{"id":691,"date":"2025-06-20T17:03:58","date_gmt":"2025-06-20T17:03:58","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=691"},"modified":"2025-09-10T14:03:30","modified_gmt":"2025-09-10T14:03:30","slug":"integration-by-parts-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integration-by-parts-learn-it-3\/","title":{"raw":"Integration by Parts: Learn It 3","rendered":"Integration by Parts: Learn It 3"},"content":{"raw":"

Integration by Parts for Definite Integrals<\/h2>\r\nNow that we have used integration by parts successfully to evaluate indefinite integrals<\/span>, we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.\r\n\r\n
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integration by parts for definite integrals<\/h3>\r\n

Let [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex] be functions with continuous derivatives on [latex]\\left[a,b\\right][\/latex]. Then<\/p>\r\n\r\n

[latex]{\\displaystyle\\int }_{a}^{b}udv={uv|}_{a}^{b}-{\\displaystyle\\int }_{a}^{b}vdu[\/latex].<\/center><\/div>\r\n<\/section>
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\u00a0Find the area of the region bounded above by the graph of [latex]y={\\tan}^{-1}x[\/latex] and below by the [latex]x[\/latex] -axis over the interval [latex]\\left[0,1\\right][\/latex].<\/div>\r\n
\r\n\r\n[reveal-answer q=\"44558879\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558879\"]\r\n
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This region is shown in Figure 1. To find the area, we must evaluate [latex]\\underset{0}{\\overset{1}{\\displaystyle\\int }}{\\tan}^{-1}xdx[\/latex].<\/p>\r\n\r\n

<\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"562\"]\"This Figure 1. To find the area of the shaded region, we have to use integration by parts.[\/caption]<\/figure>\r\n

For this integral, let\u2019s choose [latex]u={\\tan}^{-1}x[\/latex] and [latex]dv=dx[\/latex], thereby making [latex]du=\\frac{1}{{x}^{2}+1}dx[\/latex] and [latex]v=x[\/latex]. After applying the integration-by-parts formula we obtain<\/p>\r\n\r\n

[latex]\\text{Area}=x{\\tan}^{-1}{x|}_{0}^{1}-\\underset{0}{\\overset{1}{\\displaystyle\\int }}\\frac{x}{{x}^{2}+1}dx[\/latex].<\/div>\r\n \r\n

Use u<\/em>-substitution to obtain<\/p>\r\n\r\n

[latex]\\underset{0}{\\overset{1}{\\displaystyle\\int }}\\frac{x}{{x}^{2}+1}dx=\\frac{1}{2}\\text{ln}|{x}^{2}+1{|}_{0}^{1}[\/latex].<\/div>\r\n \r\n

Thus,<\/p>\r\n\r\n

[latex]\\text{Area}=x{\\tan }^{-1}x{|}_{0}^{1}-\\frac{1}{2}\\mathrm{ln}|{x}^{2}+1|{|}_{0}^{1}=\\frac{\\pi }{4}-\\frac{1}{2}\\mathrm{ln}2[\/latex].<\/div>\r\n \r\n

At this point it might not be a bad idea to do a \"reality check\" on the reasonableness of our solution. Since [latex]\\frac{\\pi }{4}-\\frac{1}{2}\\text{ln}2\\approx 0.4388[\/latex], and from Figure 1 we expect our area to be slightly less than 0.5, this solution appears to be reasonable.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section>

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Find the volume of the solid obtained by revolving the region bounded by the graph of [latex]f\\left(x\\right)={e}^{\\text{-}x}[\/latex], the x<\/em>-axis, the y<\/em>-axis, and the line [latex]x=1[\/latex] about the y<\/em>-axis.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n

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The best option to solving this problem is to use the shell method. Begin by sketching the region to be revolved, along with a typical rectangle (see the following graph).<\/p>\r\n\r\n

<\/figcaption>\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"731\"]\"This Figure 2. We can use the shell method to find a volume of revolution.[\/caption]<\/figure>\r\n

To find the volume using shells, we must evaluate [latex]2\\pi {\\displaystyle\\int }_{0}^{1}x{e}^{\\text{-}x}dx[\/latex]. To do this, let [latex]u=x[\/latex] and [latex]dv={e}^{\\text{-}x}[\/latex]. These choices lead to [latex]du=dx[\/latex] and [latex]v={\\displaystyle\\int }^{\\text{ }}{e}^{\\text{-}x}=\\text{-}{e}^{\\text{-}x}[\/latex]. Substituting into the integration-by-parts for definite integrals formula, we obtain<\/p>\r\n\r\n

[latex]\\begin{array}{cccc}\\hfill \\text{Volume}& =2\\pi \\underset{0}{\\overset{1}{\\displaystyle\\int }}x{e}^{\\text{-}x}dx=2\\pi \\left(\\text{-}x{e}^{\\text{-}x}{|}_{0}^{1}+\\underset{0}{\\overset{1}{\\displaystyle\\int }}{e}^{\\text{-}x}dx\\right)\\hfill & & \\text{Use integration by parts}.\\hfill \\\\ & =-2\\pi x{e}^{\\text{-}x}{|}_{0}^{1}-2\\pi {e}^{\\text{-}x}{|}_{0}^{1}\\hfill & & \\text{Evaluate}\\underset{0}{\\overset{1}{\\displaystyle\\int }}{e}^{\\text{-}x}dx=\\text{-}{e}^{\\text{-}x}{|}_{0}^{1}.\\hfill \\\\ & =2\\pi -\\frac{4\\pi }{e}.\\hfill & & \\text{Evaluate and simplify}.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\n<\/div>\r\n
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Analysis<\/strong><\/div>\r\n

Again, it is a good idea to check the reasonableness of our solution. We observe that the solid has a volume slightly less than that of a cylinder of radius [latex]1[\/latex] and height of [latex]\\frac{1}{e}[\/latex] added to the volume of a cone of base radius [latex]1[\/latex] and height of [latex]1-\\frac{1}{3}[\/latex]. Consequently, the solid should have a volume a bit less than<\/p>\r\n\r\n

[latex]\\pi {\\left(1\\right)}^{2}\\frac{1}{e}+\\left(\\frac{\\pi }{3}\\right){\\left(1\\right)}^{2}\\left(1-\\frac{1}{e}\\right)=\\frac{2\\pi }{3e}-\\frac{\\pi }{3}\\approx 1.8177[\/latex].<\/div>\r\n \r\n

Since [latex]2\\pi -\\frac{4\\pi }{e}\\approx 1.6603[\/latex], we see that our calculated volume is reasonable.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>

[ohm_question hide_question_numbers=1]311292[\/ohm_question]<\/section>","rendered":"

Integration by Parts for Definite Integrals<\/h2>\n

Now that we have used integration by parts successfully to evaluate indefinite integrals<\/span>, we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.<\/p>\n

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integration by parts for definite integrals<\/h3>\n

Let [latex]u=f\\left(x\\right)[\/latex] and [latex]v=g\\left(x\\right)[\/latex] be functions with continuous derivatives on [latex]\\left[a,b\\right][\/latex]. Then<\/p>\n

[latex]{\\displaystyle\\int }_{a}^{b}udv={uv|}_{a}^{b}-{\\displaystyle\\int }_{a}^{b}vdu[\/latex].<\/div>\n<\/div>\n<\/section>\n
\n
\u00a0Find the area of the region bounded above by the graph of [latex]y={\\tan}^{-1}x[\/latex] and below by the [latex]x[\/latex] -axis over the interval [latex]\\left[0,1\\right][\/latex].<\/div>\n
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