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<div id="main-column" class="span9"> <article id="integration-by-substitution-tutorial" style="margin-top: 16px;">
<h1 class="section-title">Integration by substitution (tutorial)</h1>
<ul class="breadcrumb"><li><a title="Home" href="../../../index.html"><i class="fa fa-home"></i></a><span class="divider">/</span></li><li><span class="gray">5. Calculus</span><span class="divider">/</span></li><li><span class="active">Integration by substitution (tutorial)</span></li></ul>
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<h2><u>Tutorial</u>: Integration by substitution</h2><p><img alt="" src="../../../ib/mathanalysis/analysis/calculus/images/leibniz_img1-1.jpg" style="float: right; width: 300px; height: 185px;">Throughout this tutorial you will often see the Leibniz notation for derivatives and differentials, named after the German mathematician Gottfried Wilhelm Leibniz (1646-1716). The beauty of this notation is that it provides an easy way to remember several important calculus formulas by making it seem as though the formulas were derived from algebraic manipulations of differentials. For instance, in Leibniz notation, the rule for differentiating a composite function – often referred to as the <strong>Chain Rule</strong> – is written as <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mfrac> <mrow> <mi>d</mi><mi>y</mi></mrow> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> <mo>=</mo><mfrac> <mrow> <mi>d</mi><mi>y</mi></mrow> <mrow> <mi>d</mi><mi>u</mi></mrow> </mfrac> <mo>⋅</mo><mfrac> <mrow> <mi>d</mi><mi>u</mi></mrow> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> </mrow></math></span> </span> .</p><p>This statement appears to be true because the <em>du</em>’s cancel. Even though this reasoning is <em>theoretically</em> <em><u>in</u>correct</em>, it is reasonable to consider that differentials such as <em>dx</em> and <em>dy</em> represent infinitesimally small numbers so in practice differentials can be manipulated using the rules of algebra.</p><p>In fact, the formula booklets for IB Maths HL & SL express the Chain Rule as follows.</p><p style="text-align: center;"><img alt="" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/chain_rule_1.jpg" style="width: 413px; height: 62px;"></p><p>The following is a formal definition of the Chain Rule that one may find in a standard calculus textbook.</p><p style="text-align: center;"><img alt="" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/chain_rule_2.jpg" style="width: 529px; height: 276px;"></p><h3>Using the Chain Rule</h3><p><strong>Example 1</strong>: Find <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mfrac> <mrow> <mi>d</mi><mi>y</mi></mrow> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> </mrow></math></span> </span> for <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>y</mi><mo>=</mo><msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>3</mn> </msup> </mrow></math></span> </span> .</p><p><em>Explanation/Solution</em>:</p><p>For this function, you can consider the “inside” function to be <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow><mo>=</mo><mi>u</mi><mo>=</mo><msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow></math></span> </span> and the “outside” function to be <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>f</mi><mrow><mo>(</mo> <mi>u</mi> <mo>)</mo></mrow><mo>=</mo><mi>y</mi><mo>=</mo><msup> <mi>u</mi> <mn>3</mn> </msup> </mrow></math></span> </span> . Hence, <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>y</mi><mo>=</mo><mi>f</mi><mrow><mo>(</mo> <mrow> <mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow> <mo>)</mo></mrow><mo>=</mo><mi>f</mi><mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow><mo>=</mo><msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>3</mn> </msup> </mrow></math></span> </span> . Then by the Chain Rule, you obtain <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mfrac> <mrow> <mi>d</mi><mi>y</mi></mrow> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> <mo>=</mo><munder> <munder> <mrow> <mn>3</mn><msup> <mi>u</mi> <mn>2</mn> </msup> </mrow> <mo stretchy="true">︸</mo> </munder> <mrow> <mfrac> <mrow> <mi>d</mi><mi>y</mi></mrow> <mrow> <mi>d</mi><mi>u</mi></mrow> </mfrac> </mrow> </munder> <munder> <munder> <mrow> <mrow><mo>(</mo> <mrow> <mn>2</mn><mi>x</mi></mrow> <mo>)</mo></mrow></mrow> <mo