{"id":8182,"date":"2016-12-08t11:45:36","date_gmt":"2016-12-08t19:45:36","guid":{"rendered":"\/\/www.catharsisit.com\/hs\/?p=8182"},"modified":"2016-12-02t15:40:21","modified_gmt":"2016-12-02t23:40:21","slug":"computing-definite-integral-polynomial","status":"publish","type":"post","link":"\/\/www.catharsisit.com\/hs\/ap\/computing-definite-integral-polynomial\/","title":{"rendered":"computing the definite integral of a polynomial"},"content":{"rendered":"<p>we want to focus on the definite integral of a polynomial function. these arise very commonly in calculus, so here are detailed solutions to two problems, one multiple-choice and one free-response, involving a definite integral of polynomial.<\/p>\n<p>&nbsp;<\/p>\n<h2>free-response definite integrals:<\/h2>\n<p>you will not commonly be asked to evaluate common definite integrals on the free-response, but rather you will be asked to find an area or compute a volume, which will require computing a common definite integral. suppose we want to compute the volume of the solid obtained by revolving the function <img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_5aa995b812c13793d7f85e44c6a31c35.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"r(x) = -{1\/6}(x - 6)(x+6)\" title=\"r(x) = -{1\/6}(x - 6)(x+6)\"\/> about the x-axis:<\/p>\n<p><a href=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/ap\/files\/2013\/10\/29160005\/ctdioap_img1.jpg\"><img decoding=\"async\" class=\"alignnone  wp-image-66\" alt=\"ctdioap_img1\" src=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/ap\/files\/2013\/10\/29160005\/ctdioap_img1.jpg\" width=\"512\" height=\"341\" \/><\/a><\/p>\n<p>the cross sections when cutting perpendicular to the x-axis are circles with radius given by the function\u00a0<img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_5aa995b812c13793d7f85e44c6a31c35.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"r(x) = -{1\/6}(x - 6)(x+6)\" title=\"r(x) = -{1\/6}(x - 6)(x+6)\"\/>.\u00a0the definite integral that needs to be evaluated is\u00a0<img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_968_4f3e1805611ddc4826cc00247b2e6b8a.png\" style=\"vertical-align:-32px; display: inline-block ;\" alt=\"int{-6}{6}{pi.r(x)^2 dx}\" title=\"int{-6}{6}{pi.r(x)^2 dx}\"\/>, \u00a0since this is the area of a circle multiplied by the length of the interval from -6 to 6. we compute:<\/p>\n<p><a href=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121403\/ctdioap_img2.jpg\"><img decoding=\"async\" class=\"size-full wp-image-67 aligncenter\" alt=\"ctdioap_img2\" src=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121403\/ctdioap_img2.jpg\" width=\"515\" height=\"77\" \/><\/a><\/p>\n<p>therefore to compute the integral we compute the sum of the integrals of the individual terms, since polynomials are sums of continuous functions:<\/p>\n<p><a href=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121358\/ctdioap_img3.jpg\"><img decoding=\"async\" class=\"size-full wp-image-68 aligncenter\" alt=\"ctdioap_img3\" src=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121358\/ctdioap_img3.jpg\" width=\"429\" height=\"99\" \/><\/a><\/p>\n<h3>recall the fundamental theorem of calculus (ftc):<\/h3>\n<p><b>theorem:\u00a0<\/b><i>if v(x)\u00a0<i>is a continuous function with an antiderivative v(x),\u00a0<\/i><\/i><i>then <img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_974.5_ad821d5c5e8f6c4a4230c270baf5c2ad.png\" style=\"vertical-align:-25.5px; display: inline-block ;\" alt=\"v(b) - v(a) =int{a}{b}{v(x) dx}\" title=\"v(b) - v(a) =int{a}{b}{v(x) dx}\"\/> \u00a0<i>where <\/i><i>, <\/i><i>\u00a0are in the domain of v(x).