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alt="" src="../../files/integration/area-between/main-1.png" style="float: left; width: 100px; height: 100px;"></p> <p>On this page, we'll look at how we can use integration to find areas. This is a really common question in examinations. The technique is not too hard, but there are just a couple of pitfalls to avoid which will be explained. There are 3 cases that we will consider here, the area between:</p> <ul style="margin-left: 120px;"> <li> a graph and the x axis</li> <li>a graph and the y axis</li> <li>two graphs</li> </ul> <hr class="hidden-separator"> <div class="panel panel-turquoise panel-has-colored-body"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Key Concepts</p> </div> </div> <div class="panel-body"> <p>On this page, you should learn about</p> <ul> <li>area of a region enclosed by a curve and the x axis</li> <li>area of a region enclosed by a curve and the y axis</li> <li>area between two graphs</li> </ul> </div> <div class="panel-footer"> <div><em> </em></div> </div> </div> <div class="panel panel-yellow panel-has-colored-body"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <p>Here are all the different cases that you need to know about</p> <div class="panel panel-yellow panel-has-colored-body panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Area between Curve and x-axis</p> </div> </div> <div class="panel-body"> <div> <p><img alt="" src="../../files/integration/area-between/areaxaxis1.png" style="float: left; width: 200px; height: 200px;"></p> <p>The area between the curve <strong>y=f(x)</strong> and the <strong>x axis </strong>can be found using the formula</p> <p><span class="math-tex">\(\int _{ a }^{ b }{ y } \ dx\)</span></p> <p>or using function notation:</p> <p><span class="math-tex">\(\int _{ a }^{ b }{ f(x) } \ dx\)</span></p> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Area between Curve and y-axis</p> </div> </div> <div class="panel-body"> <div> <p><img alt="" src="../../files/integration/area-between/areayaxis1.png" style="float: left; width: 200px; height: 200px;"></p> <p>The area between the curve <strong>x=f(y)</strong> and the <strong>y axis </strong>can be found using the formula</p> <p><span class="math-tex">\(\int _{ a }^{ b }{ x } \ dy\)</span></p> <p>or using function notation:</p> <p><span class="math-tex">\(\int _{ a }^{ b }{ f(y) } \ dy\)</span></p> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Areas under Axes</p> </div> </div> <div class="panel-body"> <div> <p>We should always draw a sketch of the graph before we calculate the area since any part below the axis will be calculated as a <strong>negative value</strong>. Below, we consider the area between the graph y = x²-1 and the x axis. You can check these <strong>definite integrals</strong> with your GDC:</p> <table border="0" cellpadding="0" cellspacing="0" style="width: 100%;"> <tbody> <tr> <td style="text-align: center;"><img alt="" src="../../files/integration/area-between/negarea2.png" style="width: 200px; height: 200px;"></td> <td style="text-align: center;"><img alt="" src="../../files/integration/area-between/negarea3.png" style="width: 200px; height: 200px;"></td> <td style="text-align: center;"><img alt="" src="../../files/integration/area-between/negarea1.png" style="width: 200px; height: 200px;"></td> </tr> <tr> <td style="text-align: center;"> <p><span class="math-tex">\(\int _{ 0.5}^{ 1 }{ x^2 } \ dx \approx -0.21\)</span></p> <p>The area <span class="math-tex">\(\approx 0.21\)</span></p> </td> <td style="text-align: center;"> <p><span class="math-tex">\(\int _{ 1}^{ 1.5 }{ x^2 } \ dx \approx 0.29\)</span></p> <p>The area <span class="math-tex">\(\approx 0.29\)</span></p> </td> <td style="text-align: center;"> <p><span class="math-tex">\(\int _{ 0.5}^{ 1.5}{ x^2 } \ dx \approx 0.083\)</span></p> <p>The area is NOT equal to 0.083</p> </td> </tr> </tbody> </table> <p>Clearly, the area of the third graph <strong>is NOT 0.083 </strong>and <strong>CANNOT</strong> be found by evaluating <span class="math-tex">\(\int _{ 0.5}^{ 1.5}{ x^2 } \ dx\)</span> </p> <p>The area <span class="math-tex">\(\approx 0.21+0.29\approx 0.50\)</span></p> <p>There are different ways of finding the area is this case. The simplest is to evaluate separately the part that is above to the part that is below the axis. We then make the part below a positive value.