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<li><a href="#average-value-of-a-function" id="toc-average-value-of-a-function" class="nav-link active" data-scroll-target="#average-value-of-a-function"> <span class="header-section-number">42.1</span> Average value of a function</a></li>
<li><a href="#the-mean-value-theorem-for-integrals" id="toc-the-mean-value-theorem-for-integrals" class="nav-link" data-scroll-target="#the-mean-value-theorem-for-integrals"> <span class="header-section-number">42.2</span> The mean value theorem for integrals</a></li>
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<h1 class="title d-none d-lg-block"><span class="chapter-number">42</span>&nbsp; <span class="chapter-title">Mean value theorem for integrals</span></h1>
</div>
<div class="quarto-title-meta">
</div>
</header>
<p>This section uses these add-on packages:</p>
<div class="sourceCode cell-code" id="cb1"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb1-1"><a href="#cb1-1" aria-hidden="true" tabindex="-1"></a><span class="im">using</span> <span class="bu">CalculusWithJulia</span></span>
<span id="cb1-2"><a href="#cb1-2" aria-hidden="true" tabindex="-1"></a><span class="im">using</span> <span class="bu">Plots</span></span>
<span id="cb1-3"><a href="#cb1-3" aria-hidden="true" tabindex="-1"></a><span class="im">using</span> <span class="bu">QuadGK</span></span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<hr>
<section id="average-value-of-a-function" class="level2" data-number="42.1">
<h2 data-number="42.1" class="anchored" data-anchor-id="average-value-of-a-function"><span class="header-section-number">42.1</span> Average value of a function</h2>
<p>Let <span class="math inline">\(f(x)\)</span> be a continuous function over the interval <span class="math inline">\([a,b]\)</span> with <span class="math inline">\(a &lt; b\)</span>.</p>
<p>The average value of <span class="math inline">\(f\)</span> over <span class="math inline">\([a,b]\)</span> is defined by:</p>
<p><span class="math display">\[
\frac{1}{b-a} \int_a^b f(x) dx.
\]</span></p>
<p>If <span class="math inline">\(f\)</span> is a constant, this is just the contant value, as would be expected. If <span class="math inline">\(f\)</span> is <em>piecewise</em> linear, then this is the weighted average of these constants.</p>
<section id="examples" class="level4">
<h4 class="anchored" data-anchor-id="examples">Examples</h4>
<section id="example-average-velocity" class="level5">
<h5 class="anchored" data-anchor-id="example-average-velocity">Example: average velocity</h5>
<p>The average velocity between times <span class="math inline">\(a &lt; b\)</span>, is simply the change in position during the time interval divided by the change in time. In notation, this would be <span class="math inline">\((x(b) - x(a)) / (b-a)\)</span>. If <span class="math inline">\(v(t) = x'(t)\)</span> is the velocity, then by the second part of the fundamental theorem of calculus, we have, in agreement with the definition above, that:</p>
<p><span class="math display">\[
\text{average velocity} = \frac{x(b) - x(a)}{b-a} = \frac{1}{b-a} \int_a^b v(t) dt.
\]</span></p>
<p>The average speed is the change in <em>total</em> distance over time, which is given by</p>
<p><span class="math display">\[
\text{average speed} = \frac{1}{b-a} \int_a^b \lvert v(t)\rvert dt.