\u00a0<\/i><\/i><\/p>\n<p>the ftc says that we can pick any old antiderivative <em>v(x)<\/em> for <em>v(x)<\/em>, so we need to compute a string of antiderivatives for the integrands of the terms in the sum. in the previous post we discussed but did not state:<\/p>\n<p><b>the power rule:<\/b> the derivative <img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_983.5_6f9bff368a87d90d5801c50ff4912b65.png\" style=\"vertical-align:-16.5px; display: inline-block ;\" alt=\"(x^n)\" title=\"(x^n)\"\/>&#8216;=<img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993.5_13ba2b45efe682ca14edcdcd4337dc3a.png\" style=\"vertical-align:-6.5px; display: inline-block ;\" alt=\"nx^n\" title=\"nx^n\"\/><\/p>\n<p>we used this to find that the integral <img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_983.5_99854ee6b93b7f6b63c9904ef65c0e42.png\" style=\"vertical-align:-16.5px; display: inline-block ;\" alt=\"int{ }{ }{x^n dx} = {1\/{n+1}}x^{n+1} + c\" title=\"int{ }{ }{x^n dx} = {1\/{n+1}}x^{n+1} + c\"\/>,\u00a0and since we only need one antiderivative to evaluate definite integrals, we can take \u00a0for use in this case.<\/p>\n<p>therefore we can evaluate (using the fact that\u00a0<img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_4a5f7db1b164567865cf5fffffc17dda.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"int{ }{ }{x^4 dx} = {{x^5}\/5}\" title=\"int{ }{ }{x^4 dx} = {{x^5}\/5}\"\/>,\u00a0<img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_1330d4af832c92e570e5cb575346c81a.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"int{ }{ }{x^2 dx} = {{x^3}\/3}\" title=\"int{ }{ }{x^2 dx} = {{x^3}\/3}\"\/>,\u00a0<img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_3bd934f224fbb25065c18865804cd11a.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"int{ }{ }{x^0 dx} = int{ }{ }{1 dx} =\u00a0{{x^1}\/1} = x\u00a0\" title=\"int{ }{ }{x^0 dx} = int{ }{ }{1 dx} =\u00a0{{x^1}\/1} = x\u00a0\"\/>\u00a0and the ftc):<\/p>\n<p><a href=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121354\/ctdioap_img4.jpg\"><img decoding=\"async\" class=\"size-full wp-image-69 aligncenter\" alt=\"ctdioap_img4\" src=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121354\/ctdioap_img4.jpg\" width=\"543\" height=\"137\" \/><\/a><\/p>\n<p>you can use your calculator to get 723.823 units cubed.<\/p>\n<p>&nbsp;<\/p>\n<h2>multiple-choice definite integrals:<\/h2>\n<p>here is a sample of a typical multiple-choice question asking for you to formulate a definite integral based on the same concept discussed above.<\/p>\n<p><b>question: <\/b>a solid is generated by revolving the region enclosed by the function <img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993.5_798d364fe4315e98e380ada5e580f525.png\" style=\"vertical-align:-6.5px; display: inline-block ;\" alt=\"y = 2 sqrt{x}\" title=\"y = 2 sqrt{x}\"\/>, and the lines x=2, x=3, y=1\u00a0about the x-axis. which of the following definite integrals gives the volume of the solid? (hint: draw a picture)<\/p>\n<p><a href=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121347\/ctdioap_img5.jpg\"><img decoding=\"async\" class=\"alignnone size-full wp-image-70\" alt=\"ctdioap_img5\" src=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121347\/ctdioap_img5.jpg\" width=\"140\" height=\"133\" \/><\/a><\/p>\n<p>the idea for this problem is to recognize that this solid is a difference of integrals. suppose that we had the volume of the function\u00a0<img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993.5_656b40b88fabb11b531f0129454c8b13.png\" style=\"vertical-align:-6.5px; display: inline-block ;\" alt=\"y=2 sqrt{x}\" title=\"y=2 sqrt{x}\"\/>\u00a0when bounded by the lines\u00a0\u00a0x = 2, x = 3,\u00a0and rotated about the x-axis\u2014then we would have the volume of the following solid:<\/p>\n<p><a href=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121345\/ctdioap_img6.jpg\"><img decoding=\"async\" class=\"alignnone  