</p> <p>In this case,</p> <p style="margin-left: 80px;">area = <span class="math-tex">\(|\int _{ 0.5}^{ 1}{ x^2 } \ dx|\quad+\quad\int _{ 1}^{ 1.5}{ x^2 } \ dx\)</span></p> <hr class="hidden-separator"> <p><strong>In General</strong></p> <p> <img alt="" src="../../files/integration/area-between/negarea4.png" style="float: left; width: 200px; height: 200px;">The area between the curve <strong>y=f(x)</strong> and the <strong>x axis </strong>can be found using the formula</p> <p><span class="math-tex">\(|\int _{ a}^{ b}{ y } \ dx|\quad+\quad|\int _{ b}^{ c}{ y } \ dx|\)</span></p> <p>We can use function notation</p> <p><span class="math-tex">\(|\int _{ a}^{ b}{ f(x) } \ dx|\quad+\quad|\int _{ b}^{ c}{ f(x) } \ dx|\)</span></p> <p>You will need to find the zeros (y intercepts) of your graph.</p> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Area bounded by 2 Graphs</p> </div> </div> <div class="panel-body"> <div> <p>To find the area bounded by two graphs like this, we need to subtract one area from the other:</p> <p><img alt="" src="../../files/integration/area-between/areabetween1.png" style="width: 190px; height: 163px;"><strong> equals </strong><img alt="" src="../../files/integration/area-between/areabetween2.png" style="width: 190px; height: 163px;"> <strong>subtract <img alt="" src="../../files/integration/area-between/areabetween3.png" style="width: 190px; height: 163px;"></strong></p> <p style="text-align: center;"><span class="math-tex">\(\int _{ a }^{ b }{ f(x) } \ dx\quad-\quad \int _{ a }^{ b }{ g(x) } \ dx\)</span></p> <p>In this case, you do not have to worry about graph being above or below the x axis!</p> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <hr class="hidden-separator"> <p>Here are video examples to help you understand the concepts from this page</p> <div class="panel panel-yellow panel-has-colored-body panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Area below x-axis</p> </div> </div> <div class="panel-body"> <div> <p>In this example, we see the importance of sketching the graph to see which part of the region lies above the x axis and which lies below. Finding the definite integral between 0 and 3 would find the <strong>WRONG</strong> answer here!</p> <p><em>Find the area bounded by the curve y= x<sup>3</sup> – 4x<sup>2</sup> + 3x and the x axis from x = 0 to x = 3</em></p> <p style="text-align: center"><iframe allowfullscreen="" frameborder="0" height="360" mozallowfullscreen="" src="https://player.vimeo.com/video/226799044" webkitallowfullscreen="" width="640"><br> </iframe></p> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Area between Curve and y-axis</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="505"> <p>In this example, we look at finding the area between the curve and the y axis. We also see the importance, in the second part, of thinking carefully about the geometry of the problem rather than working through it mechanically. The key <strong>command </strong>term is '<strong><em>Hence</em></strong>'.</p> <p><em>a) Find the area bounded by the curve y = <span class="math-tex">\( \sqrt{ax}\)</span>, the <strong>y axis </strong>and y = a</em></p> <p><em>b) <strong>Hence</strong>, find the area bounded by the same curve, the <strong>x axis </strong>and x = a</em></p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/226807731"></iframe></div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body panel-expandable panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Area between 2 Curves</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="506"> <p>Here is a good example that combines finding the <a href="../564.html" target="_blank">equation of a tangent</a> , finding the area between 2 curves and thinking about areas under straight lines as triangles.</p> <p><em>Find the area bounded by the curve y= e<sup>x </sup>, the tangent to the curve where x = 1 and the y axis.</em></p> <p style="text-align: center;"><iframe allowfullscreen="" frameborder="0" height="360" mozallowfullscreen="" src="https://player.vimeo.com/video/226818211" webkitallowfullscreen="" width="640"><br> </iframe></p> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> </div> </div> <div class="panel panel-has-colored-body panel-violet"> <div class="panel-heading"><a class="expander pull-right" href="#"><span class="fa fa-plus"></span></a> <div> <p>Summary</p> </div> </div> <div class="panel-body"> <div> <p><iframe align="middle" frameborder="1" height="480" scrolling="yes" src="../../files/integration/area-between/areabetweengraphsnotes.pdf" width="640"></iframe></p> <p>Print from <a href="../../files/integration/area-between/areabetweengraphsnotes.pdf" target="_blank">here</a></p> </div> </div> <div class="panel-footer"> <div> <p>text</p> </div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander pull-right" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test Yourself</p> </div> </div> <div class="panel-body"> <p>Here are quizzes that will help you practise the skills from this