\]</span></p>
<p>Let <span class="math inline">\(\bar{v}\)</span> be the average velocity. Then we have <span class="math inline">\(\bar{v} \cdot(b-a) = x(b) - x(a)\)</span>, or the change in position can be written as a constant (<span class="math inline">\(\bar{v}\)</span>) times the time, as though we had a constant velocity. This is an old intuition. <a href="http://www.math.harvard.edu/~knill/teaching/math1a_2011/exhibits/bressoud/">Bressoud</a> comments on the special case known to scholars at Merton College around <span class="math inline">\(1350\)</span> that the distance traveled by an object under uniformly increasing velocity starting at <span class="math inline">\(v_0\)</span> and ending at <span class="math inline">\(v_t\)</span> is equal to the distance traveled by an object with constant velocity of <span class="math inline">\((v_0 + v_t)/2\)</span>.</p>
</section>
<section id="example" class="level5">
<h5 class="anchored" data-anchor-id="example">Example</h5>
<p>What is the average value of <span class="math inline">\(f(x)=\sin(x)\)</span> over <span class="math inline">\([0, \pi]\)</span>?</p>
<p><span class="math display">\[
\text{average} = \frac{1}{\pi-0} \int_0^\pi \sin(x) dx = \frac{1}{\pi} (-\cos(x)) \big|_0^\pi = \frac{2}{\pi}
\]</span></p>
<p>Visually, we have:</p>
<div class="cell" data-execution_count="4">
<div class="sourceCode cell-code" id="cb2"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb2-1"><a href="#cb2-1" aria-hidden="true" tabindex="-1"></a><span class="fu">plot</span>(sin, <span class="fl">0</span>, <span class="cn">pi</span>)</span>
<span id="cb2-2"><a href="#cb2-2" aria-hidden="true" tabindex="-1"></a><span class="fu">plot!</span>(x <span class="op">-&gt;</span> <span class="fl">2</span><span class="op">/</span><span class="cn">pi</span>)</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="5">
<p><img src="mean_value_theorem_files/figure-html/cell-5-output-1.svg" class="img-fluid"></p>
</div>
</div>
</section>
<section id="example-1" class="level5">
<h5 class="anchored" data-anchor-id="example-1">Example</h5>
<p>What is the average value of the function <span class="math inline">\(f\)</span> which is <span class="math inline">\(1\)</span> between <span class="math inline">\([0,3]\)</span>, <span class="math inline">\(2\)</span> between <span class="math inline">\((3,5]\)</span> and <span class="math inline">\(1\)</span> between <span class="math inline">\((5,6]\)</span>?</p>
<p>Though not continuous, <span class="math inline">\(f(x)\)</span> is integrable as it contains only jumps. The integral from <span class="math inline">\([0,6]\)</span> can be computed with geometry: <span class="math inline">\(3\cdot 3 + 2 \cdot 2 + 1 \cdot 1 = 14\)</span>. The average then is <span class="math inline">\(14/(6-0) = 7/3\)</span>.</p>
</section>
<section id="example-2" class="level5">
<h5 class="anchored" data-anchor-id="example-2">Example</h5>
<p>What is the average value of the function <span class="math inline">\(e^{-x}\)</span> between <span class="math inline">\(0\)</span> and <span class="math inline">\(\log(2)\)</span>?</p>
<p><span class="math display">\[
\begin{align*}
\text{average} = \frac{1}{\log(2) - 0} \int_0^{\log(2)} e^{-x} dx\\
&amp;= \frac{1}{\log(2)} (-e^{-x}) \big|_0^{\log(2)}\\
&amp;= -\frac{1}{\log(2)} (\frac{1}{2} - 1)\\
&amp;= \frac{1}{2\log(2)}.
