{"id":329,"date":"2023-09-20T22:49:08","date_gmt":"2023-09-20T22:49:08","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-reverse-of-differentiation\/"},"modified":"2024-08-05T13:33:48","modified_gmt":"2024-08-05T13:33:48","slug":"antiderivatives-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/antiderivatives-learn-it-1\/","title":{"raw":"Antiderivatives: Learn It 1","rendered":"Antiderivatives: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Understand indefinite integrals and learn how to find basic antiderivatives for functions<\/li>\r\n\t<li>Use the rule for integrating functions raised to a power<\/li>\r\n\t<li>Use antidifferentiation to solve simple initial-value problems<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Finding the Antiderivative<\/h2>\r\n<p>At this point, we have seen how to calculate derivatives of many functions and have been introduced to a variety of their applications. We now ask a question that turns this process around: Given a function [latex]f[\/latex], how do we find a function with the derivative [latex]f[\/latex] and why would we be interested in such a function?<\/p>\r\n<p>We answer the first part of this question by defining antiderivatives. The <strong>antiderivative<\/strong> of a function [latex]f[\/latex] is simply a function whose derivative is [latex]f[\/latex]. But why is this concept important?<\/p>\r\n<p>Antiderivatives are essential for solving problems where you need to reverse the process of differentiation. For instance, consider rectilinear motion: If you know an object's position function [latex]s(t)[\/latex], then its velocity [latex]v(t)[\/latex] is the derivative [latex]s\u2032(t)[\/latex]. If you have the acceleration [latex]a(t)[\/latex] which is [latex]v\u2032(t)[\/latex], and need to find the velocity, you would look for an antiderivative of [latex]a(t)[\/latex].<\/p>\r\n<p>This need to find antiderivatives is not limited to physics; it arises in various fields and applications, prompting the development of methods to find antiderivatives for complex functions. These methods and more are covered in detail under the topic 'Techniques of Integration' in the second volume of this text.<\/p>\r\n<h3>The Reverse of Differentiation<\/h3>\r\n<p id=\"fs-id1165043323795\">At this point, we know how to find derivatives of various functions. We now ask the opposite question. Given a function [latex]f[\/latex], how can we find a function with derivative [latex]f[\/latex]? If we can find a function [latex]F[\/latex] with derivative [latex]f[\/latex], we call [latex]F[\/latex] an antiderivative of [latex]f[\/latex].<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<div class=\"title\">\r\n<h3 style=\"text-align: left;\">antiderivative<\/h3>\r\n<\/div>\r\n<p id=\"fs-id1165042887564\">A function [latex]F[\/latex] is an antiderivative of the function [latex]f[\/latex] if<\/p>\r\n<div id=\"fs-id1165042945928\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]F^{\\prime}(x)=f(x)[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1165042964886\">for all [latex]x[\/latex] in the domain of [latex]f[\/latex].<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Consider the function [latex]f(x)=2x[\/latex].<\/p>\r\n<p>Knowing the power rule of differentiation, we conclude that [latex]F(x)=x^2[\/latex] is an antiderivative of [latex]f[\/latex] since [latex]F^{\\prime}(x)=2x[\/latex].<\/p>\r\n<p>Are there any other antiderivatives of [latex]f[\/latex]?<\/p>\r\n<p>Yes; since the derivative of any constant [latex]C[\/latex] is zero, [latex]x^2+C[\/latex] is also an antiderivative of [latex]2x[\/latex]. Therefore, [latex]x^2+5[\/latex] and [latex]x^{2}-\\sqrt{2}[\/latex] are also antiderivatives.<\/p>\r\n<p>Are there any others that are not of the form [latex]x^2+C[\/latex] for some constant [latex]C[\/latex]?<\/p>\r\n<p>The answer is no.<\/p>\r\n<\/section>\r\n<p>From Corollary 2 of the Mean Value Theorem, we know that if [latex]F[\/latex] and [latex]G[\/latex] are differentiable functions such that [latex]F^{\\prime}(x)=G^{\\prime}(x)[\/latex], then [latex]F(x)-G(x)=C[\/latex] for some constant [latex]C[\/latex]. This fact leads to the following important theorem.