page</p> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#a2651488"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="a2651488"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> Area between graphs HL1 <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="11-593-730" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following calculates the shaded area. Select <strong>ALL</strong> correct answers</p><p><img alt="" src="../../files/integration/mixed-integration/quiz-4/q1" style="width: 400px; height: 358px;"></p></div><div class="q-answer"><p><label class="checkbox"><input type="checkbox"> <span class="math-tex">\(\int _{ 0 }^{ 4 }{ x²dx } \)</span></label></p><p><label class="checkbox"><input class="c" type="checkbox"> <span class="math-tex">\(8-\int _{ 0 }^{ 2 }{ x²dx } \)</span></label></p><p><label class="checkbox"><input class="c" type="checkbox"> <span class="math-tex">\(\int _{ 0 }^{ 4 }{ { y }^{ \frac { 1 }{ 2 } }dy } \)</span></label></p><p><label class="checkbox"><input type="checkbox"> <span class="math-tex">\(\int _{ 0 }^{ 2 }{ { y }^{ \frac { 1 }{ 2 } }dy } \)</span></label></p></div><div class="q-explanation"><p>There are 2 correct ways of finding the area of this region.</p><ul><li>You can find the area between the curve and the y axis <span class="math-tex">\(\int _{ a }^{ b }{ xdy } \)</span></li><li>You can subtract the area between the curve and the x axis from the area of 4 by 2 rectangle</li></ul></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The two functions, <strong><em>f</em></strong> and <strong><em>g</em></strong> are</p><p style="margin-left: 40px;">f(x) = cos²x</p><p style="margin-left: 40px;">g(x) = cos2x</p><p>Which one of the following will give the area of the region shaded below</p><p><img alt="" src="../../files/integration/mixed-integration/quiz-4/q2" style="width: 400px; height: 285px;"></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> <span class="math-tex">\(\int _{ \pi }^{ 0 }{ f(x)dx } -\int _{ \pi }^{ 0 }{ g(x)dx } \)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(\int _{ \pi }^{ 0 }{ \left[ f(x)-g(x) \right] dx } \)</span></label></p><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\(\int _{ 0 }^{ \pi }{ \left[ f(x)-g(x) \right] dx } \)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(\int _{ 0 }^{ \pi }{ g(x)dx } -\int _{ 0 }^{ \pi }{ f(x)dx } \)</span></label></p></div><div class="q-explanation"><p>f(x) = cos²x is the upper curve</p><p>We need to find the area under <strong><em>f</em></strong> and then subtract the area under <strong><em>g</em></strong></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Let <span class="math-tex">\(f(x)=e^{x-1}\)</span></p><p>The diagram below shows the graph of y = f(x) and the tangent at the point x = 1</p><p>Which of the following gives the area of the shaded region</p><p><img alt="" src="../../files/integration/mixed-integration/quiz-4/q3" style="width: 400px; height: 459px;"></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> <span class="math-tex">\(\int _{ 0 }^{ 1 }{ f(x)dx } -1\)</span></label></p><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\(\int _{ 0 }^{ 1 }{ f(x)dx } -0.5\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(1-\int _{ 0 }^{ 1 }{ f(x)dx } \)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(0.5-\int _{ 0 }^{ 1 }{ f(x)dx } \)</span></label></p></div><div class="q-explanation"><p>We need to find the area under the curve and subtract the area under the tangent (the triangle).</p><p>The area under the curve = <span class="math-tex">\(\int _{ 0 }^{ 1 }{ f(x)dx }\)</span></p><p>The height of the triangle is <span class="math-tex">\(f(1)=e^0=1\)</span></p><p>The area of the triangle = <span class="math-tex">\(\frac{1\times1}{2}=0.5\)</span></p><p>Hence the shaded area = <span class="math-tex">\(\int _{ 0 }^{ 1 }{ f(x)dx } -0.5\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The diagram shows the graph of y = f(x).</p><p>The area of the shaded region = 3.75.</p><p>What is the value of <span class="math-tex">\(\int _{ 1 }^{ 8 }{ { f }^{ -1 }(y)dy } \)</span></p><p><img alt="" src="../../files/integration/mixed-integration/quiz-4/q4" style="width: 400px; height: 396px;"></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> Impossible to work out from the information given</label></p><p><label class="radio"><input type="radio"> 12.25</label></p><p><label class="radio"><input class="c" type="radio"> 11.25</label></p><p><label class="radio"><input type="radio"> 13.25</label></p></div><div class="q-explanation"><p>The question is asking us to find the area between the curve and the y axis shown below</p><p><img alt="" src="../../files/integration/mixed-integration/quiz-4/q4explanation" style="width: 300px; height: 264px;"></p><p>To find this area, start with the area formed by the 