{"id":8734,"date":"2017-01-27T09:28:00","date_gmt":"2017-01-27T17:28:00","guid":{"rendered":"https:\/\/magoosh.com\/hs\/?p=8734"},"modified":"2017-01-25T20:28:30","modified_gmt":"2017-01-26T04:28:30","slug":"ap-calculus-review-indefinite-integrals","status":"publish","type":"post","link":"https:\/\/magoosh.com\/hs\/ap\/ap-calculus-review-indefinite-integrals\/","title":{"rendered":"AP Calculus Review: Indefinite Integrals"},"content":{"rendered":"

Indefinite integrals make up a substantial part of what is covered on the AP Calculus AB and BC exams. In this review article, we highlight a few concepts and techniques that you’ll need to be familiar with.<\/p>\n

What are Indefinite Integrals?<\/h2>\n

There are two kinds of integrals, the definite and indefinite integrals. This article only discusses indefinite integrals. For a more general overview, including information about definite integrals, check out this review of integrals<\/a>.<\/p>\n

An indefinite integral<\/strong> of a function f<\/em> is the most general antiderivative<\/strong> of f<\/em>.<\/p>\n

\"Indefinite<\/p>\n

Here, the function F<\/em> is any particular antiderivative for f<\/em>. That is, F<\/em>\u00a0‘<\/sup>(x<\/em>) = f<\/em>(x<\/em>). For example, F<\/em>(x<\/em>) = x<\/em>2<\/sup> is an antiderivative for f<\/em>(x<\/em>) = 2x<\/em>, since (x<\/em>2<\/sup>)’ = 2x<\/em>.<\/p>\n

The C<\/em> is the constant of integration<\/strong>. It stands for any constant, and it must be part of your answer to an indefinite integral.<\/p>\n

So for example,<\/p>\n

\"Integral<\/p>\n

What’s the Deal with the “\u00a0+\u00a0C<\/em>\u00a0” Anyway?<\/h3>\n

The reason we need to tack on that “\u00a0+\u00a0C<\/em>\u00a0” is so that we can describe absolutely every antiderivative for f<\/em>. Remember the derivative rule for constant functions:<\/p>\n

\"The<\/p>\n

Therefore, if there is a particular function F<\/em>(x<\/em>) such that F<\/em>\u00a0‘(x<\/em>) = f<\/em>(x<\/em>), then for any constant C<\/em>, we have:<\/p>\n

\"Derivative<\/p>\n

Thus the most general<\/em> antiderivative of f<\/em>(x<\/em>) would be F<\/em>(x<\/em>) + C<\/em>.<\/p>\n

Indefinite Integral Techniques<\/h2>\n

Most everyone knows that you shouldn’t use a screwdriver to pound in a nail. And hammers do not help when driving in screws. In a similar way, you should be aware that each indefinite integral problem requires its own set of tools.<\/p>\n

We’ll discuss a few integration tools, including the basic antiderivative rules, substitution, integration by parts, and partial fractions. Other more advanced tools may be covered in future Magoosh articles.<\/p>\n

Also, it’s important to realize that each technique requires quite a bit of practice before you can really get good at it. Don’t expect to become an expert on the first day.<\/p>\n

Basic Antiderivative Rules<\/h3>\n

These rules are really just derivative rules<\/a> in reverse. Here is a list of the basic antiderivative rules.<\/p>\n

\"Power<\/p>\n

\"Sum<\/p>\n

\"Constant<\/p>\n

\"Constant<\/p>\n

\"Rule<\/p>\n

\"Exponential<\/p>\n

\"Trigonometric<\/p>\n

Substitution<\/h3>\n

The Substitution Rule<\/strong>, or as it’s more commonly known, u<\/em>-substitution, is a rule that “reverses” the Chain Rule.<\/p>\n

\"Substitution<\/p>\n

This rule helps when the integrand is a composition of two functions. That is, if there is a function inside<\/em> another function. For example, we would identify (5x<\/em> + 1)8<\/sup> as a composition of the functions u<\/em> = 5x<\/em> + 1 and f<\/em>(u<\/em>) = u<\/em>8<\/sup>. So if we needed to know the indefinite integral of (5x<\/em> + 1)8<\/sup>, we could use substitution.<\/p>\n

Steps for Substitution<\/h4>\n

Substitution can be difficult because the formula requires a specific setup. However, if you follow the steps outlined below, then you’ll be sure to get it right every time.<\/p>\n

