5 min read•june 18, 2024
In AP Calculus, we encounter a variety of functions, and each may require a different approach for antidifferentiation. This skill is all about choosing the right technique or method to find the antiderivative of a given expression. Let's explore the techniques necessary for this task in detail! 🚀
If you’re an AP Calculus BC student, be sure to note that BC techniques are not incorporated in this study guide! Take a look at these three guides to ensure you’re familiar with them:
The power rule for antiderivatives is a fundamental technique used to find the antiderivative of a function raised to a power. When you have an integral in the form ∫, this rule is your go-to method. It states that if you have, where n is not equal to -1, you add 1 to the exponent and divide by the new exponent. This results in:
Where C is the constant of integration.
is a powerful technique for simplifying complex integrals. It involves substituting a portion of the expression with a single variable (usually denoted as 'u') to make the integral more manageable.
This method is particularly useful when dealing with composite functions or expressions that involve chains of functions. The basic steps for u- are as follows:
For a review on u-substitution, check out 6.9 Integrating Using Substitution!
Trigonometric functions, like sine, cosine, and tangent, frequently appear in integrals. Understanding how to handle these functions is crucial. For example, when integrating trigonometric expressions, you may use trigonometric identities or specific trigonometric integrals.
Common trigonometric integrals include:
For some practice and additional information about the integrals of trig functions, check out 6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation. The next two subtopics are covered in more detail in this guide as well!
Antidifferentiation often involves like arcsin, arccos, and arctan. Knowing the antiderivatives of these functions and how to apply them is essential.
Here are some common inverse trigonometric integrals:
Functions involving exponentials () and natural logarithms () frequently appear in calculus problems. Familiarity with their antiderivatives is necessary for solving such integrals.
Long division can be useful when dealing with rational functions, where the degree of the numerator is equal to or higher than the degree of the denominator. It helps simplify the expression before antidifferentiation.
Here is the long division method for the expression:
🥇 Step 1: Perform Long Division
Perform long division to divide the numerator by the denominator:
🥈 Step 2: Write the Integral as a Sum Rewrite the integral as a sum of two integrals:
🥉 Step 3: Integrate Each Term Separately Now, integrate each term separately:
is a technique often used with . It's crucial when you encounter expressions that can be transformed into a perfect square trinomial for easier integration.
Here's a step-by-step example of integrating a quadratic expression by completing the square, for…
🥇 Step 1: Identify the Perfect Square Trinomial Recognize that the expression can be factored as a perfect square trinomial: .
🥈 Step 2: Rewrite the Integral Rewrite the integral using the perfect square trinomial:
🥉 Step 3: Use the Power Rule for Antiderivatives Apply the power rule for antiderivatives to find the integral:
⛳ Step 4: Simplify Simplify the expression, resulting in its antiderivative:
For a review on integrating using long division and/or completing the square, check out 6.10 Integrating Functions Using Long Division and Completing the Square.
Let’s put those newly learned skills to good use! Find the antiderivative of the following expression:
Be sure to try the question out yourself before moving onto the solution below!
When approaching this problem it is best to first recognize the terms in the expression and determine which technique to use for each term.
Term 1: (Antiderivative of is )
Term 2: (Antiderivative of is ) Now, take the integral of each term individually.
Since the integral of is itself, term 1 is easy!
For term 2, the integral is . You got the two separate integrals! All that is left is to combine the results and add the term to the end.
So, the antiderivative of the given expression is . Great work! 👏
Guess what?! You made it to the end of unit 6 in AP Calculus! Ready to move on to unit seven? It’s all about differential equations. Check it out here!
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