\end{align*}
\]</span></p>
<p>Visualizing, we have</p>
<div class="cell" data-execution_count="5">
<div class="sourceCode cell-code" id="cb3"><pre class="sourceCode julia code-with-copy"><code class="sourceCode julia"><span id="cb3-1"><a href="#cb3-1" aria-hidden="true" tabindex="-1"></a><span class="fu">plot</span>(x <span class="op">-&gt;</span> <span class="fu">exp</span>(<span class="op">-</span>x), <span class="fl">0</span>, <span class="fu">log</span>(<span class="fl">2</span>))</span>
<span id="cb3-2"><a href="#cb3-2" aria-hidden="true" tabindex="-1"></a><span class="fu">plot!</span>(x <span class="op">-&gt;</span> <span class="fl">1</span><span class="op">/</span>(<span class="fl">2</span><span class="fu">*log</span>(<span class="fl">2</span>)))</span></code><button title="Copy to Clipboard" class="code-copy-button"><i class="bi"></i></button></pre></div>
<div class="cell-output cell-output-display" data-execution_count="6">
<p><img src="mean_value_theorem_files/figure-html/cell-6-output-1.svg" class="img-fluid"></p>
</div>
</div>
</section>
</section>
</section>
<section id="the-mean-value-theorem-for-integrals" class="level2" data-number="42.2">
<h2 data-number="42.2" class="anchored" data-anchor-id="the-mean-value-theorem-for-integrals"><span class="header-section-number">42.2</span> The mean value theorem for integrals</h2>
<p>If <span class="math inline">\(f(x)\)</span> is assumed integrable, the average value of <span class="math inline">\(f(x)\)</span> is defined, as above. Re-expressing gives that there exists a <span class="math inline">\(K\)</span> with</p>
<p><span class="math display">\[
K \cdot (b-a) = \int_a^b f(x) dx.
\]</span></p>
<p>When we assume that <span class="math inline">\(f(x)\)</span> is continuous, we can describe <span class="math inline">\(K\)</span> as a value in the range of <span class="math inline">\(f\)</span>:</p>
<blockquote class="blockquote">
<p><strong>The mean value theorem for integrals</strong>: Let <span class="math inline">\(f(x)\)</span> be a continuous function on <span class="math inline">\([a,b]\)</span> with <span class="math inline">\(a &lt; b\)</span>. Then there exists <span class="math inline">\(c\)</span> with <span class="math inline">\(a \leq c \leq b\)</span> with</p>
<p><span class="math inline">\(f(c) \cdot (b-a) = \int_a^b f(x) dx.\)</span>`</p>
</blockquote>
<p>The proof comes from the intermediate value theorem and the extreme value theorem. Since <span class="math inline">\(f\)</span> is continuous on a closed interval, there exists values <span class="math inline">\(m\)</span> and <span class="math inline">\(M\)</span> with <span class="math inline">\(f(c_m) = m \leq f(x) \leq M=f(c_M)\)</span>, for some <span class="math inline">\(c_m\)</span> and <span class="math inline">\(c_M\)</span> in the interval <span class="math inline">\([a,b]\)</span>. Since <span class="math inline">\(m \leq f(x) \leq M\)</span>, we must have:</p>
<p><span class="math display">\[
m \cdot (b-a) \leq K\cdot(b-a) \leq M\cdot(b-a).
\]</span></p>
<p>So in particular <span class="math inline">\(K\)</span> is in <span class="math inline">\([m, M]\)</span>. But <span class="math inline">\(m\)</span> and <span class="math inline">\(M\)</span> correspond to values of <span class="math inline">\(f(x)\)</span>, so by the intermediate value theorem, <span class="math inline">\(K=f(c)\)</span> for some <span class="math inline">\(c\)</span> that must lie in between <span class="math inline">\(c_m\)</span> and <span class="math inline">\(c_M\)</span>, which means as well that it must be in <span class="math inline">\([a,b]\)</span>.</p>
<section id="proof-of-second-part-of-fundamental-theorem-of-calculus" class="level5">
<h5 class="anchored" data-anchor-id="proof-of-second-part-of-fundamental-theorem-of-calculus">Proof of second part of Fundamental Theorem of Calculus</h5>
<p>The mean value theorem is exactly what is needed to prove formally the second part of the Fundamental Theorem of Calculus. Again, suppose <span class="math inline">\(f(x)\)</span> is continuous on <span class="math inline">\([a,b]\)</span> with <span class="math inline">\(a &lt; b\)</span>. For any <span class="math inline">\(a &lt; x &lt; b\)</span>, we define <span class="math inline">\(F(x) = \int_a^x f(u) du\)</span>. Then the derivative of <span class="math inline">\(F\)</span> exists and is <span class="math inline">\(f\)</span>.</p>
<p>Let <span class="math inline">\(h&gt;0\)</span>. Then consider the forward difference <span class="math inline">\((F(x+h) - F(x))/h\)</span>. Rewriting gives:</p>
<p><span class="math display">\[
\frac{\int_a^{x+h} f(u) du - \int_a^x f(u) du}{h} =\frac{\int_x^{x+h} f(u) du}{h} = f(\xi(h)).