8 by 2 rectangle, subtract the 1 by 1 rectangle and subtract the area given in the question</p><p><img alt="" src="../../files/integration/mixed-integration/quiz-4/q4explanationa.png" style="width: 600px; height: 289px;"></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p><img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"> Find the area of the region bounded by the graph <span class="math-tex">\(y = \frac{1}{x}\)</span>, x = 1 , x = e and the x axis</p></div><div class="q-answer"><p>Area = <input type="text" style="height: auto;" data-c="1"> <span class="review"></span></p></div><div class="q-explanation"><p><img alt="" src="../../files/integration/mixed-integration/quiz-4/q5" style="width: 300px; height: 286px;"></p><p>The area of the region can be found by working out the following integral</p><p><span class="math-tex">\(\int _{ 1 }^{ e }{ \frac { 1 }{ x } dx={ [ln|x|] }_{ 1 }^{ e } } \)</span></p><p>= lne - ln1</p><p>= 1 - 0</p><p>=1</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The area of the region bounded by the graph <span class="math-tex">\(y = \frac{1}{2x}\)</span>, x = <strong><em>a</em></strong>, x = 1 and the x axis is equal to <span class="math-tex">\(ln3\)</span></p><p>Work out the value of <strong><em>a</em></strong></p></div><div class="q-answer"><p><strong><em>a</em></strong> = <input type="text" style="height: auto;" data-c="9"> <span class="review"></span></p></div><div class="q-explanation"><p><span class="math-tex">\(\int _{ 1 }^{ a }{ \frac { 1 }{ 2x } dx } \)</span></p><p><span class="math-tex">\(=\frac { 1 }{ 2 } \int _{ 1 }^{ a }{ \frac { 1 }{ x } dx } \)</span></p><p><span class="math-tex">\(=\frac { 1 }{ 2 } { [ln|x|] }_{ 1 }^{ a }\)</span></p><p><span class="math-tex">\(=\frac { 1 }{ 2 } (lna-ln1)\)</span></p><p><span class="math-tex">\(=\frac { 1 }{ 2 } lna\)</span></p><p><span class="math-tex">\(=ln{ a }^{ \frac { 1 }{ 2 } }\)</span></p><hr class="hidden-separator"><span class="math-tex">\(ln{ a }^{ \frac { 1 }{ 2 } }=ln3 \\ { a }^{ \frac { 1 }{ 2 } }=3 \\a=9\)</span></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p><img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"> The diagram below shows the graphs of y = x² and y = 2 - x²</p><p>The area of the shaded region = <span class="math-tex">\(\frac{a}{3}\)</span></p><p>Find the value of <strong><em>a</em></strong></p><p><img alt="" src="../../files/integration/mixed-integration/quiz-4/q7" style="width: 400px; height: 340px;"></p></div><div class="q-answer"><p><em><strong>a</strong></em> = <input type="text" style="height: auto;" data-c="8"> <span class="review"></span></p></div><div class="q-explanation"><p>Area = <span class="math-tex">\(\int _{ -1 }^{ 1 }{ (2-x²)dx } -\int _{ -1 }^{ 1 }{ x²dx } \)</span></p><p>We can put the two integrals together</p><p>Area = <span class="math-tex">\(\int _{ -1 }^{ 1 }{ (2-x²-x²)dx } \)</span></p><p><span class="math-tex">\(=\int _{ -1 }^{ 1 }{ (2-2x²)dx } \)</span></p><p><span class="math-tex">\(={ \left[ 2x-\frac { 2{ x }^{ 3 } }{ 3 } \right] }_{ -1 }^{ 1 }\)</span></p><p><span class="math-tex">\(=\left[ 2-\frac { 2 }{ 3 } \right] -\left[ -2+\frac { 2 }{ 3 } \right] \)</span></p><p><span class="math-tex">\(=\frac{8}{3}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The diagram below shows the graphs of y = sinx and y = cosx.</p><p>The area of the shaded region = <span class="math-tex">\(a\sqrt{2}\)</span></p><p>Find the value of <strong><em>a</em></strong></p><p><img alt="" src="../../files/integration/mixed-integration/quiz-4/q8" style="width: 400px; height: 295px;"></p></div><div class="q-answer"><p><strong><em>a</em></strong> = <input type="text" style="height: auto;" data-c="2"> <span class="review"></span></p></div><div class="q-explanation"><p>Area = <span class="math-tex">\(\int _{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }{ sinxdx } -\int _{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }{ cosxdx } \)</span></p><p>We can put the two integrals together</p><p>Area = <span class="math-tex">\(\int _{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }{ (sinx-cosx)dx } \)</span></p><p><span class="math-tex">\(={ \left[ -cosx-sinx \right] }_{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }\)</span></p><p><span class="math-tex">\(=\left[ -cos\frac { 5\pi }{ 4 } -sin\frac { 5\pi }{ 4 } \right] -\left[ -cos\frac { \pi }{ 4 } -sin\frac { \pi }{ 4 } \right] \)</span></p><p><span class="math-tex">\(=\left[ \frac { \sqrt { 2 } }{ 2 } +\frac { \sqrt { 2 } }{ 2 } \right] -\left[ -\frac { \sqrt { 2 } }{ 2 } -\frac { \sqrt { 2 } }{ 2 } \right] \)</span></p><p><span class="math-tex">\(=\left[ \sqrt { 2 } \right] -\left[ -\sqrt { 2 } \right] \)</span></p><p><span class="math-tex">\(=2\sqrt { 2 } \)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p><img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"> Let f(x) = sinx and g(x) = sin2x</p><p>The graphs of y = f(x) and y = g(x) meet at x = 0 and x = <span class="math-tex">\(\frac{\pi}{3}\)</span> as shown below.