stretchy="true">︸</mo> </munder> <mrow> <mfrac> <mrow> <mi>d</mi><mi>u</mi></mrow> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> </mrow> </munder> </mrow></math></span> </span> and since <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>u</mi><mo>=</mo><msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow></math></span> </span> then <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mfrac> <mrow> <mi>d</mi><mi>y</mi></mrow> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> <mo>=</mo><munder> <munder> <mrow> <mn>3</mn><msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>2</mn> </msup> </mrow> <mo stretchy="true">︸</mo> </munder> <mrow> <mfrac> <mrow> <mi>d</mi><mi>y</mi></mrow> <mrow> <mi>d</mi><mi>u</mi></mrow> </mfrac> </mrow> </munder> <munder> <munder> <mrow> <mrow><mo>(</mo> <mrow> <mn>2</mn><mi>x</mi></mrow> <mo>)</mo></mrow></mrow> <mo stretchy="true">︸</mo> </munder> <mrow> <mfrac> <mrow> <mi>d</mi><mi>u</mi></mrow> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> </mrow> </munder> <mo>=</mo><mn>6</mn><mi>x</mi><msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>2</mn> </msup> </mrow></math></span> </span> .</p><h4>Questions 1 - 5</h4><p>Find the derivative of each function:</p><table align="left" border="0" cellpadding="0" cellspacing="0"><tbody><tr><td><strong>1.</strong> <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>y</mi><mo>=</mo><mfrac> <mn>1</mn> <mrow> <msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>−</mo><mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow></mrow> <mn>4</mn> </msup> </mrow> </mfrac> </mrow></math></span> </span></td><td><strong>2.</strong> <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>y</mi><mo>=</mo><mi>sin</mi><mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> </mrow> <mo>)</mo></mrow></mrow></math></span> </span></td><td><strong>3.</strong> <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>y</mi><mo>=</mo><msup> <mi>e</mi> <mrow> <mo>−</mo><mfrac bevelled="true"> <mi>x</mi> <mn>2</mn> </mfrac> </mrow> </msup> </mrow></math></span> </span></td></tr><tr><td><strong>4.</strong> <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>y</mi><mo>=</mo><msqrt> <mrow> <mi>x</mi><mo>+</mo><msqrt> <mi>x</mi> </msqrt> </mrow> </msqrt> </mrow></math></span> </span></td><td><strong>5.</strong> <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>y</mi><mo>=</mo><mn>4</mn><msup> <mrow> <mi>cos</mi></mrow> <mn>2</mn> </msup> <mrow><mo>(</mo> <mrow> <mi>π</mi><mi>x</mi></mrow> <mo>)</mo></mrow></mrow></math></span> </span></td><td></td></tr></tbody></table><hr class="hidden"><div class="pinkBg"><p><strong>Answers for Qs 1-5</strong></p><section class="tib-hiddenbox"><p><br><img alt="" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/ans_1-5_img.jpg" style="width: 513px; height: 123px;"></p></section></div><p>Differentiation and integration are <strong>inverse operations</strong>. Hence, a good starting point for investigating a particular integration method is to rewrite a differentiation rule <em>backwards</em>.</p><h3>Integration by Substitution</h3><p>The role of substitution in integration is comparable to the role of the Chain Rule in differentiation. For differentiable functions given by <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>y</mi><mo>=</mo><mi>F</mi><mrow><mo>(</mo> <mi>u</mi> <mo>)</mo></mrow><mtext> and </mtext><mi>u</mi><mo>=</mo><mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow></math></span> </span> , the Chain Rule states that</p><p style="text-align: center;"><span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mfrac> <mi>d</mi> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> <mrow><mo>[</mo> <mi>y</mi> <mo>]</mo></mrow><mo>=</mo><mfrac> <mrow> <mi>d</mi><mi>y</mi></mrow> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> <mo>=</mo><mfrac> <mrow> <mi>d</mi><mi>y</mi></mrow> <mrow> <mi>d</mi><mi>u</mi></mrow> </mfrac> <mo>⋅</mo><mfrac> <mrow> <mi>d</mi><mi>u</mi></mrow> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> </mrow></math></span> </span> or, alternatively <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mfrac> <mi>d</mi> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> <mrow><mo>[</mo> <mi>y</mi> <mo>]</mo></mrow><mo>=</mo><mfrac> <mi>d</mi> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> <mrow><mo>[</mo> <mrow> <mi>F</mi><mrow><mo>(</mo> <mrow> <mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow> <mo>)</mo></mrow></mrow> <mo>]</mo></mrow><mo>=</mo><msup> <mi>F</mi> <mo>′</mo> </msup> <mrow><mo>(</mo> <mrow> <mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow> <mo>)</mo></mrow><msup> <mi>g</mi> <mo>′</mo> </msup> <mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow></math></span> </span></p><p>Working <em>backwards</em> and letting <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>f</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow><mo>=</mo><msup> <mi>F</mi> <mo>′</mo> </msup> <mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow></math></span> </span> , i.e. <strong><em>F </em>is the antiderivative of<em> f</em></strong>, it follows that</p><p><span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <mi>f</mi><mrow><mo>(</mo> <mrow> <mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow> <mo>)</mo></mrow><msup> <mi>g</mi> <mo>′</mo> </msup> <mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow><mi>d</mi><mi>x</mi><mo>=</mo><mi>F</mi><mrow><mo>(</mo> <mrow> <mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow> <mo>)</mo></mrow><mo>+</mo><mi>C</mi></mrow> </mrow> </mstyle></mrow></math></span> </span></p><p>If <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>u</mi><mo>=</mo><mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow></math></span> </span> , then <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>d</mi><mi>u</mi><mo>=</mo><msup> <mi>g</mi> <mo>′</mo> </msup> <mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow><mi>d</mi><mi>x</mi></mrow></math></span> </span> and <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <mi>f</mi><mrow><mo>(</mo> <mi>u</mi> <mo>)</mo></mrow><mi>d</mi><mi>u</mi><mo>=</mo><mi>F</mi><mrow><mo>(</mo> <mi>u</mi> <mo>)</mo></mrow><mo>+</mo><mi>C</mi></mrow> </mrow> </mstyle></mrow></math></span> </span></p><p>Note that in the integral <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <mi>f</mi><mrow><mo>(</mo> <mrow> <mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow> <mo>)</mo></mrow><msup> <mi>g</mi> <mo>′</mo> </msup> <mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow><mi>d</mi><mi>x</mi></mrow> </mrow> </mstyle></mrow></math></span> </span> , the integrand <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>f</mi><mrow><mo>(</mo> <mrow> <mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow> <mo>)</mo></mrow><msup> <mi>g</mi> <mo>′</mo> </msup> <mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow><mi>d</mi><mi>x</mi></mrow></math></span> </span> is a composite function made up of an “outside” function <em>f</em> and an “inside” function <em>g</em>. Also the derivative of the “inside” function, <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <msup> <mi>g</mi> <mo>′</mo> </msup> <mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow></math></span> </span> , is a factor of the integrand.</p><p style="text-align: center;"><img alt="" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/int_sub_img3.jpg" style="width: 545px; height: 160px;"></p><h3>Using Integration by Substitution</h3><p>In order to apply this technique of integration – essentially performing the <strong>Chain Rule <em>backwards</em></strong> – one must recognize a pattern in the integrand as illustrated above. If the integrand follows the pattern <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>f</mi><mrow><mo>(</mo> <mrow> <mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow> <mo>)</mo></mrow><msup> <mi>g</mi> <mo>′</mo> </msup> <mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow><mi>d</mi><mi>x</mi></mrow></math></span> </span>(can also be represented by <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>f</mi><mrow><mo>(</mo> <mi>u</mi> <mo>)</mo></mrow><mi>d</mi><mi>u</mi></mrow></math></span> </span> ), then it will be possible to apply <strong>integration by substitution</strong> – which usually involves performing a <strong><em>u</em>-substitution</strong>. This <em>change of variable </em>technique uses the Leibniz notation for the differential. That is, if <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>u</mi><mo>=</mo><mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow><mo>,</mo><mtext> then </mtext><mi>d</mi><mi>u</mi><mo>=</mo><msup> <mi>g</mi> <mo>′</mo> </msup> <mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow><mi>d</mi><mi>x</mi></mrow></math></span> </span> and then the integral takes the form:</p><p style="text-align: center;"><span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <mi>f</mi><mrow><mo>(</mo> <mrow> <mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow> <mo>)</mo></mrow><msup> <mi>g</mi> <mo>′</mo> </msup> <mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow><mi>d</mi><mi>x</mi><mo>=</mo><mstyle display=""> <mrow><mo>∫</mo> <mrow> <mi>f</mi><mrow><mo>(</mo> <mi>u</mi> <mo>)</mo></mrow><mi>d</mi><mi>u</mi><mo>=</mo><mi>F</mi><mrow><mo>(</mo> <mi>u</mi> <mo>)</mo></mrow><mo>+</mo><mi>C</mi></mrow> </mrow> </mstyle></mrow> </mrow> </mstyle></mrow></math></span> </span> (recall that <em>F</em> is the antiderivative of <em>f </em>)</p><p><strong>Example 2</strong>: Evaluate <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <msup> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> </mrow> </mstyle></mrow> <mn>2</mn> </msup> <mrow><mo>(</mo> <mrow> <mn>2</mn><mi>x</mi></mrow> <mo>)</mo></mrow><mi>d</mi><mi>x</mi></mrow></math></span> </span></p><p><em>Explanation/Solution:</em> </p><p>Let <em>u</em> be the “inside” function, <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>u</mi><mo>=</mo><msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow></math></span> </span> ; then <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>d</mi><mi>u</mi><mo>=</mo><mn>2</mn><mi>x</mi><mtext> </mtext><mi>d</mi><mi>x</mi></mrow></math></span> </span> . Substitute to obtain the following:</p><p><span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>2</mn> </msup> <mrow><mo>(</mo> <mrow> <mn>2</mn><mi>x</mi></mrow> <mo>)</mo></mrow><mi>d</mi><mi>x</mi><mo>=</mo><mstyle display=""> <mrow><mo>∫</mo> <mrow> <msup> <mi>u</mi> <mn>2</mn> </msup> <mi>d</mi><mi>u</mi></mrow> </mrow> </mstyle></mrow> </mrow> </mstyle></mrow></math></span> </span> Now continue by finding the antiderivative <em>F</em> of <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>f</mi><mrow><mo>(</mo> <mi>u</mi> <mo>)</mo></mrow><mo>=</mo><msup> <mi>u</mi> <mn>2</mn> </msup> </mrow></math></span> </span></p><p><span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <msup> <mi>u</mi> <mn>2</mn> </msup> <mi>d</mi><mi>u</mi><mo>=</mo><mfrac> <mn>1</mn> <mn>3</mn> </mfrac> <msup> <mi>u</mi> <mn>3</mn> </msup> <mo>+</mo><mi>C</mi></mrow> </mrow> </mstyle></mrow></math></span> </span> re-substituting <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>u</mi><mo>=</mo><msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow></math></span> </span> gives <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <msup> <mi>u</mi> <mn>2</mn> </msup> <mi>d</mi><mi>u</mi><mo>=</mo><mfrac> <mn>1</mn> <mn>3</mn> </mfrac> <msup> <mi>u</mi> <mn>3</mn> </msup> <mo>+</mo><mi>C</mi></mrow> </mrow> </mstyle><mo>=</mo><mfrac> <mn>1</mn> <mn>3</mn> </mfrac> <msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>3</mn> </msup> <mo>+</mo><mi>C</mi></mrow></math></span> </span></p><h4>Question 6</h4><p>Use the Chain Rule to verify that the derivative of <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mfrac> <mn>1</mn> <mn>3</mn> </mfrac> <msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>3</mn> </msup> <mo>+</mo><mi>C</mi></mrow></math></span> </span> is the integrand of the original integral in Example 2.</p><div class="pinkBg"><p><strong>Answer for Q 6</strong></p><section class="tib-hiddenbox"><p><span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mfrac> <mi>d</mi> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> <mrow><mo>[</mo> <mrow> <mfrac> <mn>1</mn> <mn>3</mn> </mfrac> <msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>3</mn> </msup> </mrow> <mo>]</mo></mrow><mo>=</mo><mfrac> <mn>1</mn> <mn>3</mn> </mfrac> <mrow><mo>[</mo> <mrow> <mn>3</mn><msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>2</mn> </msup> <mo>⋅</mo><mn>2</mn><mi>x</mi></mrow> <mo>]</mo></mrow><mo>=</mo><mn>2</mn><mi>x</mi><msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>2</mn> </msup> </mrow></math></span> </span></p></section></div><p>Many integrands contain the essential part (i.e. the variable part) of <em>du</em>, but are missing a constant multiple. In such cases, one can multiply and divide by the necessary constant multiple.