Integration By Parts Formula Definite Integral

The integration by parts formula is a powerful technique in calculus that allows us to integrate products of functions. While it is generally used for indefinite integrals, it can also be applied to definite integrals. This article will delve into the integration by parts formula, focusing on its application to definite integrals, providing a comprehensive understanding of its theory, application, and nuances.

Understanding Integration by Parts

Integration by parts is derived from the product rule for differentiation. It essentially reverses this rule to allow us to integrate more complex functions. The formula for integration by parts is:

$\int u , dv = uv - \int v , du$

Where:

• u is a function we choose to differentiate.
• dv is a function we choose to integrate.
• du is the derivative of u.
• v is the integral of dv.

The key to successfully using integration by parts lies in the strategic selection of u and dv. The goal is to choose u and dv such that the integral on the right-hand side, ∫ v du, is simpler than the original integral ∫ u dv.

The LIATE Acronym

A helpful mnemonic device for choosing u is the acronym LIATE:

• Logarithmic functions (e.g., ln(x), log(x))
• Inverse trigonometric functions (e.g., arctan(x), arcsin(x))
• Algebraic functions (e.g., x, x², polynomials)
• Trigonometric functions (e.g., sin(x), cos(x))
• Exponential functions (e.g., eˣ, 2ˣ)

The function that appears higher on the list is generally a good choice for u. However, this is just a guideline, and sometimes experimentation is necessary to find the most effective choice.

Integration by Parts for Definite Integrals

When dealing with definite integrals, the integration by parts formula is slightly modified to include the limits of integration. The formula becomes:

$\int_{a}^{b} u , dv = [uv]{a}^{b} - \int{a}^{b} v , du$

Where:

• a and b are the lower and upper limits of integration, respectively.
• [uv]_{a}^{b} represents uv evaluated at b minus uv evaluated at a.

Steps to Apply Integration by Parts to Definite Integrals

Here's a step-by-step guide on how to apply integration by parts to definite integrals:

1. Choose u and dv: Select appropriate functions for u and dv based on the LIATE rule or by considering which choice will simplify the integral.
2. Calculate du and v: Differentiate u to find du and integrate dv to find v. Remember to include the differential dx with du (e.g., du = u'(x) dx).
3. Apply the formula: Substitute u, v, du, and dv into the integration by parts formula for definite integrals: $\int_{a}^{b} u , dv = [uv]{a}^{b} - \int{a}^{b} v , du$
4. Evaluate uv at the limits of integration: Calculate uv at the upper limit b and subtract the value of uv at the lower limit a.
5. Evaluate the remaining integral: Integrate ∫v du with respect to x and evaluate it at the limits of integration a and b.
6. Combine the results: Add the result from step 4 and the negative of the result from step 5 to obtain the final answer.

Example 1: A Simple Definite Integral

Let's evaluate the definite integral:

$\int_{0}^{\pi/2} x \cos(x) , dx$

1. Choose u and dv: Using LIATE, we choose u = x and dv = cos(x) dx.
2. Calculate du and v:
   • du = dx
   • v = sin(x)
3. Apply the formula: $\int_{0}^{\pi/2} x \cos(x) , dx = [x \sin(x)]{0}^{\pi/2} - \int{0}^{\pi/2} \sin(x) , dx$
4. Evaluate uv at the limits: $[x \sin(x)]_{0}^{\pi/2} = \left(\frac{\pi}{2} \sin\left(\frac{\pi}{2}\right)\right) - (0 \cdot \sin(0)) = \frac{\pi}{2} \cdot 1 - 0 = \frac{\pi}{2}$
5. Evaluate the remaining integral: $\int_{0}^{\pi/2} \sin(x) , dx = [-\cos(x)]_{0}^{\pi/2} = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) = -0 + 1 = 1$
6. Combine the results: $\int_{0}^{\pi/2} x \cos(x) , dx = \frac{\pi}{2} - 1$

Therefore, the value of the definite integral is (π/2) - 1.

Example 2: A More Complex Definite Integral

Let's evaluate the definite integral:

$\int_{1}^{e} \ln(x) , dx$

1. Choose u and dv: In this case, we choose u = ln(x) and dv = dx.
2. Calculate du and v:
   • du = (1/x) dx
   • v = x
3. Apply the formula: $\int_{1}^{e} \ln(x) , dx = [x \ln(x)]{1}^{e} - \int{1}^{e} x \cdot \frac{1}{x} , dx$
4. Evaluate uv at the limits: $[x \ln(x)]_{1}^{e} = (e \ln(e)) - (1 \ln(1)) = e \cdot 1 - 1 \cdot 0 = e$
5. Evaluate the remaining integral: $\int_{1}^{e} 1 , dx = [x]_{1}^{e} = e - 1$
6. Combine the results: $\int_{1}^{e} \ln(x) , dx = e - (e - 1) = 1$

Therefore, the value of the definite integral is 1.