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">General Form of an Antiderivative<\/h3>\r\n<p id=\"fs-id1165042884523\">Let [latex]F[\/latex] be an antiderivative of [latex]f[\/latex] over an interval [latex]I[\/latex]. Then,<\/p>\r\n<ol id=\"fs-id1165043009392\">\r\n\t<li>for each constant [latex]C[\/latex], the function [latex]F(x)+C[\/latex] is also an antiderivative of [latex]f[\/latex] over [latex]I[\/latex];<\/li>\r\n\t<li>if [latex]G[\/latex] is an antiderivative of [latex]f[\/latex] over [latex]I[\/latex], there is a constant [latex]C[\/latex] for which [latex]G(x)=F(x)+C[\/latex] over [latex]I[\/latex].<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1165042987370\">In other words, the most general form of the antiderivative of [latex]f[\/latex] over [latex]I[\/latex] is [latex]F(x)+C[\/latex].<\/p>\r\n<\/section>\r\n<p>We use this fact and our knowledge of derivatives to find all the antiderivatives for several functions.<\/p>\r\n<section class=\"textbox example\">\r\n<p>For each of the following functions, find all antiderivatives.<\/p>\r\n<ol id=\"fs-id1165043115403\" style=\"list-style-type: lower-alpha;\">\r\n\t<li>[latex]f(x)=3x^2[\/latex]<\/li>\r\n\t<li>[latex]f(x)=\\dfrac{1}{x}[\/latex]<\/li>\r\n\t<li>[latex]f(x)= \\cos x[\/latex]<\/li>\r\n\t<li>[latex]f(x)=e^x[\/latex]<\/li>\r\n<\/ol>\r\n<p>[reveal-answer q=\"46129\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"46129\"]<\/p>\r\n<p>a. Because<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^3)=3x^2[\/latex]<\/p>\r\n<p>then [latex]F(x)=x^3[\/latex] is an antiderivative of [latex]3x^2[\/latex]. Therefore, every antiderivative of [latex]3x^2[\/latex] is of the form [latex]x^3+C[\/latex] for some constant [latex]C[\/latex], and every function of the form [latex]x^3+C[\/latex] is an antiderivative of [latex]3x^2[\/latex].<\/p>\r\n<p>b. Let [latex]f(x)=\\ln |x|[\/latex]. For [latex]x&gt;0, \\, f(x)=\\ln (x)[\/latex] and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\ln x)=\\dfrac{1}{x}[\/latex]<\/p>\r\n<p>For [latex]x&lt;0, \\, f(x)=\\ln (\u2212x)[\/latex] and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\ln (\u2212x))=-\\dfrac{1}{\u2212x}=\\dfrac{1}{x}[\/latex]<\/p>\r\n<p>Therefore,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\ln |x|)=\\dfrac{1}{x}[\/latex]<\/p>\r\n<p>Thus, [latex]F(x)=\\ln |x|[\/latex] is an antiderivative of [latex]\\frac{1}{x}[\/latex]. Therefore, every antiderivative of [latex]\\frac{1}{x}[\/latex] is of the form [latex]\\ln |x|+C[\/latex] for some constant [latex]C[\/latex] and every function of the form [latex]\\ln |x|+C[\/latex] is an antiderivative of [latex]\\frac{1}{x}[\/latex].<\/p>\r\n<p>c. We have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\sin x)= \\cos x[\/latex],<\/p>\r\n<p>so [latex]F(x)= \\sin x[\/latex] is an antiderivative of [latex] \\cos x[\/latex]. Therefore, every antiderivative of [latex] \\cos x[\/latex] is of the form [latex] \\sin x+C[\/latex] for some constant [latex]C[\/latex] and every function of the form [latex] \\sin x+C[\/latex] is an antiderivative of [latex] \\cos x[\/latex].<\/p>\r\n<p>d. Since<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}(e^x)=e^x[\/latex],<\/p>\r\n<p>then [latex]F(x)=e^x[\/latex] is an antiderivative of [latex]e^x[\/latex]. Therefore, every antiderivative of [latex]e^x[\/latex] is of the form [latex]e^x+C[\/latex] for some constant [latex]C[\/latex] and every function of the form [latex]e^x+C[\/latex] is an antiderivative of [latex]e^x[\/latex].<\/p>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/j81IZAEfwhI?controls=0&amp;start=47&amp;end=158&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.10Antiderivatives47to158_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.10 Antiderivatives\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]5318[\/ohm_question]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Understand indefinite integrals and learn how to find basic antiderivatives for functions<\/li>\n<li>Use the rule for integrating functions raised to a power<\/li>\n<li>Use antidifferentiation to solve simple initial-value problems<\/li>\n<\/ul>\n<\/section>\n<h2>Finding the Antiderivative<\/h2>\n<p>At this point, we have seen how to calculate derivatives of many functions and have been introduced to a variety of their applications. We now ask a question that turns this process around: Given a function [latex]f[\/latex], how do we find a function with the derivative [latex]f[\/latex] and why would we be interested in such a function?