</p><p>Find the area of the shaded region</p><p><img alt="" src="../../files/integration/mixed-integration/quiz-4/q9" style="width: 400px; height: 295px;"></p></div><div class="q-answer"><p>Area = <input type="text" style="height: auto;" data-c="0.25"> <span class="review"></span></p></div><div class="q-explanation"><p>Area = <span class="math-tex">\(\int _{ 0 }^{ \frac { \pi }{ 3 } }{ (sin2x-sinx)dx } \)</span></p><p><span class="math-tex">\(={ \left[ -\frac { cos2x }{ 2 } +cosx \right] }_{ 0 }^{ \frac { \pi }{ 3 } }\)</span></p><p><span class="math-tex">\(=\left[ -\frac { cos\frac { 2\pi }{ 3 } }{ 2 } +cos\frac { \pi }{ 3 } \right] -\left[ -\frac { cos0 }{ 2 } +cos0 \right] \)</span></p><p><span class="math-tex">\(=\left[ \frac { 1 }{ 4 } +\frac { 1 }{ 2 } \right] -\left[ -\frac { 1 }{ 2 } +1 \right] \)</span></p><p><span class="math-tex">\(=\frac { 1 }{ 4 } \)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The diagram below shows the graph of <span class="math-tex">\(y = \frac{2x}{x^2-1}\)</span></p><p>The shaded region has an area <span class="math-tex">\(lna\)</span></p><p>Find the value of <strong><em>a</em></strong></p><p><img alt="" src="../../files/integration/mixed-integration/quiz-4/q10" style="width: 400px; height: 352px;"></p></div><div class="q-answer"><p><strong><em>a</em></strong> = <input type="text" style="height: auto;" data-c="5"> <span class="review"></span></p></div><div class="q-explanation"><p>Area = <span class="math-tex">\(\int _{ 2 }^{ 4 }{ \frac { 2x }{ x²-1 } dx } \)</span></p><p>We might recognise that this integral is in the form <span class="math-tex">\(=\int { \frac { f'(x) }{ f(x) } dx=ln|f(x)|+C } \)</span></p><p>Or, we could use Integration by substitution with the substitution u = x² - 1</p><p><span class="math-tex">\(\int _{ 2 }^{ 4 }{ \frac { 2x }{ x²-1 } dx } ={ \left[ ln|x²-1| \right] }_{ 2 }^{ 4 }\)</span></p><p>= ln 15 - ln 3</p><p><span class="math-tex">\(=ln\frac { 15 }{ 3 } \)</span></p><p>= ln3</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div> </div> </div> <div class="modal-footer slide-quiz-actions"> <div class=""> <div class="pull-left pull-xs-none mb-xs-3"> <button class="btn btn-default d-xs-none btn-prev"> <i class="fa fa-arrow-left"></i> Prev </button> </div> <div class="pull-right pull-xs-none"> <button class="btn btn-success btn-xs-block text-xs-center btn-results" style="display: none"> <i class="fa fa-bar-chart"></i> Check Results </button> <button class="btn btn-default d-xs-none btn-next"> Next <i class="fa fa-arrow-right"></i> </button> <button class="btn btn-default btn-xs-block text-xs-center btn-close" data-dismiss="modal" style="display: none"> Close </button> </div> </div> </div> </div> </div></div> <hr class="hidden-separator"> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#9de5a723"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="9de5a723"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> Area between Graphs HL2 <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="11-164-730" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p><p>The function y = f(x) is plotted below</p><p><img alt="" src="../../files/integration/area-between/quiz1.png" style="width: 397px; height: 235px;"></p><p>The area bounded by the function y = f(x), the x axis, x = -1 and x = 2 is given by</p><p><span class="math-tex">\(\int _{ b }^{ a }{ f(x)dx } \quad +\quad \left| \int _{ d }^{ c }{ f(x)dx } \right| \)</span></p><p>Find <em><strong>a</strong></em> , <em><strong>b</strong></em> , <em><strong>c</strong></em> & <em><strong>d</strong></em></p></div><div class="q-answer"><p><em><strong>a</strong></em> = <input type="text" style="height: auto;" data-c="1"> <span class="review"></span></p><p><em><strong>b</strong></em> = <input type="text" style="height: auto;" data-c="-1"> <span class="review"></span></p><p><em><strong>c</strong></em> = <input type="text" style="height: auto;" data-c="2"> <span class="review"></span></p><p><em><strong>d</strong></em> = <input type="text" style="height: auto;" data-c="1"> <span class="review"></span></p></div><div class="q-explanation"><p>The important thing about this question is that the absolute symbol around the 2nd integral makes the result a positive value.</p><p>Answers <em><strong>c</strong></em> and <em><strong>d</strong></em> could be reversed.