Example 3: Repeated Integration by Parts

Sometimes, a single application of integration by parts is not sufficient. In such cases, you may need to apply the formula multiple times. Consider the following definite integral:

$\int_{0}^{1} x^2 e^x , dx$

1. Choose u and dv: We choose u = x² and dv = eˣ dx.
2. Calculate du and v:
   • du = 2x dx
   • v = eˣ
3. Apply the formula: $\int_{0}^{1} x^2 e^x , dx = [x^2 e^x]{0}^{1} - \int{0}^{1} 2x e^x , dx$
4. Evaluate uv at the limits: $[x^2 e^x]_{0}^{1} = (1^2 e^1) - (0^2 e^0) = e - 0 = e$
5. Evaluate the remaining integral: We need to apply integration by parts again to evaluate ∫ 2x eˣ dx.
   • Let u₁ = 2x and dv₁ = eˣ dx.
   • Then du₁ = 2 dx and v₁ = eˣ. $\int_{0}^{1} 2x e^x , dx = [2x e^x]{0}^{1} - \int{0}^{1} 2 e^x , dx$ $[2x e^x]{0}^{1} = (2(1) e^1) - (2(0) e^0) = 2e - 0 = 2e$ $\int{0}^{1} 2 e^x , dx = [2 e^x]{0}^{1} = 2e^1 - 2e^0 = 2e - 2$ So, $\int{0}^{1} 2x e^x , dx = 2e - (2e - 2) = 2$
6. Combine the results: $\int_{0}^{1} x^2 e^x , dx = e - 2$

Therefore, the value of the definite integral is e - 2.

Common Pitfalls and Considerations

• Incorrect Choice of u and dv: Selecting the wrong u and dv can lead to a more complicated integral. Always consider the LIATE rule and experiment if necessary.
• Forgetting the Limits of Integration: When dealing with definite integrals, it is crucial to remember to evaluate uv and the resulting integral at the limits of integration.
• Sign Errors: Pay close attention to the signs when applying the formula and evaluating the terms. A simple sign error can lead to an incorrect answer.
• Cyclic Integration: In some cases, repeated application of integration by parts may lead back to the original integral. This requires a slightly different approach, where you solve for the integral algebraically.
• Discontinuities: Ensure that the functions u, v, du, and dv are continuous over the interval of integration. If there are discontinuities, the integration by parts formula may not be directly applicable, and you might need to split the integral into smaller intervals.

Advanced Techniques and Applications

Tabular Integration

Tabular integration, also known as the "Tic-Tac-Toe" method, is a shortcut for applying integration by parts repeatedly when one of the functions can be repeatedly differentiated to zero. It is particularly useful for integrals of the form ∫ xⁿ f(x) dx, where f(x) is a function like eˣ, sin(x), or cos(x).

Here's how it works:

1. Create a table with three columns: "Sign," "u (Differentiate)," and "dv (Integrate)."
2. List the signs alternating between + and - starting with +.
3. List u and its derivatives in the "Differentiate" column until you reach zero.
4. List dv and its integrals in the "Integrate" column, matching the number of rows in the "Differentiate" column.
5. Multiply diagonally and add the products, following the signs.

For example, let's revisit the integral ∫₀¹ x² eˣ dx using tabular integration:

The integral becomes:

$\int_{0}^{1} x^2 e^x , dx = [x^2 e^x - 2x e^x + 2 e^x]_{0}^{1}$

Evaluating at the limits:

$[(1^2 e^1 - 2(1) e^1 + 2 e^1) - (0^2 e^0 - 2(0) e^0 + 2 e^0)] = [e - 2e + 2e] - [0 - 0 + 2] = e - 2$

Which matches the result we obtained earlier.

Applications in Physics and Engineering

Integration by parts has numerous applications in physics and engineering. Some notable examples include:

• Calculating Centers of Mass: Determining the center of mass of an object often involves integrating the product of the position and density functions.
• Solving Differential Equations: Integration by parts is used to solve certain types of differential equations, particularly those involving products of functions.
• Fourier Analysis: In signal processing and Fourier analysis, integration by parts is used to evaluate integrals involving trigonometric functions and exponential functions.
• Probability Theory: Calculating expected values and variances of continuous random variables often requires integration by parts.
• Electromagnetism: Determining the magnetic field due to a current distribution can involve integrals that are solved using integration by parts.

Integration by Parts with Trigonometric Functions

When integrating products of trigonometric functions and other functions, it is often useful to apply integration by parts multiple times. In some cases, this can lead to a cyclic pattern, where the original integral reappears on the right-hand side. In such situations, you can solve for the integral algebraically.

For example, consider the integral:

$\int e^x \cos(x) , dx$

1. Apply integration by parts:
   • Let u = cos(x) and dv = eˣ dx.
   • Then du = -sin(x) dx and v = eˣ. $\int e^x \cos(x) , dx = e^x \cos(x) - \int e^x (-\sin(x)) , dx = e^x \cos(x) + \int e^x \sin(x) , dx$
2. Apply integration by parts again:
   • Let u₁ = sin(x) and dv₁ = eˣ dx.
   • Then du₁ = cos(x) dx and v₁ = eˣ. $\int e^x \sin(x) , dx = e^x \sin(x) - \int e^x \cos(x) , dx$
3. Substitute back into the original equation: $\int e^x \cos(x) , dx = e^x \cos(x) + \left(e^x \sin(x) - \int e^x \cos(x) , dx\right)$
4. Solve for the integral: $2 \int e^x \cos(x) , dx = e^x \cos(x) + e^x \sin(x)$ $\int e^x \cos(x) , dx = \frac{1}{2} e^x (\cos(x) + \sin(x)) + C$

Conclusion

The integration by parts formula is a fundamental tool in calculus, particularly when dealing with integrals involving products of functions. Understanding how to apply this formula to definite integrals, choosing appropriate functions for u and dv, and recognizing common pitfalls are essential for mastering this technique. Through careful application and practice, one can effectively tackle a wide range of integration problems in mathematics, physics, engineering, and other related fields. By understanding the theoretical underpinnings and practical applications of integration by parts, you can expand your problem-solving capabilities and deepen your appreciation for the elegance and power of calculus.