</p><p><strong>Example 3</strong>: Evaluate <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <mi>x</mi><msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>2</mn> </msup> <mi>d</mi><mi>x</mi></mrow> </mrow> </mstyle></mrow></math></span> </span></p><p><span class="tib-mathml"><math><mrow> <mo mathvariant="italic"></mo></mrow></math></span> <em>Explanation/Solution</em>:</p><p>This is similar to the integral in Example 2, except that the integrand is missing a factor of 2. Recognizing that 2<em>x</em> is the derivative of <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow></math></span> </span> , you can let <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>u</mi><mo>=</mo><msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow></math></span> </span> and since <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>d</mi><mi>u</mi><mo>=</mo><mn>2</mn><mi>x</mi><mtext> </mtext><mi>d</mi><mi>x</mi></mrow></math></span> </span> then it follows that <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mi>d</mi><mi>u</mi><mo>=</mo><mi>x</mi><mtext> </mtext><mi>d</mi><mi>x</mi></mrow></math></span> </span> and then substituting gives the following:</p><p style="text-align: center;"><span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <mi>x</mi><msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>2</mn> </msup> <mi>d</mi><mi>x</mi></mrow> </mrow> </mstyle><mo>=</mo><msup> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mi>u</mi> </mrow> </mstyle></mrow> <mn>2</mn> </msup> <mo>⋅</mo><mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mi>d</mi><mi>u</mi><mo>=</mo><mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <msup> <mi>u</mi> <mn>2</mn> </msup> <mi>d</mi><mi>u</mi><mo>=</mo><mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <mrow><mo>[</mo> <mrow> <mfrac> <mn>1</mn> <mn>3</mn> </mfrac> <msup> <mi>u</mi> <mn>3</mn> </msup> </mrow> <mo>]</mo></mrow></mrow> </mrow> </mstyle><mo>+</mo><mi>C</mi><mo>=</mo><mfrac> <mn>1</mn> <mn>6</mn> </mfrac> <msup> <mrow> <mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>3</mn> </msup> <mo>+</mo><mi>C</mi></mrow></math></span> </span></p><p><strong>Example 4</strong>: Evaluate <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <msup> <mrow> <mi>sin</mi></mrow> <mn>2</mn> </msup> <mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow><mi>cos</mi><mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow><mi>d</mi><mi>x</mi></mrow> </mrow> </mstyle></mrow></math></span> </span></p><p><em>Explanation/Solution</em>:</p><p>The key recognition is that <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <msup> <mrow> <mi>sin</mi></mrow> <mn>2</mn> </msup> <mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow><mo>=</mo><msup> <mrow> <mrow><mo>(</mo> <mrow> <mi>sin</mi><mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow></mrow> <mo>)</mo></mrow></mrow> <mn>2</mn> </msup> </mrow></math></span> </span> and that the derivative of <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>sin</mi><mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow></mrow></math></span> </span> is very close (only off by a constant) to <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>cos</mi><mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow></mrow></math></span> </span> . Therefore, let <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>u</mi><mo>=</mo><mi>sin</mi><mn>3</mn><mi>x</mi><mo>,</mo><mtext> then </mtext><mi>d</mi><mi>u</mi><mo>=</mo><mrow><mo>(</mo> <mrow> <mi>cos</mi><mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow><mrow><mo>(</mo> <mn>3</mn> <mo>)</mo></mrow><mi>d</mi><mi>x</mi></mrow></math></span> </span> and <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mfrac> <mtext>1</mtext> <mtext>3</mtext> </mfrac> <mi>d</mi><mi>u</mi><mo>=</mo><mi>cos</mi><mn>3</mn><mi>x</mi><mtext> </mtext><mi>d</mi><mi>x</mi></mrow></math></span> </span> . Substituting <span class="tib-mathml"><span class="tib-mathml"><math> <mi>u</mi></math></span> </span> and <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mfrac> <mn>1</mn> <mn>3</mn> </mfrac> <mi>u</mi></mrow></math></span> </span> in the given integral gives the following:</p><p style="text-align: center;"><span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <msup> <mrow> <mi>sin</mi></mrow> <mn>2</mn> </msup> <mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow><mi>cos</mi><mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow><mi>d</mi><mi>x</mi></mrow> </mrow> </mstyle><mo>=</mo><mstyle display=""> <mrow><mo>∫</mo> <mrow> <msup> <mi>u</mi> <mn>2</mn> </msup> <mo>⋅</mo><mfrac> <mn>1</mn> <mn>3</mn> </mfrac> <mi>d</mi><mi>u</mi><mo>=</mo><mfrac> <mn>1</mn> <mn>3</mn> </mfrac> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <msup> <mi>u</mi> <mn>2</mn> </msup> <mi>d</mi><mi>u</mi></mrow> </mrow> </mstyle></mrow> </mrow> </mstyle><mo>=</mo><mfrac> <mn>1</mn> <mn>3</mn> </mfrac> <mrow><mo>(</mo> <mrow> <mfrac> <mn>1</mn> <mn>3</mn> </mfrac> <msup> <mi>u</mi> <mn>3</mn> </msup> </mrow> <mo>)</mo></mrow><mo>+</mo><mi>C</mi><mo>=</mo><mfrac> <mn>1</mn> <mn>9</mn> </mfrac> <msup> <mrow> <mi>sin</mi></mrow> <mn>3</mn> </msup> <mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow><mo>+</mo><mi>C</mi></mrow></math></span> </span></p><h4>Question 7</h4><p>Verify the result for Example 4 above by finding the derivative of <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mfrac> <mn>1</mn> <mn>9</mn> </mfrac> <msup> <mrow> <mi>sin</mi></mrow> <mn>3</mn> </msup> <mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow></mrow></math></span> </span> .