Integration by Parts
When performing integration in calculus, there are certain problems that might now be solved using basic integration methods.
More complicated integration problems might be solvable using a method called “integration by parts”.
Integration by parts is a technique in calculus that is based on the idea of “reversing” the product rule used in differentiation. While many integrals can be solved through direct methods or substitution, some expressions involve the product of two different types of functions—such as polynomials with exponential, logarithmic, or trigonometric terms—that are not easily integrated in a straightforward way. Integration by parts provides a systematic approach to handle these cases.
The method is based on splitting an integrand into two parts: one that becomes simpler when differentiated, and another that is manageable when integrated. By applying this method, the original integral is transformed into a new one that is often easier to evaluate. This process sometimes needs to be applied more than once or combined with other techniques, but it can significantly reduce the complexity of otherwise challenging problems.
In practice, integration by parts is especially useful in problems from physics, engineering, and environmental science, where expressions involving exponential functions or trigonometric functions frequently appear. Applying this method of integration by parts not only strengthens problem-solving skills but also helps with an understanding of how integration connects to differentiation.
The video below provides more details and explanations for the topic of integration by parts together with several worked-out examples.
Integration by Parts Method
Integration by parts is a technique in calculus that is based on the idea of applying the inverse of the product rule used in differentiation. While many integrals can be solved through direct methods or substitution, some integrals involve the product of two different types of functions—such as polynomials with exponential, logarithmic, or trigonometric terms—that are not easily integrated in a straightforward way. Integration by parts provides a method to address these types of integrals.
Derivation from the Product Rule
The product rule in differentiation states that the derivative of the product of two functions is given by:
\[ \frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} \]
Rearranging this expression and integrating both sides leads to the integration by parts formula:
\[ \int u\,dv = uv – \int v\,du \]
This formula provides the foundation for solving integrals involving products of functions.
Guidelines for Choosing u and dv (LIATE Rule)
A critical step in applying integration by parts is choosing which part of the integrand to label as u and which as dv. A common guideline is the LIATE rule, which provides an ordered listing for choosing the “u” expression on the left hand side of the integration by parts formula:
- L – Logarithmic functions (ln x, log base a of x)
- I – Inverse trigonometric functions (arctan x, arcsin x, etc.)
- A – Algebraic functions (polynomials such as x, x², x³)
- T – Trigonometric functions (sin x, cos x, etc.)
- E – Exponential functions (ex, ax)
According to this rule, the function that appears earlier in the list should generally be chosen as u, while the remaining part of the integrand is taken as the dv portion.
Example 1: Integrating a Polynomial and an Exponential
Evaluate the integral:
\[ \int x e^x\,dx \]
Step 1: Apply LIATE. The polynomial term x is chosen as u, and \( e^x dx \) becomes dv.
\( u = x,\; dv = e^x dx \)
\( du = dx,\; v = e^x \)
Step 2: Apply the integration by parts formula:
\[ \int u\,dv = uv – \int v\,du \]
\[ \int x e^x\,dx = x e^x – \int e^x\,dx = x e^x – e^x + C \]
Thus, the solution is:
\[ \int x e^x\,dx = e^x (x – 1) + C \]
Example 2: Integrating a Polynomial and a Trigonometric Function
Evaluate:
\[ \int x \sin(x)\,dx \]
Step 1: Choose u and dv using LIATE. Polynomial x is u, and sin(x) dx is dv.
\( u = x,\; dv = \sin(x)\,dx \)
\( du = dx,\; v = -\cos(x) \)
Step 2: Apply the formula:
\[ \int u\,dv = uv – \int v\,du \]
\[ \int x \sin(x)\,dx = -x \cos(x) + \int \cos(x)\,dx = -x \cos(x) + \sin(x) + C \]
Thus, the solution is:
\[ \int x \sin(x)\,dx = -x \cos(x) + \sin(x) + C \]
Example 3: Logarithmic Function
Evaluate:
\[ \int \ln(x)\,dx \]
Step 1: Rewrite as the product of functions: ln(x) * 1.
Choose \( u = \ln(x) \) (logarithmic) and \( dv = dx \).
\( u = \ln(x),\; dv = dx \)
\( du = \frac{1}{x}dx,\; v = x \)
Step 2: Apply formula:
\[ \int \ln(x)\,dx = x \ln(x) – \int x \cdot \frac{1}{x}\,dx \]
\[ = x \ln(x) – \int 1\,dx = x \ln(x) – x + C \]
Applications of Integration by Parts
Integration by parts has wide-ranging applications beyond pure mathematics. In business, it is used in computing present and future values of certain financial instruments when exponential growth or decay is involved. For instance, integrating functions that model compound interest or depreciation may require this technique. In environmental science, integration by parts can be applied when solving integrals that describe pollutant decay, population models, or oscillating phenomena in ecosystems. Physics and engineering also heavily rely on this method in analyzing oscillatory motion, wave behavior, and problems involving exponential decay.
Example 4: Business Application
Suppose we want to compute the integral:
\[ \int t e^{-rt}\,dt \quad (where\ r > 0) \]
This integral can represent the present value of a time-dependent cash flow that decays exponentially.
Step 1: Choose \( u = t \), \( dv = e^{-rt} dt \).
\( u = t,\; dv = e^{-rt} dt \)
\( du = dt,\; v = \left(-\frac{1}{r}\right)e^{-rt} \)
Step 2: Apply the formula:
\[ \int t e^{-rt}\,dt = \left(-\frac{t}{r}\right)e^{-rt} + \int \frac{1}{r}e^{-rt}\,dt \]
\[ = \left(-\frac{t}{r}\right)e^{-rt} – \frac{1}{r^2}e^{-rt} + C \]
This solution is important in finance and business for evaluating certain models of cash flow.
Example 5: Environmental Science Application
Consider a pollutant concentration modeled by the function:
\[ C(t) = t e^{-kt} \]
To find the total pollutant over time, we compute:
\[ \int t e^{-kt}\,dt \]
Using integration by parts as before:
\( u = t,\; dv = e^{-kt} dt \)
\( du = dt,\; v = \left(-\frac{1}{k}\right)e^{-kt} \)
\[ \int t e^{-kt}\,dt = \left(-\frac{t}{k}\right)e^{-kt} + \int \frac{1}{k}e^{-kt}\,dt \]
\[ = \left(-\frac{t}{k}\right)e^{-kt} – \frac{1}{k^2}e^{-kt} + C \]
This integral provides valuable insights into how pollutant concentrations decrease over time and can be used in environmental policy and regulation.
Integration by parts is an important tool for integrating more complex functions. By understanding the steps, and carefully practicing its application across different examples—polynomials, exponentials, trigonometric, logarithmic functions, and applied scenarios—students can develop confidence in applying the integration by parts method. The method often requires creativity in selecting u and dv, guided by the LIATE rule, and may need multiple iterations of the formula.
Further Applications of Integration by Parts
Beyond mathematics, physics, and finance, integration by parts finds applications in statistics, economics, and engineering. For instance, in probability theory, expected values involving random variables and exponential distributions often require integration by parts. In economics, consumer and producer surplus problems that involve integrating functions of demand and supply curves may also require this method. In engineering, integration by parts is frequently used when analyzing signal processing problems, such as Fourier transforms, where integrals of products of trigonometric and exponential functions are common.
