$[-/1\mathrm{.}7$ Points$]$

SCALCET$96\mathrm{.}5\mathrm{.}026.$

Let $f_{\text{a}\text{v}\text{g}}[a,b]$ denote the average value of $f$ on the interval $a,b.$

Show that if $f_{\text{a}\text{v}\text{g}}[a,b]=(\frac{c-a}{b-a})f_{\text{a}\text{v}\text{g}}[a,c](\frac{b-c}{b-a})f_{\text{a}\text{v}\text{g}}[c,b].a,b,c{f}_{\text{a}\text{v}\text{g}}[a,b]f_{\text{a}\text{v}\text{g}}=(,){\displaystyle \underset{a}{\overset{b}{}}}f(x)dx{f}_{\text{a}\text{v}\text{g}}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{c}{}}}f(x)dx(,){\displaystyle \underset{c}{\overset{b}{}}}f(x)dx{f}_{\text{a}\text{v}\text{g}}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{c}{}}}f(x)dx(,){\displaystyle \underset{c}{\overset{b}{}}}f(x)dx$

$=\frac{c-a}{b-a}[(,){\displaystyle \underset{a}{\overset{c}{}}}f(x)dx](\frac{1}{b-a}{\displaystyle \underset{c}{\overset{b}{}}}f(x)dx]f_{\text{a}\text{v}\text{g}}[a,b]=(\frac{c-a}{b-a})f_{\text{a}\text{v}\text{g}}[a,c](\frac{b-c}{b-a})f_{\text{a}\text{v}\text{g}}[c,b]a,$ this integral can $be$ represented $as$ a sum $of$ two integrals over adjacent intervals $as$ follows.

$f_{\text{a}\text{v}\text{g}}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{c}{}}}f(x)dx(,){\displaystyle \underset{c}{\overset{b}{}}}f(x)dx$

$To$ transform these integrals into the desired result, $we$ perform some algebraic manipulation.

$f_{\text{a}\text{v}\text{g}}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{c}{}}}f(x)dx(,){\displaystyle \underset{c}{\overset{b}{}}}f(x)dx$

$=\frac{c-a}{b-a}[(,){\displaystyle \underset{a}{\overset{c}{}}}f(x)dx](\frac{1}{b-a}{\displaystyle \underset{c}{\overset{b}{}}}f(x)dx]$

Thus, $f_{\text{a}\text{v}\text{g}}[a,b]=(\frac{c-a}{b-a})f_{\text{a}\text{v}\text{g}}[a,c](\frac{b-c}{b-a})f_{\text{a}\text{v}\text{g}}[c,b].$

Need Help?$a.By$ the definition $of$ the average value $of{f}_{\text{a}\text{v}\text{g}}[a,b],$

$f_{\text{a}\text{v}\text{g}}=(,){\displaystyle \underset{a}{\overset{b}{}}}f(x)dx$

$By$ the property $of$ integrals, and since $a,$ this integral can $be$ represented $as$ a sum $of$ two integrals over adjacent intervals $as$ follows.

$f_{\text{a}\text{v}\text{g}}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{c}{}}}f(x)dx(,){\displaystyle \underset{c}{\overset{b}{}}}f(x)dx$

$To$ transform these integrals into the desired result, $we$ perform some algebraic manipulation.

$f_{\text{a}\text{v}\text{g}}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{c}{}}}f(x)dx(,){\displaystyle \underset{c}{\overset{b}{}}}f(x)dx$

$=\frac{c-a}{b-a}[(,){\displaystyle \underset{a}{\overset{c}{}}}f(x)dx](\frac{1}{b-a}{\displaystyle \underset{c}{\overset{b}{}}}f(x)dx]$

Thus, $f_{\text{a}\text{v}\text{g}}[a,b]=(\frac{c-a}{b-a})f_{\text{a}\text{v}\text{g}}[a,c](\frac{b-c}{b-a})f_{\text{a}\text{v}\text{g}}[c,b].$

Need Help?$a,$ then

$f_{\text{a}\text{v}\text{g}}[a,b]=(\frac{c-a}{b-a})f_{\text{a}\text{v}\text{g}}[a,c](\frac{b-c}{b-a})f_{\text{a}\text{v}\text{g}}[c,b].$

Let $a,b,cbe$ any three real numbers such that $a.By$ the definition $of$ the average value $of{f}_{\text{a}\text{v}\text{g}}[a,b],$

$f_{\text{a}\text{v}\text{g}}=(,){\displaystyle \underset{a}{\overset{b}{}}}f(x)dx$

$By$ the property $of$ integrals, and since $a,$ this integral can $be$ represented $as$ a sum $of$ two integrals over adjacent intervals $as$ follows.

$f_{\text{a}\text{v}\text{g}}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{c}{}}}f(x)dx(,){\displaystyle \underset{c}{\overset{b}{}}}f(x)dx$

$To$ transform these integrals into the desired result, $we$ perform some algebraic manipulation.

$f_{\text{a}\text{v}\text{g}}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{c}{}}}f(x)dx(,){\displaystyle \underset{c}{\overset{b}{}}}f(x)dx$

$=\frac{c-a}{b-a}[(,){\displaystyle \underset{a}{\overset{c}{}}}f(x)dx](\frac{1}{b-a}{\displaystyle \underset{c}{\overset{b}{}}}f(x)dx]$

Thus, $f_{\text{a}\text{v}\text{g}}[a,b]=(\frac{c-a}{b-a})f_{\text{a}\text{v}\text{g}}[a,c](\frac{b-c}{b-a})f_{\text{a}\text{v}\text{g}}[c,b].$

Need Help?