</p><p>When using <em>u</em>-substitution with a definite integral (examples 2 – 4 were <strong>indefinite</strong> integrals), it is often convenient to determine the limits of integration for the variable <em>u</em> rather than to convert the antiderivative back to the variable <em>x</em> and evaluate at the original limits. Symbolically, this can be expressed as</p><p style="text-align: center;"><span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow> <msubsup> <mo>∫</mo> <mrow> <mtext> </mtext><mi>a</mi></mrow> <mrow> <mtext> </mtext><mi>b</mi></mrow> </msubsup> <mrow> <mi>f</mi><mrow><mo>(</mo> <mrow> <mi>g</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow></mrow> <mo>)</mo></mrow></mrow> </mrow> </mstyle><msup> <mi>g</mi> <mo>′</mo> </msup> <mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow><mtext> </mtext><mi>d</mi><mi>x</mi><mo>=</mo><mstyle display=""> <mrow> <msubsup> <mo>∫</mo> <mrow> <mtext> </mtext><mi>g</mi><mrow><mo>(</mo> <mi>a</mi> <mo>)</mo></mrow></mrow> <mrow> <mtext> </mtext><mi>g</mi><mrow><mo>(</mo> <mi>b</mi> <mo>)</mo></mrow></mrow> </msubsup> <mrow> <mi>f</mi><mrow><mo>(</mo> <mi>u</mi> <mo>)</mo></mrow><mtext> </mtext><mi>d</mi><mi>u</mi></mrow> </mrow> </mstyle></mrow></math></span> </span> Example 5 below illustrates this.</p><div class="pinkBg"><p><strong>Answer for Q 7</strong></p><section class="tib-hiddenbox"><p><span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mfrac> <mi>d</mi> <mrow> <mi>d</mi><mi>x</mi></mrow> </mfrac> <mrow><mo>[</mo> <mrow> <mfrac> <mn>1</mn> <mn>9</mn> </mfrac> <msup> <mrow> <mi>sin</mi></mrow> <mn>3</mn> </msup> <mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow></mrow> <mo>]</mo></mrow><mo>=</mo><mfrac> <mn>1</mn> <mn>9</mn> </mfrac> <mrow><mo>[</mo> <mrow> <mn>3</mn><msup> <mrow> <mrow><mo>(</mo> <mrow> <mi>sin</mi><mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow></mrow> <mo>)</mo></mrow></mrow> <mn>2</mn> </msup> <mo>⋅</mo><mi>cos</mi><mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow><mo>⋅</mo><mn>3</mn></mrow> <mo>]</mo></mrow><mo>=</mo><msup> <mrow> <mi>sin</mi></mrow> <mn>2</mn> </msup> <mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow><mo>⋅</mo><mi>cos</mi><mrow><mo>(</mo> <mrow> <mn>3</mn><mi>x</mi></mrow> <mo>)</mo></mrow></mrow></math></span> </span></p></section></div><p><strong>Example 5</strong>: Evaluate <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow> <msubsup> <mo>∫</mo> <mrow> <mtext> </mtext><mn>0</mn></mrow> <mrow> <mtext> </mtext><mn>1</mn></mrow> </msubsup> <mrow> <mi>x</mi><msqrt> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>3</mn></mrow> </msqrt> <mtext> </mtext><mi>d</mi><mi>x</mi></mrow> </mrow> </mstyle></mrow></math></span> </span></p><p><em>Explanation/Solution</em>:</p><p>Let <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>u</mi><mo>=</mo><msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>3</mn><mo>,</mo><mtext> then </mtext><mi>d</mi><mi>u</mi><mo>=</mo><mn>2</mn><mi>x</mi><mtext> </mtext><mi>d</mi><mi>x</mi><mtext> and </mtext><mfrac> <mtext>1</mtext> <mtext>2</mtext> </mfrac> <mi>d</mi><mi>u</mi><mo>=</mo><mi>x</mi><mtext> </mtext><mi>d</mi><mi>x</mi></mrow></math></span> </span></p><p>Before substituting, determine the new upper and lower limits of integration.</p><p><u>lower limit</u>: when <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>x</mi><mo>=</mo><mn>0</mn><mo>,</mo><mtext> </mtext><mi>u</mi><mo>=</mo><msup> <mn>0</mn> <mn>2</mn> </msup> <mo>+</mo><mn>3</mn><mo>=</mo><mn>3</mn></mrow></math></span> </span> <u>upper limi</u>t: <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>x</mi><mo>=</mo><mn>1</mn><mo>,</mo><mtext> </mtext><mi>u</mi><mo>=</mo><msup> <mn>1</mn> <mn>2</mn> </msup> <mo>+</mo><mn>3</mn><mo>=</mo><mn>4</mn></mrow></math></span> </span></p><p>Now, you can substitute to obtain</p><p style="text-align: center;"><img alt="" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/int_sub_img4.jpg" style="width: 511px; height: 133px;"></p><h4>Question 8</h4><p>Explain how you can check the result for Example 5 above using your GDC.