{"id":8687,"date":"2017-01-27t16:00:35","date_gmt":"2017-01-28t00:00:35","guid":{"rendered":"\/\/www.catharsisit.com\/hs\/?p=8687"},"modified":"2017-01-27t11:47:38","modified_gmt":"2017-01-27t19:47:38","slug":"ap-calculus-exam-review-integrals","status":"publish","type":"post","link":"\/\/www.catharsisit.com\/hs\/ap\/ap-calculus-exam-review-integrals\/","title":{"rendered":"ap calculus exam review: integrals"},"content":{"rendered":"
the ap calculus exams include a substantial number of problems involving integrals. in this short review, you’ll see the major topics that you’ll need to be aware of.<\/p>\n
there are two kinds of integrals, the indefinite<\/em> and definite<\/em> integrals. they differ in many ways, however the notation is almost the same.<\/p>\n the numbers a<\/em> and b<\/em> in the definite integral notation are called the limits<\/strong> or bounds<\/strong> of integration.<\/p>\n an indefinite integral<\/strong> is another name for the antiderivative<\/strong>. in fact, the indefinite integral stands for every possible antiderivative! that’s where the ”\u00a0+\u00a0c<\/em>\u00a0” comes in.<\/p>\n here, the function f<\/em> is an 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 but there are many other antiderivatives! because the derivative of any constant is 0, it’s also true that (x<\/em>2<\/sup> + 1)’ = 2x<\/em>, or (x<\/em>2<\/sup> – 65.32236)’ = 2x<\/em>.<\/p>\n therefore, the most general<\/em> antiderivative of 2x<\/em> would be x<\/em>2<\/sup> + c<\/em>, where c<\/em> is an as-yet-undetermined constant.<\/p>\n this is why you always tack on that ”\u00a0+\u00a0c<\/em>\u00a0” at the end of your answer.<\/p>\n now let’s talk about finding that function f<\/em>(x<\/em>). that’s the hard part. there are various rules and formulas that can help to find the integral (antiderivative). perhaps the most important rules include the power rule and the rules for breaking down sums, differences, and constant multiples.<\/p>\n for example, these rules allow you to integrate any polynomial. let’s see an example:<\/p>\n notice near the end, we used \u222b 7dx<\/em> = 7x<\/em>. this is because the derivative of 7x<\/em> is equal to 7. it works the same way for any constant term.<\/p>\n there are many more integral formulas, techniques, and methods that you must master to score high on the ap calculus exam. for more information, check out this list of basic antiderivative formulas<\/a> as well as this detailed review of integration techniques<\/a>.<\/p>\n definite integrals measure the area under a curve (among other things). the limits of integration, a<\/em> and b<\/em>, specify the left and right boundaries for the area under the graph of f<\/em>(x<\/em>) and above the x<\/em>-axis.<\/p>\n to evaluate a definite integral, use the fundamental theorem of calculus<\/strong>:<\/p>\n in this formula, f<\/em> is the same thing it was before: an antiderivative for f<\/em>. so the process goes like this:<\/p>\n find the area under y<\/em> = 9 – x<\/em>2<\/sup> between x<\/em> = -1 and x<\/em> = 2.<\/p>\n we first must set up the problem as a definite integral. then evaluate the integral to find the area.<\/p>\n disclaimer:<\/strong> the idea of “area under a function” really only makes sense if f<\/em>(x<\/em>) \u2265 0 on the interval. in fact, it’s more correct to say that the definite integral computes the net area<\/em> \u2014 that is, area above the x<\/em>-axis minus area below the x<\/em>-axis.<\/p>\n here’s an example, using only the concept of net area to solve the integral problem.<\/p>\n find how in the world can we evaluate the integral if we’re not even given a formula for f<\/em>(x<\/em>)? well, we are given the graph. this problem relies on our ability to interpret the definite integral as a net area.<\/p>\n just find the area of each triangles. remember, a<\/em> = (1\/2)bh<\/em>.<\/p>\n the triangle above the axis counts as positive, while the one below the axis counts as negative.<\/p>\n therefore, the final answer is: 2 + (-3) = -1.<\/p>\n not only is it essential to know how to compute integrals, but you also must know what they’re good for. we’ve already talked about definite integrals and area. here are a few additional applications that you should be aware of.<\/p>\n if s<\/em>(t<\/em>) is a position<\/em> function, then you know from differential calculus that the derivative, v<\/em>(t<\/em>) = s<\/em>\u00a0‘(t<\/em>) is the velocity<\/em> function. moreover, the second derivative, a<\/em>(t<\/em>) = s<\/em>\u00a0”(t<\/em>) is the acceleration<\/em> function.<\/p>\n you can use integrals to go the other way. in other words, if you are given acceleration, you can find a velocity function, and if given velocity, then you can find a position function.<\/p>\n typically further information may be required to nail down exactly what the appropriate constants are.<\/p>\n the length of a curve y<\/em> = f<\/em>(x<\/em>) between x<\/em> = a<\/em> and x<\/em> = b<\/em> may be computed using the following definite integral.<\/p>\n solids of revolution can be quite tricky. however, it’s good to know that definite integrals are just the right tool for computing their volumes. the basic formulas are as follows:<\/p>\n there are many topics related to integrals on the ap calculus exams. each topic requires practice and study, but hopefully this review will get you started. remember the key points:<\/p>\n the ap calculus exams include many problems involving integrals. in this short review, you’ll see the major topics that you’ll need to be aware of.<\/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-8687","post","type-post","status-publish","format-standard","hentry","category-ap","tag-ap-calculus"],"acf":[],"yoast_head":"\n<\/p>\n
indefinite integrals<\/h2>\n
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don’t forget the ”\u00a0+\u00a0c<\/em>\u00a0“<\/h3>\n
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finding antiderivatives<\/h3>\n
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more formulas?<\/h3>\n
definite integrals and the fundamental theorem of calculus<\/h2>\n
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an example definite integral<\/h3>\n
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net area<\/h3>\n
, based on the graph below.<\/p>\n
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applications of integrals<\/h2>\n
position, velocity, and acceleration<\/h3>\n
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length of a curve<\/h3>\n
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volume of solids of revolution<\/h3>\n
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summary<\/h2>\n
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