    \n
  1. Identify a part of the function that you will try to substitute, and write it down: u<\/em> = g<\/em>(x<\/em>). It may not be obvious what to pick, so don’t be afraid of a little trial and error at first.<\/li>\n
  2. Take the differential<\/strong> of your substitution. That is, find the derivative of g<\/em> and write it in the form, du<\/em> = g<\/em>\u00a0‘(x<\/em>)\u00a0dx<\/em>.<\/li>\n
  3. Substitute both u<\/em> and du<\/em> into the original integral. This may involve solving the differential for dx<\/em> and then replacing the dx<\/em> in the integral.<\/li>\n
  4. If the new integral involves only u<\/em> and du<\/em>, then simplify and integrate using standard methods.<\/li>\n
  5. Finally, plug u<\/em> = g<\/em>(x<\/em>) back in so that your answer is in terms of the original variable x<\/em>.<\/li>\n<\/ol>\n

    Using the Substitution Rule<\/h4>\n

    \"Substitution<\/p>\n

    First we must decide what to substitute. Experience tells us to look for expressions within parentheses.<\/p>\n

    Don’t forget to take the differential. I find it helpful to solve for dx<\/em>.<\/p>\n

    \"Substitution<\/p>\n

    Now you can replace 5x<\/em> + 1 by u<\/em> and dx<\/em> by (1\/5)du<\/em>. Then integrate and finally plug back in u = 5x<\/em> + 1<\/em>.<\/p>\n

    \"Substitution<\/p>\n

    Integration by Parts<\/h3>\n

    Integration by Parts<\/strong> (IBP) is a powerful method that may be used when there are certain kinds of products in the integrand. In fact, you can think of IBP as a way to “reverse” the Product Rule.<\/p>\n

    Suppose u<\/em> and v<\/em> are differentiable functions of x<\/em>. Then the IBP formula states that:<\/p>\n

    \"Integration<\/p>\n

    Example Using IBP<\/h4>\n

    Typically we use IBP when there are products of powers of x<\/em>, exponential functions, and\/or trigonometric functions in the integrand.<\/p>\n

    \"IBP<\/p>\n

    Here we will choose u<\/em> = 3x<\/em>, and dv<\/em> = cos x<\/em> dx<\/em>. Again, experience guides our choices. If you had chosen the functions the other way around, then the integral would have gotten more complicated.<\/p>\n

    Now find du<\/em> by taking a derivative, and v<\/em> by integrating.<\/p>\n

    \"IBP,<\/p>\n

    Next, we use the IBP formula to rewrite the original integral in a different way. Try to track where u<\/em>, v<\/em>, du<\/em>, and dv<\/em> show up in the problem. (Hint:<\/em> They’re color coded.)<\/p>\n

    \"IBP<\/p>\n

    Partial Fractions<\/h3>\n

    Last but not least, let’s talk about the method of Partial Fractions<\/strong> (PF). We can use PF whenever the integrand is a rational function<\/em> whose denominator has a\u00a0degree\u00a0of at least 2. The main idea is to break apart the fraction into a sum of\u00a0simpler fractions.<\/p>\n

    In this short review, there is not enough time to explain all of the details. So if you’re interested in learning more, check out this article<\/a>.<\/p>\n

    Instead, let’s see a quick example of the technique in action.<\/p>\n

    \"Partial<\/p>\n

    The key is to use your algebra skills to factor the denominator and split into two fractions, solving for the unknown constants in each numerator.<\/p>\n

    \"Partial<\/p>\n

    It can be determined that A<\/em> = -2 and B<\/em> = 3 in this example. Again, because this article is just review, we leave some of the details to you.<\/p>\n

    Now we can work out the problem completely.<\/p>\n

    \"Example<\/p>\n

    Summary<\/h2>\n

    Indefinite integral problems come in many different types on the AP Calculus Exams. Remember that an indefinite integral is the most general antiderivative of a function.<\/p>\n

    Among the wide range of techniques available, most problems can be handled by one or more of the following methods.<\/p>\n

      \n
    • Basic antiderivative formulas, including the Power Rule and rules for special kinds of functions (such as trigonometric and exponential).<\/li>\n
    • Substitution<\/li>\n
    • Integration by parts<\/li>\n
    • Partial fractions<\/li>\n<\/ul>\n

      After much practice, you will be able to choose the best technique for each integral problem. Just like a good carpenter, using the right tool makes the job easy. You may even come to enjoy the challenge of indefinite integrals!<\/p>\n","protected":false},"excerpt":{"rendered":"

      Indefinite integrals make up a substantial part of what is covered on the AP Calculus AB and BC exams. In this review article, we highlight a few concepts and techniques that you’ll need to be familiar with. What are Indefinite Integrals? There are two kinds of integrals, the definite and indefinite integrals. This article only […]<\/p>\n","protected":false},"author":223,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24932],"class_list":["post-8734","post","type-post","status-publish","format-standard","hentry","category-ap","tag-ap-calculus"],"acf":[],"yoast_head":"\nAP Calculus Review: Indefinite Integrals - Magoosh Blog | High School<\/title>\n<meta name=\"description\" content=\"Indefinite integrals make up a substantial part of the AP Calculus AB and BC exams. 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