{"id":8734,"date":"2017-01-27t09:28:00","date_gmt":"2017-01-27t17:28:00","guid":{"rendered":"\/\/www.catharsisit.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":"\/\/www.catharsisit.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],"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. click here to learn the concepts and techniques that you'll need.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link 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