you have two functions multiplied together within an integral. For these products, direct integration isn't straightforward, so we apply this strategic method.
Choosing which function to differentiate and which to integrate is key to effectively applying integration by parts. A common mnemonic to help with this decision is "LIATE," which stands for:

• Logarithmic functions (L)
• Inverse trigonometric functions (I)
• Algebraic functions (A)
• Trigonometric functions (T)
• Exponential functions (E)

These are often listed in the order of priority for choosing a "u" function. The selection process can simplify the integral or make it more complex if chosen poorly.

Product Rule of Differentiation

The product rule of differentiation forms the foundation for the integration by parts formula. In calculus, the product rule is a technique used to differentiate the product of two functions.
If you have two functions, say, \(f(x)\) and \(g(x)\), their differentiation is given by:
\[ \frac{d}{dx}[f(x) \, g(x)] = f'(x) \, g(x) + f(x) \, g'(x) \]
This rule allows us to find the derivative of a product by taking the derivative of each function one at a time while keeping the other intact.

Integration by parts takes this differentiation concept and reverses it for integral calculation. By starting with the product rule and manipulating it, we derive the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Here, one function (\(u\)) is differentiated, and the other (\(dv\)) is integrated, to help resolve an otherwise tricky integral.

Calculus Formulas

In calculus, mastering various formulas is pivotal to solving problems efficiently. Among these, the formula for integration by parts stands out. Developed from the product rule of differentiation, it aids in integrating functions that are products of one another.
The formula for integration by parts is given by:
\[ \int u \, dv = uv - \int v \, du \]
This formula should be committed to memory as it recurrently appears when integrating products of two functions. Understanding how and when to apply it can radically simplify the process of integration.

When working with calculus formulas, it's beneficial to also understand their derivations and connections. For instance, noting how integration by parts mirrors the structure of the product rule of differentiation can enhance comprehension and application.
Practice is crucial with these formulas, as applying them enhances fluency in calculus and increases problem-solving efficiency in more complex scenarios.