\]</span></p>
<p>The value <span class="math inline">\(\xi(h)\)</span> is just the <span class="math inline">\(c\)</span> corresponding to a given value in <span class="math inline">\([x, x+h]\)</span> guaranteed by the mean value theorem. We only know that <span class="math inline">\(x \leq \xi(h) \leq x+h\)</span>. But this is plenty - it says that <span class="math inline">\(\lim_{h \rightarrow 0+} \xi(h) = x\)</span>. Using the fact that <span class="math inline">\(f\)</span> is continuous and the known properties of limits of compositions of functions this gives <span class="math inline">\(\lim_{h \rightarrow 0+} f(\xi(h)) = f(x)\)</span>. But this means that the (right) limit of the secant line expression exists and is equal to <span class="math inline">\(f(x)\)</span>, which is what we want to prove. Repeating a similar argument when <span class="math inline">\(h &lt; 0\)</span>, finishes the proof.</p>
<p>The basic notion used is simply that for small <span class="math inline">\(h\)</span>, this expression is well approximated by the left Riemann sum taken over <span class="math inline">\([x, x+h]\)</span>:</p>
<p><span class="math display">\[
f(\xi(h)) \cdot h = \int_x^{x+h} f(u) du.
\]</span></p>
</section>
</section>
<section id="questions" class="level2" data-number="42.3">
<h2 data-number="42.3" class="anchored" data-anchor-id="questions"><span class="header-section-number">42.3</span> Questions</h2>
<section id="question" class="level6">
<h6 class="anchored" data-anchor-id="question">Question</h6>
<p>Between <span class="math inline">\(0\)</span> and <span class="math inline">\(1\)</span> a function is constantly <span class="math inline">\(1\)</span>. Between <span class="math inline">\(1\)</span> and <span class="math inline">\(2\)</span> the function is constantly <span class="math inline">\(2\)</span>. What is the average value of the function over the interval <span class="math inline">\([0,2]\)</span>?</p>
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<section id="question-1" class="level6">
<h6 class="anchored" data-anchor-id="question-1">Question</h6>
<p>Between <span class="math inline">\(0\)</span> and <span class="math inline">\(2\)</span> a function is constantly <span class="math inline">\(1\)</span>. Between <span class="math inline">\(2\)</span> and <span class="math inline">\(3\)</span> the function is constantly <span class="math inline">\(2\)</span>. What is the average value of the function over the interval <span class="math inline">\([0,3]\)</span>?</p>
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<section id="question-2" class="level6">
<h6 class="anchored" data-anchor-id="question-2">Question</h6>
<p>What integral will show the intuition of the Merton College scholars that the distance traveled by an object under uniformly increasing velocity starting at <span class="math inline">\(v_0\)</span> and ending at <span class="math inline">\(v_t\)</span> is equal to the distance traveled by an object with constant velocity of <span class="math inline">\((v_0 + v_t)/2\)</span>.</p>
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\((v(0) + v(t))/2 \cdot \int_0^t du = (v(0) + v(t))/2 \cdot t\)
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\(\int_0^t (v(0) + v(u))/2 du = v(0)/2\cdot t + x(u)/2\ \big|_0^t\)
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\(\int_0^t v(u) du = v^2/2 \big|_0^t\)
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<section id="question-3" class="level6">
<h6 class="anchored" data-anchor-id="question-3">Question</h6>
<p>Find the average value of <span class="math inline">\(\cos(x)\)</span> over the interval <span class="math inline">\([-\pi/2, \pi/2]\)</span>.</p>
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<section id="question-4" class="level6">
<h6 class="anchored" data-anchor-id="question-4">Question</h6>
<p>Find the average value of <span class="math inline">\(\cos(x)\)</span> over the interval <span class="math inline">\([0, \pi]\)</span>.</p>
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<section id="question-5" class="level6">
<h6 class="anchored" data-anchor-id="question-5">Question</h6>