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p><p>The graph below shows the area bounded by y = x² and y = <span class="math-tex">\(\sqrt{x}\)</span></p><p><img alt="" src="../../files/integration/area-between/quiz2.png" style="width: 199px; height: 156px;"></p><p>Which expression represents the area?</p></div><div class="q-answer"><p><label class="radio"><input type="radio"> <span class="math-tex">\(\int _{ 0 }^{ 1 }{ \sqrt { x } dx \ +} \int _{ 0 }^{ 1 }{ { x }^{ 2 }dx } \)</span></label> <label class="radio"></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\( \int _{ 0 }^{ 1 }{ { x }^{ 2 }dx \ - } \int _{ 0 }^{ 1 }{ \sqrt { x } dx }\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(\int _{ 1 }^{ 0 }{ \sqrt { x } dx\ - } \int _{ 1 }^{ 0 }{ { x }^{ 2 }dx } \)</span></label></p><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\(\int _{ 0 }^{ 1 }{ \sqrt { x } dx- } \int _{ 0 }^{ 1 }{ { x }^{ 2 }dx } \)</span></label></p></div><div class="q-explanation"></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p> The area between the graph y = lnx, the x axis and the lines x = <span class="math-tex">\(\frac{1}{e}\)</span>and x = e is in the form <span class="math-tex">\(2-\frac{a}{e}\)</span></p><p>Find a</p></div><div class="q-answer"><p> <input type="text" style="height: auto;" data-c="2"> <span class="review"></span></p></div><div class="q-explanation"><p>This question requires the integral of lnx. This is found using integration by parts. See <a href="../635/integration-by-parts.html#eg3" target="_blank"><img class="sibico" src="../../../img/sibico/video.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="Video"> integration of lnx</a> in "special integrals"</p><p><img alt="" src="../../files/integration/area-between/quiz3a.png" style="width: 338px; height: 257px;"></p><p>Area = <span class="math-tex">\(\left| \int _{ \frac { 1 }{ e } }^{ 1 }{ lnx }\ dx \right| \quad +\quad \int _{ 1 }^{ e }{ lnx } \ dx\)</span></p><p>= <span class="math-tex">\({ \left| { [xlnx-x] }_{ \frac { 1 }{ e } }^{ 1 } \right| \quad +\quad \left[ xlnx\quad -\quad x \right] }_{ 1 }^{ e }\)</span></p><p>= <span class="math-tex">\({ \left| { [1ln1-1]\quad -\quad [\frac { 1 }{ e } ln(\frac { 1 }{ e } )-\frac { 1 }{ e } ] } \right| \quad +\quad ([elne\quad -\quad e]-[1ln1\quad -\quad 1]) }\)</span></p><p>= <span class="math-tex">\({ \left| { [0-1]\quad -\quad [\frac { 1 }{ e } (-1)-\frac { 1 }{ e } ] } \right| \quad +\quad ([e(1)\quad -\quad e]-[0\quad -\quad 1]) }\)</span></p><p>= <span class="math-tex">\({ \left| { [-1] - [-\frac { 2 }{ e } ] } \right| \quad +\quad ([0]-[- 1]) }\)</span></p><p>= <span class="math-tex">\({ \left| { -1 +\frac { 2 }{ e } } \right| \quad +\quad 1 }\)</span></p><p>= <span class="math-tex">\({ { 1 -\frac { 2 }{ e } } \quad +\quad 1 }\)</span></p><p>= <span class="math-tex">\({ { 2 -\frac { 2 }{ e } } }\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div> </div> </div> <div class="modal-footer slide-quiz-actions"> <div class=""> <div class="pull-left pull-xs-none mb-xs-3"> <button class="btn btn-default d-xs-none btn-prev"> <i class="fa fa-arrow-left"></i> Prev </button> </div> <div class="pull-right pull-xs-none"> <button class="btn btn-success btn-xs-block text-xs-center btn-results" style="display: none"> <i class="fa fa-bar-chart"></i> Check Results </button> <button class="btn btn-default d-xs-none btn-next"> Next <i class="fa fa-arrow-right"></i> </button> <button class="btn btn-default btn-xs-block text-xs-center btn-close" data-dismiss="modal" style="display: none"> Close </button> </div> </div> </div> </div> </div></div> <p> </p> </div> <div class="panel-footer"> <div> <p>text</p> </div> </div> </div> <div class="panel panel-has-colored-body panel-default"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Exam-style Questions</p> </div> </div> <div class="panel-body"> <div class="panel panel-has-colored-body panel-default panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 1</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="280"> <p><img class="sibico" src="../../../img/sibico/hl-green.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL easy"> <img class="sibico" src="../../../img/sibico/sl-orange.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="SL moderate"> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p> <p>Let f(x)=sinx, for <span class="math-tex">\(0\le x\le 2\pi \)</span></p> <p>The following diagram shows the graph of <strong><em>f</em></strong></p> <p><img alt="" src="../../files/integration/area-between/esq1.png" style="width: 498px; height: 190px;"></p> <p>The shaded region R is enclosed by the graph of <strong><em>f</em></strong>, the line x=a , where a<<span class="math-tex">\(\pi\)</span> and the x-axis.</p> <p>The area of R is <span class="math-tex">\(\left( 1-\frac { \sqrt { 3 } }{ 2 } \right) \)</span>. Find the value of <strong><em>b</em></strong>.