Lesson 6.5: Integration Techniques: Substitution and By Parts

Introduction

In this lesson, we are going to dive into two powerful techniques for integration: substitution and integration by parts. These techniques will help you tackle more complex integrals that you may encounter in your studies.

Learning Objectives

By the end of this lesson, you should be able to:

Perform integration by substitution, including changing limits for definite integrals.
Use integration by parts, including repeated applications where necessary.
Effectively choose the appropriate technique for a given integrand.
Adjust limits correctly when using substitution in definite integrals.
Apply integration by parts, including the strategy of "twice and rearrange."

Hook

Imagine solving a puzzle. Every piece fits into a specific spot based on its shape. Integration techniques are similar — each integral requires the right method to solve it. Let's unlock these techniques!

H2: Integration by Substitution

Integration by substitution is a method that can simplify complex integrals by changing variables. The basic idea is to transform an integral into a more manageable form. Here’s how it works:

The Process

Choose a substitution: Select a substitution for a part of the integral that simplifies it. Often, you'll want to let $ u $ equal a function inside the integral.
Compute the differential: Calculate $ du $ to relate $ dx $ and $ du $. This step may require you to express $ dx $ in terms of $ du $.
Change limits if necessary: If you're evaluating a definite integral, you'll need to adjust the limits of integration based on your substitution.
Rewrite the integral: Substitute into the integral and simplify it.
Integrate: Solve the new integral in terms of $ u $.
Back-substitute: Finally, replace $ u $ with the original variable to get your answer back in terms of $ x $.

Example 1:

Evaluate the integral $ \int (3x^2 + 2) \cdot 3x \, dx $ using substitution.

Solution:

Let $ u = 3x^2 + 2 $. Then $ du = 6x \, dx $ or $ \frac{du}{6} = x \, dx $.
The integral becomes:

$$ \int u \cdot \frac{du}{6} = \frac{1}{6} \int u \, du = \frac{1}{6} \cdot \frac{u^2}{2} + C = \frac{(3x^2 + 2)^2}{12} + C $$.

Example 2: Definite Integral

Evaluate the definite integral $ \int_0^1 (4x^3 + 2) \, dx $ with substitution.

Solution:

First, let’s find the antiderivative without substitution: $ \int (4x^3 + 2) \, dx = x^4 + 2x + C $.
Next, evaluate it: $ [1^4 + 2(1)] - [0^4 + 2(0)] = 3 - 0 = 3 $.
Alternatively, use $ u = 4x^3 + 2 $, then compute:

$$ u \text{ changes from } 2 \text{ to } 6. $$

Adjust limits: from $ x = 0 $, $ u = 2 $ and $ x = 1 $, $ u = 6 $.

$$ \int_2^6 u \cdot \frac{du}{12} = \left[\frac{u^2}{24}

ight]_2^6 = $\frac{6^2 - 2^2}{24}$ = $\frac{36 - 4}{24}$ = $\frac{32}{24}$ = $\frac{4}{3}$ $$.

H2: Integration by Parts

Integration by parts is based on the product rule for differentiation. The formula can be stated as:

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

where $ u $ is a function that we differentiate, and $ dv $ is a function that we integrate.

The Process

Choose $ u $ and $ dv $: Select which part of the integrand you will differentiate (set as $ u $) and which part to integrate (set as $ dv $).
Differentiate and Integrate: Compute $ du $ and $ v $ accordingly.
Substitute into the formula: Use the integration by parts formula to rewrite the integral.
Simplify if necessary and integrate: This often leads to a simpler integral.
Repeat as needed: Continue to apply the integration by parts if necessary until you arrive at a solvable integral.

Example 3:

Evaluate $ \int x e^x \, dx $ using integration by parts.

Solution:

Let $ u = x $ (then $ du = dx $) and $ dv = e^x \, dx $ (then $ v = e^x $).
Apply the formula:

$$ \int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C $$

Example 4: Repeated Integration by Parts

Evaluate $ \int x^2 \ln(x) \, dx $ using integration by parts twice.

Solution:

First, set $ u = \ln(x) $ (thus $ du = \frac{1}{x}dx $), and $ dv = x^2 dx $ (so, $ v = \frac{x^3}{3} $).
The first application gives:

$$ \int x^2 \ln(x) \, dx = \frac{x^3}{3} \ln(x) - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx $$

Simplifying gives you:

$$ = \frac{x^3}{3} \ln(x) - \frac{1}{3} \int x^2 \, dx = \frac{x^3}{3} \ln(x) - \frac{1}{3} \cdot \frac{x^3}{3} + C = \frac{x^3}{3} \ln(x) - \frac{x^3}{9} + C $$

H2: Conclusion

In this lesson, we learned two essential techniques for integration: substitution and integration by parts. Both methods require practice to master, especially in determining which technique to use based on the integrand you are given. Remember, the goal is to simplify complex integrals into manageable ones, allowing you to find their antiderivatives with ease.

H1: Study Notes

Integration by Substitution: Use substitution to simplify integrals.
Integration by Parts: Apply the formula $ \int u \, dv = uv - \int v \, du $.
Choose wisely: Identify both $ u $ and $ dv $ carefully for effective integration by parts.
Definite Integrals: Adjust limits when using substitution.
Repeated Applications: Integration by parts can be applied multiple times as necessary to solve the integral.