wp-image-71\" alt=\"ctdioap_img6\" src=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121345\/ctdioap_img6.jpg\" width=\"482\" height=\"339\" \/><\/a><\/p>\n<p>given this volume, we would only need to subtract the volume of the following figure, derived by rotating y=1\u00a0bounded by x=2, x=3,\u00a0about the x-axis:<\/p>\n<p><a href=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121330\/ctdioap_img7.jpg\"><img decoding=\"async\" class=\"alignnone  wp-image-72\" alt=\"ctdioap_img7\" src=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121330\/ctdioap_img7.jpg\" width=\"496\" height=\"288\" \/><\/a><\/p>\n<p>from the upper volume, with radius <img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_985.5_64eee64b78413f5bdfd770b5eb43a872.png\" style=\"vertical-align:-14.5px; display: inline-block ;\" alt=\"r_1 = 2sqrt{x}\" title=\"r_1 = 2sqrt{x}\"\/>:<\/p>\n<p><a href=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121312\/ctdioap_img8.jpg\"><img decoding=\"async\" class=\"alignnone  wp-image-73\" alt=\"ctdioap_img8\" src=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121312\/ctdioap_img8.jpg\" width=\"488\" height=\"327\" \/><\/a><\/p>\n<p>therefore we need to subtract the two integrals, however using the integral laws we can express this in the form <img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_983.5_faac5e813440f5e8c107dda14a2c630c.png\" style=\"vertical-align:-16.5px; display: inline-block ;\" alt=\"pi.({r_1}^2 - {r_2}^2) dx\" title=\"pi.({r_1}^2 - {r_2}^2) dx\"\/>,\u00a0which we follow up by substitution of our names for <img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_985.5_f4d668514f2652dd003ea5f2fbd46a9a.png\" style=\"vertical-align:-14.5px; display: inline-block ;\" alt=\"r_1\" title=\"r_1\"\/>,<img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_985.5_a7227958d60e57a69d2eec173d11ebda.png\" style=\"vertical-align:-14.5px; display: inline-block ;\" alt=\"r_2\" title=\"r_2\"\/>:<\/p>\n<p><a href=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121258\/ctdioap_img9.jpg\"><img decoding=\"async\" class=\"size-full wp-image-74 aligncenter\" alt=\"ctdioap_img9\" src=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121258\/ctdioap_img9.jpg\" width=\"498\" height=\"100\" \/><\/a><\/p>\n<p>so the answer is a.<br \/>\nto compute the value of the integral we see that<br \/>\n<a href=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121253\/ctdioap_img10.jpg\"><img decoding=\"async\" class=\"size-full wp-image-75 aligncenter\" alt=\"ctdioap_img10\" src=\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/sites\/10\/2013\/10\/27121253\/ctdioap_img10.jpg\" width=\"482\" height=\"152\" \/><\/a><\/p>\n<p>this has the value <img decoding=\"async\" src=\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993.5_8161a91c6f2da1629d0649d34608a685.png\" style=\"vertical-align:-6.5px; display: inline-block ;\" alt=\"28.2743 units^2\" title=\"28.2743 units^2\"\/>.<\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>we want to focus on the definite integral of a polynomial function. these arise very commonly in calculus, so here are detailed solutions to two problems, one multiple-choice and one free-response, involving a definite integral of polynomial. &nbsp; free-response definite integrals: you will not commonly be asked to evaluate common definite integrals on the free-response, [&hellip;]<\/p>\n","protected":false},"author":48,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24929],"class_list":["post-8182","post","type-post","status-publish","format-standard","hentry","category-ap","tag-ap-calculus"],"acf":[],"yoast_head":"<!-- this site is optimized with the yoast seo premium plugin v21.7 (yoast seo v21.7) - 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