</p> <p> </p> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <p> </p> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"><content> </content>Area = <span class="math-tex">\(\int _{ a }^{ \pi }{ \sin { x } \ dx } \)</span></section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p><span class="fa fa-print" style="color:rgb(0, 0, 0);font-size:14px;"></span> Print from <a href="../../files/integration/area-between/esq-sol-areabetweengraphs1.pdf" target="_blank">here</a></p> <p><iframe align="middle" frameborder="0" height="480" scrolling="yes" src="../../files/integration/area-between/esq-sol-areabetweengraphs1.pdf" width="640"></iframe></p> </section> <h4> </h4> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-default panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 2</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="269"> <p><img class="sibico" src="../../../img/sibico/hl-orange.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL moderate"> <img class="sibico" src="../../../img/sibico/sl-red.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="SL difficult"> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p> <p><em>Show that the area bounded by the graphs of y = f(x) and y = g(x) in the interval <span class="math-tex">\(0\le x\le \pi \)</span> is given by 2 - <span class="math-tex">\(\frac { \pi }{ 2} \)</span></em></p> <p><em>f(x) = sinx </em></p> <p><em>g(x) = sin²x </em></p> <p><img alt="" src="../../files/integration/area-between/esqdiagram3.png" style="width: 530px; height: 355px;"></p> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <ol> <li>You need to find the area between the two graphs y = sinx and y=sin²x</li> <li>The tricky part is integrating sin²x. To do this you will need to use the trig identity for cos2x</li> </ol> </section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p><span class="fa fa-print" style="color:rgb(0, 0, 0);font-size:14px;"></span> Print from <a href="../../files/integration/area-between/esq_ans_areabetweengraphs2.pdf" target="_blank">here</a></p> <p><iframe align="middle" frameborder="0" height="480" scrolling="yes" src="../../files/integration/area-between/esq_ans_areabetweengraphs2.pdf" width="640"></iframe></p> </section> <h4> </h4> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-expandable panel-default"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 3</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="282"> <p><img class="sibico" src="../../../img/sibico/hl-orange.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL moderate"> <img class="sibico" src="../../../img/sibico/sl-red.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="SL difficult"> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p> <p>Let <span class="math-tex">\(f(x)=ln\left( \frac { x }{ x-1 } \right) \)</span> for x>1</p> <p>a) Find f ' (x)</p> <p>b) <strong>Hence</strong>, show that the area bounded by g(x) = <span class="math-tex">\(\frac { 1 }{ x(x-1) } \)</span> , the x axis, x = 2 and x = e is given by <span class="math-tex">\(ln\left( \frac { 2e-2 }{ e } \right) \)</span></p> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p><content> </content>a) use log laws to make this function easier to differentiate (log<span class="math-tex">\(\frac{a}{b}\)</span>=loga - logb)</p> <p>b) <em><strong>'Hence'</strong></em> is important here. This suggests that the way to do part b) follows from what you have found in part a). You should notice that the function to integrate is the same as the answer to part a)</p> </section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p><span class="fa fa-print" style="color:rgb(0, 0, 0);font-size:14px;"></span> Print from <a href="../../files/integration/area-between/esq-sol-areabetweengraphs3.pdf" target="_blank">here</a></p> <p><iframe align="middle" frameborder="0" height="480" scrolling="yes" src="../../files/integration/area-between/esq-sol-areabetweengraphs3.pdf" width="640"></iframe></p> </section> </div> </div> </div> </div> <div class="panel panel-has-colored-body panel-default panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 4</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="278"> <p><img class="sibico" src="../../../img/sibico/hl-red.