</p><div class="pinkBg"><p><strong>Answer for Q 8</strong></p><section class="tib-hiddenbox"><p><br><img alt="" class="noborder" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/ans_8_img.jpg" style="width: 515px; height: 131px;"></p></section></div><h4>Questions 9 & 10</h4><p>Evaluate each integral in two different ways. Explain any difference in the forms of the answers.</p><p><strong>9.</strong> <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <msup> <mrow> <mrow><mo>(</mo> <mrow> <mn>2</mn><mi>x</mi><mo>−</mo><mn>1</mn></mrow> <mo>)</mo></mrow></mrow> <mn>2</mn> </msup> <mtext> </mtext><mi>d</mi><mi>x</mi></mrow> </mrow> </mstyle></mrow></math></span> </span> <strong>10.</strong> <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <mi>sin</mi><mi>x</mi><mi>cos</mi><mi>x</mi><mtext> </mtext><mi>d</mi><mi>x</mi></mrow> </mrow> </mstyle></mrow></math></span> </span></p><div class="pinkBg"><p><strong>Answers for Qs 9 & 10</strong></p><section class="tib-hiddenbox"><p><br><img alt="" class="noborder" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/ans_9-10_img.jpg" style="width: 626px; height: 391px;"></p></section></div><h4>Questions 11-14</h4><p>For each definite integral: (a) Evaluate for two different sets of limits: (i) from <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>x</mi><mo>=</mo><mo>−</mo><mtext> </mtext><mi>a</mi><mtext> to </mtext><mi>x</mi><mo>=</mo><mo>+</mo><mtext> </mtext><mi>a</mi></mrow></math></span> </span> ; and (ii) from <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>x</mi><mo>=</mo><mn>0</mn><mtext> to </mtext><mi>x</mi><mo>=</mo><mo>+</mo><mtext> </mtext><mi>a</mi></mrow></math></span> </span> . Evaluate the definite integrals by <em>u</em>-substitution and integrate with limits for <em>u</em> (as in Example 5). (b) Verify each answer by graphing the function on your GDC and evaluating the definite integral on your GDC.</p><p><strong>11.</strong> <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <mi>x</mi><msqrt> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mo>+</mo><mn>3</mn></mrow> </msqrt> <mtext> </mtext><mi>d</mi><mi>x</mi></mrow> </mrow> </mstyle><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>a</mi><mo>=</mo><mn>3</mn></mrow></math></span> </span> <strong>12.</strong> <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <msup> <mrow> <mi>sin</mi></mrow> <mn>2</mn> </msup> <mi>x</mi><mi>cos</mi><mi>x</mi><mtext> </mtext><mi>d</mi><mi>x</mi><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>a</mi><mo>=</mo><mfrac> <mi>π</mi> <mn>2</mn> </mfrac> </mrow> </mrow> </mstyle></mrow></math></span> </span></p><p><strong>13.</strong> <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <msup> <mrow> <mi>cos</mi></mrow> <mn>2</mn> </msup> <mi>x</mi></mrow> </mrow> </mstyle><mi>sin</mi><mi>x</mi><mtext> </mtext><mi>d</mi><mi>x</mi><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>a</mi><mo>=</mo><mfrac> <mi>π</mi> <mn>2</mn> </mfrac> </mrow></math></span> </span> <strong>14.</strong> <span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mstyle display=""> <mrow><mo>∫</mo> <mrow> <msup> <mi>x</mi> <mn>2</mn> </msup> <mi>cos</mi><mrow><mo>(</mo> <mrow> <msup> <mi>x</mi> <mn>3</mn> </msup> </mrow> <mo>)</mo></mrow><mtext> </mtext><mi>d</mi><mi>x</mi></mrow> </mrow> </mstyle><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>a</mi><mo>=</mo><mn>1</mn></mrow></math></span> </span></p><div class="pinkBg"><p><strong>Answer for Q 11</strong></p><section class="tib-hiddenbox"><p><br><img alt="" class="noborder" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/ans_q_11_img.jpg" style="width: 625px; height: 482px;"></p></section></div><div class="pinkBg"><p><strong>Answer for Q 12</strong></p><section class="tib-hiddenbox"><p><br><img alt="" class="noborder" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/ans_q_12_img.jpg" style="width: 625px; height: 480px;"></p></section></div><div class="pinkBg"><p><strong>Answer for Q 13</strong></p><section class="tib-hiddenbox"><p><br><img alt="" class="noborder" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/ans_q_13_img.jpg" style="width: 625px; height: 491px;"></p></section></div><div class="pinkBg"><p><strong>Answer for Q 14</strong></p><section class="tib-hiddenbox"><p><br><img alt="" class="noborder" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/ans_q_14_img.jpg" style="width: 610px; height: 495px;"></p></section></div><h4>Question 15</h4><p>From your work on questions 11 – 14, form some conclusions/conjectures regarding definite integrals and the symmetry of certain types of functions.