OBJECTIVES

When you finish this section, you should be able to:

1. Use properties of the definite integral (p. 369)
2. Work with the Mean Value Theorem for Integrals (p. 372)
3. Find the average value of a function (p. 373)

NOTE

From now on, we will refer to Parts 1 and 2 of the FundamentalTheorem of Calculus simply as the Fundamental Theorem ofCalculus.

THEOREM The Integral of the Sum of Two Functions

If two functions \(f\) and \(g\) are continuous on the closedinterval \([a,b]\), then\[\begin{equation*}\boxed{\bbox[5pt]{\int_{a}^{b}[ f(x)+g(x)] \,dx=\int_{a}^{b}f(x)\,dx+\int_{a}^{b}g(x)\,dx }}\tag{1}\end{equation*}\]

IN WORDS

The integral of a sum equals the sum of the integrals.

Proof

Since \(f\) and \(g\) are continuous on \([a,b],\) then the definiteintegral of each function exists. Let \(F\) and\(G\) be an antiderivative of \(f\) and \(g,\) respectively, on\((a,b) \). Then \(F' =f\) and \(G^{\prime} =g\) on \((a,b) \).

Also, since \((F + G)' = F^\prime + G' = f + g,\)then \(F+G\) is an antiderivative of \(f+g\) on \((a,b) .\) Now\[\begin{eqnarray*}\int_{a}^{b}[f(x)+g(x)]\,dx=\bigg[F(x)+G(x)\bigg] _{a}^{b} &=&[F(b)+G(b)]-[F(a)+G(a)] \\&=&[F(b)-F(a)]+[G(b)-G(a)] \\&=&{\int_{a}^{b}{f(x)\,dx+{\int_{a}^{b}{g(x)\,dx}}}} \end{eqnarray*}\]

EXAMPLE 1 Using Property (1) of the Definite Integral

\[\begin{eqnarray*}\int_{0}^{1}(x^{2}+e^{x})\,dx &=& \int_{0}^{1}x^{2}dx+\int_{0}^{1}e^{x}\,dx=\left[ \dfrac{x^{3}}{3}\right] _{0}^{1}+ \Big[ e^{x}\Big] _{0}^{1} \\[4pt]&=&\left[ \dfrac{1^{3}}{3}-0\right] +\Big[ e^{1}-e^{0}\Big] =\dfrac{1}{3}+e-1=e-\dfrac{2}{3} \end{eqnarray*}\]

NOW WORK

Problem 13.

THEOREM The Integral of a Constant Times a Function

Suppose a function \(f\) is continuous on the closed interval \([a,b]\). If \(k\) is a constant, then\[\begin{equation*}\boxed{\bbox[5pt]{\int_{a}^{b}kf(x)\,dx=k\int_{a}^{b}f(x)\,dx }}\tag{2}\end{equation*}\]

IN WORDS

A constant factor can be factored out of an integral.

EXAMPLE 2 Using Property (2) of the Definite Integral

\[\begin{eqnarray*}{\int_{1}^{e}}\dfrac{{3}}{x}{\,dx=3} {{\,{\int_{1}^{e}}}}\dfrac{1}{x}{dx=3} \Big[ \ln \vert x\vert \Big] _{1}^{e}={3}\left( {\ln \,}e-\ln1\right) =3(1-0) =3 \end{eqnarray*}\]

NOW WORK

Problem 15.

THEOREM

Suppose each of the functions \(f_{{1}}\), \(f_{{2}}, \ldots , f_{n}\) is continuous on the closed interval \([a,b]\). If \(k_{{1}}\), \(k_{{2}}, \ldots\), \(k_{n}\) are constants, then \[\begin{equation*}\boxed{\bbox[5pt]{\begin{array}{@{\hspace*{-6pt}}lll@{}}&&\int_{a}^{b}\left[ k_{{1}}f_{{1}}(x)+k_{{2}}f_{{2}}(x)+\cdots+\,k_{n} f_{n}(x)\right]\,dx \\[9pt]&&  = k_{{1}}\int_{a}^{b}f_{{1}}(x)\,dx + k_{{2}} \int_{a}^{b}f_{{2}}(x)\,dx+\cdots + k_{n}\int_{a}^{b}f_{n}(x)\,dx \end{array}}}\tag{3}\end{equation*}\]

EXAMPLE 3 Using Property (3) of the Definite Integral

Find \(\int_{1}^{2}\dfrac{3x^{3}-6x^{2}-5x+4}{2x}dx\).