<\/p>\n<p>We answer the first part of this question by defining antiderivatives. The <strong>antiderivative<\/strong> of a function [latex]f[\/latex] is simply a function whose derivative is [latex]f[\/latex]. But why is this concept important?<\/p>\n<p>Antiderivatives are essential for solving problems where you need to reverse the process of differentiation. For instance, consider rectilinear motion: If you know an object&#8217;s position function [latex]s(t)[\/latex], then its velocity [latex]v(t)[\/latex] is the derivative [latex]s\u2032(t)[\/latex]. If you have the acceleration [latex]a(t)[\/latex] which is [latex]v\u2032(t)[\/latex], and need to find the velocity, you would look for an antiderivative of [latex]a(t)[\/latex].<\/p>\n<p>This need to find antiderivatives is not limited to physics; it arises in various fields and applications, prompting the development of methods to find antiderivatives for complex functions. These methods and more are covered in detail under the topic &#8216;Techniques of Integration&#8217; in the second volume of this text.<\/p>\n<h3>The Reverse of Differentiation<\/h3>\n<p id=\"fs-id1165043323795\">At this point, we know how to find derivatives of various functions. We now ask the opposite question. Given a function [latex]f[\/latex], how can we find a function with derivative [latex]f[\/latex]? If we can find a function [latex]F[\/latex] with derivative [latex]f[\/latex], we call [latex]F[\/latex] an antiderivative of [latex]f[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\">\n<div class=\"title\">\n<h3 style=\"text-align: left;\">antiderivative<\/h3>\n<\/div>\n<p id=\"fs-id1165042887564\">A function [latex]F[\/latex] is an antiderivative of the function [latex]f[\/latex] if<\/p>\n<div id=\"fs-id1165042945928\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]F^{\\prime}(x)=f(x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042964886\">for all [latex]x[\/latex] in the domain of [latex]f[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Consider the function [latex]f(x)=2x[\/latex].<\/p>\n<p>Knowing the power rule of differentiation, we conclude that [latex]F(x)=x^2[\/latex] is an antiderivative of [latex]f[\/latex] since [latex]F^{\\prime}(x)=2x[\/latex].<\/p>\n<p>Are there any other antiderivatives of [latex]f[\/latex]?<\/p>\n<p>Yes; since the derivative of any constant [latex]C[\/latex] is zero, [latex]x^2+C[\/latex] is also an antiderivative of [latex]2x[\/latex]. Therefore, [latex]x^2+5[\/latex] and [latex]x^{2}-\\sqrt{2}[\/latex] are also antiderivatives.<\/p>\n<p>Are there any others that are not of the form [latex]x^2+C[\/latex] for some constant [latex]C[\/latex]?<\/p>\n<p>The answer is no.<\/p>\n<\/section>\n<p>From Corollary 2 of the Mean Value Theorem, we know that if [latex]F[\/latex] and [latex]G[\/latex] are differentiable functions such that [latex]F^{\\prime}(x)=G^{\\prime}(x)[\/latex], then [latex]F(x)-G(x)=C[\/latex] for some constant [latex]C[\/latex]. This fact leads to the following important theorem.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">General Form of an Antiderivative<\/h3>\n<p id=\"fs-id1165042884523\">Let [latex]F[\/latex] be an antiderivative of [latex]f[\/latex] over an interval [latex]I[\/latex]. Then,<\/p>\n<ol id=\"fs-id1165043009392\">\n<li>for each constant [latex]C[\/latex], the function [latex]F(x)+C[\/latex] is also an antiderivative of [latex]f[\/latex] over [latex]I[\/latex];<\/li>\n<li>if [latex]G[\/latex] is an antiderivative of [latex]f[\/latex] over [latex]I[\/latex], there is a constant [latex]C[\/latex] for which [latex]G(x)=F(x)+C[\/latex] over [latex]I[\/latex].