</p><div class="pinkBg"><p><strong>Answer for Q 15</strong></p><section class="tib-hiddenbox"><p><br><img alt="" class="noborder" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/ans_q_15_img.jpg" style="width: 621px; height: 151px;"></p></section></div><h4>Question 16</h4><p>Find an equation for the function <em>f</em> that has the indicated derivative and whose graph passes through the given point.<br><span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <msup> <mi>f</mi> <mo>′</mo> </msup> <mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow><mo>=</mo><mi>x</mi><msqrt> <mrow> <mn>1</mn><mo>−</mo><msup> <mi>x</mi> <mn>2</mn> </msup> </mrow> </msqrt> <mtext> point </mtext><mrow><mo>(</mo> <mrow> <mn>0</mn><mo>,</mo><mtext> </mtext><mfrac> <mn>7</mn> <mn>3</mn> </mfrac> </mrow> <mo>)</mo></mrow></mrow></math></span> </span></p><div class="pinkBg"><p><strong>Answer for Q 16</strong></p><section class="tib-hiddenbox"><p><span class="tib-mathml"><span class="tib-mathml"><math> <mrow> <mi>f</mi><mrow><mo>(</mo> <mi>x</mi> <mo>)</mo></mrow><mo>=</mo><mfrac> <mrow> <mo>−</mo><msqrt> <mrow> <msup> <mrow> <mrow><mo>(</mo> <mrow> <mn>1</mn><mo>−</mo><msup> <mi>x</mi> <mn>2</mn> </msup> </mrow> <mo>)</mo></mrow></mrow> <mn>3</mn> </msup> </mrow> </msqrt> <mo>+</mo><mn>8</mn></mrow> <mn>3</mn> </mfrac> </mrow></math></span> </span></p></section></div><h4>Questions 17-22</h4><p>Determine whether the statement is true or false. If it is true, show mathematical working and/or reasoning to verify the statement; and if it is false, explain why or give an example that shows it is false.</p><p><img alt="" class="noborder" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/int_sub_img5.jpg" style="width: 648px; height: 157px;"></p><div class="pinkBg"><p><strong>Answers for Qs 17-19</strong></p><section class="tib-hiddenbox"><p><br><img alt="" class="noborder" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/ans_q_17-19_img.jpg" style="width: 625px; height: 390px;"></p></section></div><div class="pinkBg"><p><strong>Answers for Qs 20-22</strong></p><section class="tib-hiddenbox"><p><br><img alt="" class="noborder" src="../../../ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/ans_q_20-22_img.jpg" style="width: 625px; height: 353px;"></p></section></div><p><span style="color:#FF0000;">download tutorial here:</span></p><p><a href="/media/ib/mathhlsl/files/Teaching%20Materials/Calculus/integration-by-sub/integration-by-substitution-tutorial_v1.pdf" target="_blank" title="Student handouts"><img class="ico" src="../../../img/icons/student-handout.png"> Integration by Substitution tutorial v1</a> </p><script>document.querySelectorAll('.tib-teacher-only').forEach(e => e.remove());</script>
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$('img.ico[src="/img/icons/comments.png"]').each((function(){var t=$(this).attr("title");$(this).removeAttr("title"),$(this).wrap('<a class="tib-popover" href="#" data-content="'+t+'" data-togle="popover" data-placement="top" />')})),$('img.ico[src="../../../img/icons/comments.png"]').each((function(){var t=$(this).attr("title");$(this).removeAttr("title"),$(this).wrap('<a class="tib-popover" href="#" data-content="'+t+'" data-togle="popover" data-placement="top" />')})),$(".tib-popover").popover({html:!0,trigger:"hover",delay:{show:300,hide:300},placement:function(t,e){var o=$(e).position();return o.top>200?"top":o.left<515?"right":o.top<200?"bottom":o.left>515?"left":"top"}}).click((function(t){t.preventDefault()}));var carouselTime=6500;$("div.carousel.slide").carousel({interval:carouselTime}),$(".tib-indicators > img").click((function(){var t=$(this).index(),e;$(this).closest(".carousel.slide").carousel(t)})),setShowResultsListeners($("#container")),addMarksThreads($("#modal-std-write"));
$('a.btn.showhider').click(function(e) {
var showHiderId = $( this ).attr('rel');
if( $('#'+showHiderId).find('iframe').length > 0 ) {
$('#'+showHiderId+' iframe').each(function() {
if ( $(this).attr('src').indexOf('.pdf') > 0 ) {
this.contentWindow.location.reload(true);
}
});
}
});
});
</script>
</body>
</html>