Solution The function \(f(x)=\dfrac{3x^{3}-6x^{2}-5x+4}{2x}\) is continuous on the closed interval \([1,2] \). Using algebra and the properties of the definite integral, we get\[\begin{eqnarray*}\int_{1}^{2}\dfrac{3x^{3}-6x^{2}-5x+4}{2x}dx &=&\int_{1}^{2}\left[\dfrac{3}{2} x^{2}- 3x- \dfrac{5}{2}+\dfrac{2}{x}\right] dx\\&=& \int_{1}^{2} \dfrac{3}{2} x^{2}dx-\int_{1}^{2} 3x\, dx-\int_{1}^{2} \dfrac{5}{2} \, dx+\int_{1}^{2}\dfrac{2}{x}~dx \\[4pt]&=& \dfrac{3}{2}\int_{1}^{2}x^{2}~dx-3\int_{1}^{2}x\, dx-\dfrac{5}{2}\int_{1}^{2}~dx+2\int_{1}^{2}\dfrac{1}{x}~dx \\&=& \dfrac{3}{2} \left[ \dfrac{x^{3}}{3}\right] _{1}^{2} -3 \left[ \dfrac{x^{2}}{2}\right] _{1}^{2} -\dfrac{5}{2} \Big[x\Big]_{1}^{2} +2 \Big[\ln \vert x\vert \Big]_{1}^{2} \\&=&\dfrac{1}{2} ( 8-1) - 3\left( 2-\dfrac{1}{2}\right)-\dfrac{5}{2}(2-1) + 2( \ln 2- \ln 1 ) \\&=&-\dfrac{7}{2}+2\ln2\approx -2.114\end{eqnarray*}\]

NOW WORK

Problem 27.

THEOREM

If a function \(f\) is continuous on an interval containing thenumbers \(a\), \(b\), and \(c\), then \[\begin{equation*}\boxed{\bbox[5pt]{\int_{a}^{b}f(x)\,dx=\int_{a}^{c}f(x)\,dx+\int_{c}^{b}f(x)\,dx }}\tag{4}\end{equation*}\]

EXAMPLE 4 Using Property (4) of the Definite Integral

1. If \(f\) is continuous on the closed interval \([2,7]\), then \[\int_{2}^{7}f(x)\,dx=\int_{2}^{4}f(x)\,dx+\int_{4}^{7}f(x)\,dx \]
2. If \(g\) is continuous on the closed interval \([3,25]\), then \[\int_{3}^{10}g(x)\,dx=\int_{3}^{25}g (x)\,dx+ \int_{25}^{10}g(x) dx \]

NOW WORK

Problem 43.

EXAMPLE 5 Using Property (4) of the Definite Integral

Find the area \(A\) under the graph of\[f(x) =\left\{ \begin{array}{c@{ }c@{ }ccc}x^{2} & \hbox{if} & 0 & \leq x &\lt10 \\ 100 & \hbox{if} & 10 &\leq x & \leq 15\end{array}\right.\]

from \(0\) to \(15.\)

Solution See Figure 5.25. Since \(f\) is nonnegative on the closed interval\([0,15] ,\) then \(\int_{0}^{15}f(x)\, dx\) equals the area \(A\) under the graph of \(f\)from \(0\) to \(15.\) Since \(f\) is continuous on \([0,15],\)\[\begin{eqnarray*}\int_{0}^{15}f(x)\, dx &=& \int_{0}^{10}f(x)\, dx+\int_{10}^{15}f(x)\,dx=\int_{0}^{10}x^{2}dx+\int_{10}^{15}100\,dx \\[4.5pt]&=& \left[\dfrac{x^{3}}{3}\right] _{0}^{10}+ \Big[100x\Big]_{10}^{15}=\dfrac{1000}{3}+500=\dfrac{2500}{3}\end{eqnarray*}\]

Figure 5.25: \(A=\int^{15}_0 f(x)\, dx\)

The area under the graph of \(f\) is approximately 833.33 square units.

NOW WORK

Problem 35.

THEOREM Bounds on an Integral

If a function \(f\) is continuous on a closed interval \([a,b]\)and if \(m\) and \(M\) denote the absolute minimum value and the absolute maximum value,respectively, of \(f\) on \([a,b]\), then\[\begin{equation*}\boxed{\bbox[5pt]{m(b-a)\leq \int_{a}^{b}f(x)\,dx\leq M\,(b-a) }}\tag{5}\end{equation*}\]

EXAMPLE 6 Using Property (5) of Definite Integrals

1. Find an upper estimate and a lower estimate for the area \(A\) under thegraph of \(f(x) =\sin x\) from \(0\) to \(\pi \).
2. Find the actual area under the graph.

Solution The graph of \(f\) is shown in Figure 5.27.Since \(f(x) \geq 0\) for all \(x\) in the closed interval \([ 0,\pi ] \), the area \(A\) under its graph is given by thedefinite integral, \(\int_{0}^{\pi }\sin x~dx\).

Figure 5.27: \(f(x) = \sin x, 0 \leq x \leq \pi\).

(a) From the Extreme Value Theorem, \(f\) has anabsolute minimum value and an absolute maximum value on the interval \([0,\pi] .\)The absolute maximum of \(f\) occurs at \(x=\dfrac{\pi}{2},\) and its value is \(f \left( \dfrac{\pi }{2}\right) =\sin \dfrac{\pi }{2}=1.\) The absoluteminimum occurs at \(x=0\) and at \(x=\pi \); the absolute minimumvalue is \(f (0) =\sin 0=0=f (\pi)\). Using the inequalities in (5), the area under the graph of \(f\) is bounded as follows:\[0\leq \int_{0}^{\pi }\sin x~dx\leq \pi \]

(b) The actual area under the graph is\[A= \int_{0}^{\pi}\sin x\,dx= \bigg[-\cos x\bigg] _{0}^{\pi }=-\cos \pi+\cos 0=1+1=2\hbox{ square units} \]

NOW WORK

Problem 53.

