Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 6.1 : Average Function Value
3. Find \({f_{{\rm{avg}}}}\) for \(f\left( x \right) = 4{x^2} - x + 5\) on \(\left[ { - 2,3} \right]\) and determine the value(s) of \(c\) in \(\left[ { - 2,3} \right]\) for which \(f\left( c \right) = {f_{{\rm{avg}}}}\).
Show All Steps Hide All Steps
Start SolutionFirst, we need to use the formula for the notes in this section to find \({f_{{\rm{avg}}}}\).
\[{f_{{\rm{avg}}}} = \frac{1}{{3 - \left( { - 2} \right)}}\int_{{ - \,2}}^{3}{{4{x^2} - x + 5\,dx}} = \frac{1}{5}\left. {\left( {\frac{4}{3}{x^3} - \frac{1}{2}{x^2} + 5x} \right)} \right|_{ - 2}^3 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{83}}{6}}}\] Show Step 2Note that for the second part of this problem we are really just asking to find the value of \(c\) that satisfies the Mean Value Theorem for Integrals.
There really isn’t much to do here other than solve \(f\left( c \right) = {f_{{\rm{avg}}}}\).
\[\begin{align*}4{c^2} - c + 5 & = \frac{{83}}{6}\\ 4{c^2} - c - \frac{{53}}{6} & = 0 \hspace{0.5in} \Rightarrow \hspace{0.5in}c = \frac{{1 \pm \sqrt {1 - 4\left( 4 \right)\left( { - \frac{{53}}{6}} \right)} }}{{2\left( 4 \right)}} = \frac{{1 \pm \sqrt {\frac{{427}}{3}} }}{{2\left( 4 \right)}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 1.3663,\,\,\,1.6163}}\end{align*}\]So, unlike the example from the notes both of the numbers that we found here are in the interval and so are both included in the answer.