derivative rules<\/a> in reverse. here is a list of the basic antiderivative rules.<\/p>\n <\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
<\/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 <\/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>\nsteps 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
\nidentify 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>\ntake 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>\nsubstitute 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>\nif the new integral involves only u<\/em> and du<\/em>, then simplify and integrate using standard methods.<\/li>\nfinally, 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>\nusing the substitution rule<\/h4>\n <\/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 <\/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 <\/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>\nsuppose u<\/em> and v<\/em> are differentiable functions of x<\/em>. then the ibp formula states that:<\/p>\n <\/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 <\/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>\nnow find du<\/em> by taking a derivative, and v<\/em> by integrating.<\/p>\n <\/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 <\/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>\nin 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>\ninstead, let’s see a quick example of the technique in action.<\/p>\n
<\/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
<\/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>\nnow we can work out the problem completely.<\/p>\n
<\/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
\nbasic 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>\nafter 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":"\n
ap calculus review: indefinite integrals - magoosh blog | high school<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n