<\/li>\n<\/ol>\n<p id=\"fs-id1165042987370\">In other words, the most general form of the antiderivative of [latex]f[\/latex] over [latex]I[\/latex] is [latex]F(x)+C[\/latex].<\/p>\n<\/section>\n<p>We use this fact and our knowledge of derivatives to find all the antiderivatives for several functions.<\/p>\n<section class=\"textbox example\">\n<p>For each of the following functions, find all antiderivatives.<\/p>\n<ol id=\"fs-id1165043115403\" style=\"list-style-type: lower-alpha;\">\n<li>[latex]f(x)=3x^2[\/latex]<\/li>\n<li>[latex]f(x)=\\dfrac{1}{x}[\/latex]<\/li>\n<li>[latex]f(x)= \\cos x[\/latex]<\/li>\n<li>[latex]f(x)=e^x[\/latex]<\/li>\n<\/ol>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q46129\">Show Solution<\/button><\/p>\n<div id=\"q46129\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. Because<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^3)=3x^2[\/latex]<\/p>\n<p>then [latex]F(x)=x^3[\/latex] is an antiderivative of [latex]3x^2[\/latex]. Therefore, every antiderivative of [latex]3x^2[\/latex] is of the form [latex]x^3+C[\/latex] for some constant [latex]C[\/latex], and every function of the form [latex]x^3+C[\/latex] is an antiderivative of [latex]3x^2[\/latex].<\/p>\n<p>b. Let [latex]f(x)=\\ln |x|[\/latex]. For [latex]x>0, \\, f(x)=\\ln (x)[\/latex] and<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\ln x)=\\dfrac{1}{x}[\/latex]<\/p>\n<p>For [latex]x<0, \\, f(x)=\\ln (\u2212x)[\/latex] and<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\ln (\u2212x))=-\\dfrac{1}{\u2212x}=\\dfrac{1}{x}[\/latex]<\/p>\n<p>Therefore,<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\ln |x|)=\\dfrac{1}{x}[\/latex]<\/p>\n<p>Thus, [latex]F(x)=\\ln |x|[\/latex] is an antiderivative of [latex]\\frac{1}{x}[\/latex]. Therefore, every antiderivative of [latex]\\frac{1}{x}[\/latex] is of the form [latex]\\ln |x|+C[\/latex] for some constant [latex]C[\/latex] and every function of the form [latex]\\ln |x|+C[\/latex] is an antiderivative of [latex]\\frac{1}{x}[\/latex].<\/p>\n<p>c. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}(\\sin x)= \\cos x[\/latex],<\/p>\n<p>so [latex]F(x)= \\sin x[\/latex] is an antiderivative of [latex]\\cos x[\/latex]. Therefore, every antiderivative of [latex]\\cos x[\/latex] is of the form [latex]\\sin x+C[\/latex] for some constant [latex]C[\/latex] and every function of the form [latex]\\sin x+C[\/latex] is an antiderivative of [latex]\\cos x[\/latex].<\/p>\n<p>d. Since<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{d}{dx}(e^x)=e^x[\/latex],<\/p>\n<p>then [latex]F(x)=e^x[\/latex] is an antiderivative of [latex]e^x[\/latex]. Therefore, every antiderivative of [latex]e^x[\/latex] is of the form [latex]e^x+C[\/latex] for some constant [latex]C[\/latex] and every function of the form [latex]e^x+C[\/latex] is an antiderivative of [latex]e^x[\/latex].<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/j81IZAEfwhI?controls=0&amp;start=47&amp;end=158&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.10Antiderivatives47to158_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.10 Antiderivatives&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm5318\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5318&theme=lumen&iframe_resize_id=ohm5318&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":28,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.10 Antiderivatives\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":292,"module-header":"learn_it","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"4.10 Antiderivatives","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/329"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/329\/revisions"}],"predecessor-version":[{"id":4563,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/329\/revisions\/4563"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/292"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/329\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=329"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=329"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=329"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=329"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}