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL difficult"> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p> <p>The graph below shows the function <strong>f(x) = sinx </strong>where a < b <span class="math-tex">\(\le \frac { \pi }{ 2 } \)</span></p> <p><img alt="" src="../../files/integration/area-between/esq3.png" style="width: 400px; height: 233px;"></p> <p>a) Write an expression for the area A, bounded by the curve y = f(x), the x axis, x = a and x = b</p> <p>b) <strong>Hence</strong>, without using any further integration, show that the area B = <span class="math-tex">\(bsinb - asina - cosb + cosa\)</span></p> <p>c) Area B can <strong><u>also </u></strong>be found by finding the area bounded by the curve x = f(y), the y axis, y = f(a) and y = f(b). Find the area B using this method and <strong>show</strong> that your answer is the same as the one you found in part b)</p> <p> </p> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <p> </p> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p><content> </content>b) Area B can be found by considering the 2 rectangles. You can find the area of these two rectangles without doing any further integration. For example, the height of the larger rectangle is <em>sinb</em></p> <p>c) Find the area bounded by x = arcsiny, y= sina, y= sinb and the y axis.</p> <p>You can use integration by parts to find <span class="math-tex">\(\int { 1\bullet \arcsin { y}_\ dy } \)</span> . This method is similar to integrating <a href="../635/integration-by-parts.html#eg3" target="_blank">lnx</a></p> <p>You will then be required to find <span class="math-tex">\(\int { \frac { y }{ \sqrt { 1-{ y }^{ 2 } } } dy } \)</span> which you can do using the method of <a href="../635/index.htm" target="_blank">integration by substitution</a> or recognition.</p> </section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p><img alt="" src="../../files/integration/area-between/esq3.png" style="width: 400px; height: 233px;"></p> <p>a) Area A</p> <p><span class="math-tex">\(= \int _{ a }^{ b }{ sinx } dx\\ ={ [-cosx] }_{ a }^{ b }\\ =-cosb\ +\ cosa\)</span></p> <p>b)</p> <p><img alt="" src="../../files/integration/area-between/esq3---partb.png" style="width: 400px; height: 233px;"></p> <p>Area B = area large rectangle - area small rectangle - area A</p> <p>= bsinB - a sinA - (-cosb + cosa)</p> <p>= bsinB - a sinA + cosb - cosa</p> <p>c) Graph is y = sinx which is equivalent to <strong>x = arcsiny</strong> for <span class="math-tex">\(0\le \ x\le \frac { \pi }{ 2} \ \)</span></p> <p>Area B <span class="math-tex">\(=\int _{ sina }^{ sinb }{ \arcsin { y } }\ dy\)</span></p> <p>Let's consider <span class="math-tex">\(\int { \arcsin { y } }\ dy\)</span></p> <p>We can use integration by parts</p> <p><span class="math-tex">\(\int {u\frac{{dv}}{{dx}}}\ dx = uv - \int {\frac{{du}}{{dx}}}\ vdx\)</span></p> <p>since our variable is y , this becomes</p> <p><span class="math-tex">\(\int {u\frac{{dv}}{{dy}}} \ dy = uv - \int {\frac{{du}}{{dy}}} v \ dy\)</span></p> <p>We make our integral a product of two parts (this is like <a href="../635/integration-by-parts.html#eg3" target="_blank">integrating <em>lnx</em>)</a></p> <p><span class="math-tex">\(\int 1\cdot { \arcsin { y } } \ dy\)</span></p> <p style="margin-left: 160px;"><span class="math-tex">\(u = arcsiny \quad \frac { dv }{ dy } = 1\\ \frac { du }{ dy } = \frac { 1 }{ \sqrt { 1-{ y }^{ 2 } } }\quad v = y\\\)</span></p> <p><span class="math-tex">\(=y \cdot arcsiny - \int \frac { y }{ \sqrt { 1-{ y }^{ 2 } } } \ dy\)</span></p> <p>To find <span class="math-tex">\(\int \frac { y }{ \sqrt { 1-{ y }^{ 2 } } } \ dy\)</span> , we need to use integration by substitution (or recognition)</p> <p style="margin-left: 200px;">Let w = 1 - y<sup>2</sup></p> <p style="margin-left: 200px;"><span class="math-tex">\( \frac { dw }{dy } = -2y\)</span></p> <p><span class="math-tex">\(=y \cdot arcsiny + \int \frac { 1 }{ 2 } { w }^{ -\frac { 1 }{ 2 } }\ dw\)</span></p> <p><span class="math-tex">\(=y \cdot arcsiny + { w }^{ \frac { 1 }{ 2 } } \ +c\)</span></p> <p><span class="math-tex">\(=y \cdot arcsiny + \sqrt {1-{ y }^2 } \ +c\)</span></p> <p>So, let's go back to the area of B</p> <p>Area B <span class="math-tex">\(=\int _{ sina }^{ sinb }{ \arcsin { y } }\ dy\)</span></p> <p><span class="math-tex">\(={ \left[ { \ y \cdot arcsiny + \sqrt {1-{ y }^2 } } \ \right] }_{ sina }^{ sinb }\)</span></p> <p><span class="math-tex">\(=[\ sinb\cdot b+\sqrt { 1-sin²b } \ ]-[ \ sina\cdot a+\sqrt { 1-sin²a } \ ]\)</span></p> <p style="margin-left: 40px;">Since, 1-sin²x = cos²x</p> <p><span class="math-tex">\(=[\ sinb\cdot b+cosb\ ]-[ \ sina\cdot a+cosa \ ]\)</span></p> <p><span class="math-tex">\(= b\cdot sinb+cosb- \ a\cdot sina-cosa \)</span></p> <p>= bsinB - a sinA + cosb - cosa</p> </section> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="page-container panel-self-assessment" data-id="730"> <div class="panel-heading">MY PROGRESS</div> <div class="panel-body understanding-rate"> <div class="msg"></div> <label class="label-lg">Self-assessment</label><p>How much of <strong>Area between Graphs HL</strong> have you understood?</p><div 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