<p>Find the average value of <span class="math inline">\(f(x) = e^{-2x}\)</span> between <span class="math inline">\(0\)</span> and <span class="math inline">\(2\)</span>.</p>
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<section id="question-6" class="level6">
<h6 class="anchored" data-anchor-id="question-6">Question</h6>
<p>Find the average value of <span class="math inline">\(f(x) = \sin(x)^2\)</span> over the <span class="math inline">\(0\)</span>, <span class="math inline">\(\pi\)</span>.</p>
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<section id="question-7" class="level6">
<h6 class="anchored" data-anchor-id="question-7">Question</h6>
<p>Which is bigger? The average value of <span class="math inline">\(f(x) = x^{10}\)</span> or the average value of <span class="math inline">\(g(x) = \lvert x \rvert\)</span> over the interval <span class="math inline">\([0,1]\)</span>?</p>
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That of \(f(x) = x^{10}\).
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That of \(g(x) = \lvert x \rvert\).
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<section id="question-8" class="level6">
<h6 class="anchored" data-anchor-id="question-8">Question</h6>
<p>Define a family of functions over the interval <span class="math inline">\([0,1]\)</span> by <span class="math inline">\(f(x; a,b) = x^a \cdot (1-x)^b\)</span>. Which has a greater average, <span class="math inline">\(f(x; 2,3)\)</span> or <span class="math inline">\(f(x; 3,4)\)</span>?</p>
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\(f(x; 3,4)\)
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\(f(x; 2,3)\)
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<section id="question-9" class="level6">
<h6 class="anchored" data-anchor-id="question-9">Question</h6>
<p>Suppose the average value of <span class="math inline">\(f(x)\)</span> over <span class="math inline">\([a,b]\)</span> is <span class="math inline">\(100\)</span>. What is the average value of <span class="math inline">\(100 f(x)\)</span> over <span class="math inline">\([a,b]\)</span>?</p>
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<section id="question-10" class="level6">
<h6 class="anchored" data-anchor-id="question-10">Question</h6>
<p>Suppose <span class="math inline">\(f(x)\)</span> is continuous and positive on <span class="math inline">\([a,b]\)</span>.</p>
<ul>
<li>Explain why for any <span class="math inline">\(x &gt; a\)</span> it must be that:</li>
</ul>
<p><span class="math display">\[
F(x) = \int_a^x f(x) dx &gt; 0
\]</span></p>
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Because the mean value theorem says this is \(f(c) (x-a)\) for some \(c\) and both terms are positive by the assumptions
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Because the definite integral is only defined for positive area, so it is always positive
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<li>Explain why <span class="math inline">\(F(x)\)</span> is increasing.</li>
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By the fundamental theorem of calculus, part I, \(F'(x) = f(x) &gt; 0\), hence \(F(x)\) is increasing
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By the intermediate value theorem, as \(F(x) &gt; 0\), it must be true that \(F(x)\) is increasing
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By the extreme value theorem, \(F(x)\) must reach its maximum, hence it must increase.
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<h6 class="anchored" data-anchor-id="question-11">Question</h6>
<p>For <span class="math inline">\(f(x) = x^2\)</span>, which is bigger: the average of the function <span class="math inline">\(f(x)\)</span> over <span class="math inline">\([0,1]\)</span> or the geometric mean which is the exponential of the average of the logarithm of <span class="math inline">\(f\)</span> over the same interval?</p>
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The exponential of the average of \(\log(f)\)
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The average of \(f\)
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