6.1-bo'lim : Funktsiyaning o'rtacha qiymati
3. \(\left[ { - 2,3} \right]\) da \(f\left( x \right) = 4{x^2} - x + 5\) uchun \({f_{{\rm{avg}}}}\) ni toping va \(f\left( c \right) = {f_{{\rm{avg}}}}\) uchun \(\left[ { - 2,3} \right]\) da \(c\) qiymati(lar)ini aniqlang.
Yechim
Birinchidan, \({f_{{\rm{avg}}}}\) ni topish uchun ushbu bo'limdagi eslatmalar formulasidan foydalanishimiz kerak.
\[{f_{{\rm{avg}}}} = \frac{1}{{3 - \left( { - 2} \right)}}\int_{{ - \,2}}^{3}{{4{x^2} - x + 5\,dx}} = \frac{1}{5}\left. {\left( {\frac{4}{3}{x^3} - \frac{1}{2}{x^2} + 5x} \right)} \right|_{ - 2}^3 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{83}}{6}}}\]2-qadam
E'tibor bering, ushbu muammoning ikkinchi qismi uchun biz faqat integrallar uchun o'rtacha qiymat teoremasini qondiradigan \(c\) qiymatini topishni so'raymiz.
Bu erda, albatta, \(f\left( c \right) = {f_{{\rm{avg}}}}\) ni hal qilishdan boshqa qiladigan ko'p narsa yo'q.
\[\begin{align*}4{c^2} - c + 5 & = \frac{{83}}{6}\\ 4{c^2} - c - \frac{{53}}{6} & = 0 \hspace{0.5in} \Rightarrow \hspace{0.5in}c = \frac{{1 \pm \sqrt {1 - 4\left( 4 \right)\left( { - \frac{{53}}{6}} \right)} }}{{2\left( 4 \right)}} = \frac{{1 \pm \sqrt {\frac{{427}}{3}} }}{{2\left( 4 \right)}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 1.3663,\,\,\,1.6163}}\end{align*}\]Shunday qilib, eslatmadagi misoldan farqli o'laroq, biz bu erda topgan raqamlarning ikkalasi ham intervalda va ikkalasi ham javobga kiritilgan.