THEOREM Mean Value Theorem for Integrals

If a function \(f\) is continuous on a closed interval \([a,b]\), thereis a real number \(u\), \(a\,\leq \,u\,\leq \,b\), for which \[\boxed{\bbox[5pt]{\int_{a}^{b}f(x)\,dx=f(u)(b-a) }}\tag{6}\]

Proof

Let \(f\) be a function that is continuous on a closed interval \([a,b] \).

If \(f\) is a constant function, say, \(f(x) =k,\) on \([a,b]\), then \[\int_{a}^{b}f(x)\,dx= \int_{a}^{b}k\,dx =k(b-a) =f(u)(b-a) \]

for any choice of \(u\) in \([a,b]\).

If \(f\) is not a constant function on \([a,b]\), then by the Extreme ValueTheorem, \(f\) has an absolute maximum and an absolute minimum on \([a,b]\). Suppose \(f\) assumes its absolute minimum at the number \(c\) so that \(f(c)=m\); and suppose \(f\) assumes its absolute maximum at the number \(C\)so that \(f (C)\,=M\). Then by the Bounds on an Integral Theorem (5), we have\[m(b-a)\leq {\int_{a}^{b}{f(x)\,dx\leq M(b-a)}}\qquad \hbox{for all } x\hbox{ in }[a,b]\]

We divide each part by \((b-a) \) and replace \(m\) by \(f(c)\) and\(M\) by \(f (C)\). Then \[f(c)\leq \dfrac{{1}}{{b-a}}\int_{a}^{b}f(x)\,dx\leq f(C)\]

Since \(\dfrac{1}{b-a}\int_{a}^{b}f(x)\,dx\) is a real number between \(f(c)\) and \(f (C)\), it follows from the Intermediate Value Theorem that there is a real number \(u\) between \(c\) and \(C\), for which \[f(u)\,={\dfrac{{1}}{{b-a}}}{\int_{a}^{b}{f(x)\,dx}}\]

That is, there is a real number \(u\), \(a\leq u\leq b\), for which \[{\int_{a}^{b}{f(x)\,dx}}=f(u)\,(b-a) \]

EXAMPLE 7 Using the Mean Value Theorem for Integrals

Find the number(s) \(u\) guaranteed by the Mean Value Theorem forIntegrals for \(\int_{2}^{6}x^{2}dx.\)

Solution The Mean Value Theorem for Integrals states there is anumber \(u,\) \(2\leq u\leq 6,\) for which \[\int_{2}^{6}x^{2}dx=f(u) (6-2) =4u^{2}\qquad\qquad{\color{#0066A7}{\hbox{\(f(u)=u^2\)}}}\]

We integrate to obtain \[\int_{2}^{6}x^{2}dx=\left[\dfrac{x^{3}}{3}\right] _{2}^{6}=\dfrac{1}{3}(216-8) =\dfrac{208}{3}\]

Then\[\begin{eqnarray*}\begin{array}{rl@{\qquad}l@{\hskip-2pc}}\dfrac{208}{3} &= 4u^{2} \\u^{2} &= \dfrac{52}{3} & {\color{#0066A7}{2\leq u\leq 6}} \\u &= \sqrt{\dfrac{52}{3}}\approx 4.163 &{\color{#0066A7}{\hbox{Disregard the negative solutionsince \(u>0\).}}}\end{array}\end{eqnarray*}\]

NOW WORK

Problem 55.

IN WORDS

The average value \(\bar{y}=\dfrac{1}{b-a}\int_{a}^{b}{f(x)\,dx}\) of a function \(f\) equals thevalue \({f(u)}\) in the Mean Value Theorem for Integrals.

DEFINITION Average Value of a Function over an Interval

Let \(f\) be a function that is continuous on the closed interval \([a,b] .\) The average value \({\boldsymbol {\bar{y}}}\) of\({\boldsymbol f}\) over \(\boldsymbol{[a,b]}\) is\[\boxed{\bbox[5pt]{\bar{y}=\dfrac{1}{b-a} \int_{a}^{b}f(x)\,dx }}\tag{8}\]

EXAMPLE 8 Finding the Average Value of a Function

Find the average value of \(f(x)=3x-8\) on the closed interval \([0,2] .\)

Solution The average value of \(f(x)=3x-8\) on theclosed interval \([0,2]\) is given by \[\begin{eqnarray*}\bar{y} &=&\dfrac{1}{b-a}\int_{a}^{b}f(x)\,dx=\dfrac{1}{2-0}{\int_{0}^{2}}(3x-8) dx\\&=& \dfrac{1}{2}\left[\dfrac{3x^{2}}{2}-8x \right] _{0}^{2} =\dfrac{1}{2}(6-16) =-5\end{eqnarray*}\]

The average value of \(f\) on \([0,2] \) is \(\bar{y}=-5.\)

NOW WORK

Problem 61.