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Chapter 7: Problem 4

Explain how integration by parts is used to evaluate a definite integral.

Short Answer

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Question: Explain the process of evaluating a definite integral using integration by parts. Answer: To evaluate a definite integral using integration by parts, follow these steps: 1. Recall the Integration by Parts Formula for definite integrals: \[ \int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du\] 2. Identify u and dv from the given integral, considering the LIATE acronym as a guideline for choosing u. 3. Compute du by differentiating u, and obtain v by integrating dv. 4. Plug u, v, du, and dv into the Integration by Parts Formula: \[ \int_a^b u\, dv = [uv]_a^b - \int_a^b v\, du\] 5. Evaluate the remaining integral, which might require another integration by parts or other integration techniques. 6. Substitute the limits of integration (a and b) in the evaluated expression: \[ [uv]_a^b - \text{ Evaluated Integral }\] The result obtained after the last step is the solution to the definite integral.

Step by step solution

01

Recall the Integration by Parts Formula

To evaluate a definite integral using integration by parts, you need to recall the formula. It is derived from the product rule for differentiation: \[ \int u \, dv = uv - \int v\, du\] Where u and v are continuously differentiable functions. For definite integrals, the formula becomes: \[ \int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du\] Here, the limits of integration are added, and the result after differentiation will be evaluated at the upper and lower limits.
02

Identify u and dv

Choose the two functions u and dv from the given integral, such that their product forms the integrand. Remember the acronym "LIATE" (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential) as a guideline for choosing u. The first category discovered in the integrand will typically be the best choice.
03

Compute du and v

Differentiate u to obtain du and integrate dv to obtain v. It's essential to perform these operations correctly, as they are required for the next steps.
04

Plug u, v, du, and dv into the Integration by Parts Formula

With all the required components (u, v, du, and dv) calculated, substitute them into the integration by parts formula for definite integrals: \[ \int_a^b u\, dv = [uv]_a^b - \int_a^b v\, du\] Now, you have a new integral to evaluate.
05

Evaluate the Remaining Integral

Evaluate the new integral you obtained in the previous step. In some cases, the new integral may still be complex and may require integrating by parts again or using other integration techniques. After evaluating the resulting integral, continue with the next step.
06

Substitute the Limits of Integration

Now, to get the final result for the definite integral, substitute the limits of integration (a and b) in the evaluated expression: \[ [uv]_a^b - \text{ Evaluated Integral }\] Substitute the upper limit b first, then subtract the result obtained by substituting the lower limit a. The result you obtain after this step will be the solution to the definite integral. These are the steps to evaluate a definite integral using integration by parts. This method can be applied to various integrals where the integrand is the product of two functions, and it can be combined with other integration techniques when needed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
A definite integral is a way to calculate the area under a curve, between two points, on a given interval. Specifically, it represents the accumulation of quantities, like area, volume, and many physical properties, over a certain range. When you're working with definite integrals, it's crucial to understand that they have specific limits, typically denoted as \[ \int_a^b f(x) \, dx \] where \( a \) and \( b \) are the lower and upper limits of integration, respectively. These limits define the starting and ending points on the x-axis.
  • The function \( f(x) \) inside the integral is known as the integrand.
  • The result of a definite integral is a specific numerical value, unlike indefinite integrals which include a constant of integration.
Understanding definite integrals is key when applying integration by parts because it allows you to find precise values over specific intervals.
Product Rule for Differentiation
The product rule for differentiation is a fundamental concept necessary for understanding integration by parts. This rule applies when you're differentiating a product of two functions.In mathematical terms, if you have two functions, \( u(x) \) and \( v(x) \), the product rule states:\[ (uv)' = u'v + uv' \]
  • \( u' \) is the derivative of \( u \) with respect to \( x \).
  • \( v' \) is the derivative of \( v \) with respect to \( x \).
This concept is crucial because integration by parts is derived from rearranging the product rule. It's like "undo-ing" the differentiation of products to help find the integral of a product of functions.
Differentiable Functions
Differentiable functions are functions that have a derivative at every point in their domain. This means that the function is smooth, without any sharp corners or discontinuities. In the context of integration by parts, both functions \( u \) and \( v \) need to be differentiable.Why is this important?
  • Both \( u \) and \( v \) must be differentiable for you to compute \( du = u' \, dx \) and \( dv \) correctly.
  • Differentiability ensures that you can confidently apply the product rule in reverse, which is how integration by parts operates.
  • When functions are differentiable, their derivatives \( u' \) and \( v' \) offer vital expressions for constructing new integrals.
Without differentiable functions, the crucial calculations in integration by parts, including \( du \) and \( v \), couldn't be computed accurately.
Integration Techniques
Integration techniques are various strategies used to solve integrals. Integration by parts is one of these useful methods. It's particularly handy when you deal with the integral of a product of two functions.Each technique serves specific kinds of functions:
  • Integration by Parts: Ideal for integrating products of algebraic and exponential functions.
  • Substitution: Used when the integral contains a composite function, similar to using the chain rule in reverse.
  • Partial Fraction Decomposition: Effective for rational functions, where the integrand can be expressed as a sum of simpler fractions.
Integration by parts involves selecting \( u \) and \( dv \) wisely to simplify the integral. By doing this, the original complex integral is reduced to a simpler one which is then easier to evaluate. Remembering handy acronyms like "LIATE" can guide you in choosing \( u \) in these scenarios.

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